hypothesis testing for mean standard deviation unknown

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8.4.3 Hypothesis Testing for the Mean

$\quad$ $H_0$: $\mu=\mu_0$, $\quad$ $H_1$: $\mu \neq \mu_0$.

$\quad$ $H_0$: $\mu \leq \mu_0$, $\quad$ $H_1$: $\mu > \mu_0$.

$\quad$ $H_0$: $\mu \geq \mu_0$, $\quad$ $H_1$: $\mu \lt \mu_0$.

Two-sided Tests for the Mean:

Therefore, we can suggest the following test. Choose a threshold, and call it $c$. If $|W| \leq c$, accept $H_0$, and if $|W|>c$, accept $H_1$. How do we choose $c$? If $\alpha$ is the required significance level, we must have

As discussed above, we let \begin{align}%\label{} W(X_1,X_2, \cdots,X_n)=\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}}. \end{align} Note that, assuming $H_0$, $W \sim N(0,1)$. We will choose a threshold, $c$. If $|W| \leq c$, we accept $H_0$, and if $|W|>c$, accept $H_1$. To choose $c$, we let \begin{align} P(|W| > c \; | \; H_0) =\alpha. \end{align} Since the standard normal PDF is symmetric around $0$, we have \begin{align} P(|W| > c \; | \; H_0) = 2 P(W>c | \; H_0). \end{align} Thus, we conclude $P(W>c | \; H_0)=\frac{\alpha}{2}$. Therefore, \begin{align} c=z_{\frac{\alpha}{2}}. \end{align} Therefore, we accept $H_0$ if \begin{align} \left|\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}} \right| \leq z_{\frac{\alpha}{2}}, \end{align} and reject it otherwise.
We have \begin{align} \beta (\mu) &=P(\textrm{type II error}) = P(\textrm{accept }H_0 \; | \; \mu) \\ &= P\left(\left|\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}} \right| \lt z_{\frac{\alpha}{2}}\; | \; \mu \right). \end{align} If $X_i \sim N(\mu,\sigma^2)$, then $\overline{X} \sim N(\mu, \frac{\sigma^2}{n})$. Thus, \begin{align} \beta (\mu)&=P\left(\left|\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}} \right| \lt z_{\frac{\alpha}{2}}\; | \; \mu \right)\\ &=P\left(\mu_0- z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}} \leq \overline{X} \leq \mu_0+ z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}}\right)\\ &=\Phi\left(z_{\frac{\alpha}{2}}+\frac{\mu_0-\mu}{\sigma / \sqrt{n}}\right)-\Phi\left(-z_{\frac{\alpha}{2}}+\frac{\mu_0-\mu}{\sigma / \sqrt{n}}\right). \end{align}
Let $S^2$ be the sample variance for this random sample. Then, the random variable $W$ defined as \begin{equation} W(X_1,X_2, \cdots, X_n)=\frac{\overline{X}-\mu_0}{S / \sqrt{n}} \end{equation} has a $t$-distribution with $n-1$ degrees of freedom, i.e., $W \sim T(n-1)$. Thus, we can repeat the analysis of Example 8.24 here. The only difference is that we need to replace $\sigma$ by $S$ and $z_{\frac{\alpha}{2}}$ by $t_{\frac{\alpha}{2},n-1}$. Therefore, we accept $H_0$ if \begin{align} |W| \leq t_{\frac{\alpha}{2},n-1}, \end{align} and reject it otherwise. Let us look at a numerical example of this case.

$\quad$ $H_0$: $\mu=170$, $\quad$ $H_1$: $\mu \neq 170$.

Let's first compute the sample mean and the sample standard deviation. The sample mean is \begin{align}%\label{} \overline{X}&=\frac{X_1+X_2+X_3+X_4+X_5+X_6+X_7+X_8+X_9}{9}\\ &=165.8 \end{align} The sample variance is given by \begin{align}%\label{} {S}^2=\frac{1}{9-1} \sum_{k=1}^9 (X_k-\overline{X})^2&=68.01 \end{align} The sample standard deviation is given by \begin{align}%\label{} S&= \sqrt{S^2}=8.25 \end{align} The following MATLAB code can be used to obtain these values: x=[176.2,157.9,160.1,180.9,165.1,167.2,162.9,155.7,166.2]; m=mean(x); v=var(x); s=std(x); Now, our test statistic is \begin{align} W(X_1,X_2, \cdots, X_9)&=\frac{\overline{X}-\mu_0}{S / \sqrt{n}}\\ &=\frac{165.8-170}{8.25 / 3}=-1.52 \end{align} Thus, $|W|=1.52$. Also, we have \begin{align} t_{\frac{\alpha}{2},n-1} = t_{0.025,8} \approx 2.31 \end{align} The above value can be obtained in MATLAB using the command $\mathtt{tinv(0.975,8)}$. Thus, we conclude \begin{align} |W| \leq t_{\frac{\alpha}{2},n-1}. \end{align} Therefore, we accept $H_0$. In other words, we do not have enough evidence to conclude that the average height in the city is different from the average height in the country.

Let us summarize what we have obtained for the two-sided test for the mean.

One-sided Tests for the Mean:

As before, we define the test statistic as \begin{align}%\label{} W(X_1,X_2, \cdots,X_n)=\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}}. \end{align} If $H_0$ is true (i.e., $\mu \leq \mu_0$), we expect $\overline{X}$ (and thus $W$) to be relatively small, while if $H_1$ is true, we expect $\overline{X}$ (and thus $W$) to be larger. This suggests the following test: Choose a threshold, and call it $c$. If $W \leq c$, accept $H_0$, and if $W>c$, accept $H_1$. How do we choose $c$? If $\alpha$ is the required significance level, we must have \begin{align} P(\textrm{type I error}) &= P(\textrm{Reject }H_0 \; | \; H_0) \\ &= P(W > c \; | \; \mu \leq \mu_0) \leq \alpha. \end{align} Here, the probability of type I error depends on $\mu$. More specifically, for any $\mu \leq \mu_0$, we can write \begin{align} P(\textrm{type I error} \; | \; \mu) &= P(\textrm{Reject }H_0 \; | \; \mu) \\ &= P(W > c \; | \; \mu)\\ &=P \left(\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}}> c \; | \; \mu\right)\\ &=P \left(\frac{\overline{X}-\mu}{\sigma / \sqrt{n}}+\frac{\mu-\mu_0}{\sigma / \sqrt{n}}> c \; | \; \mu\right)\\ &=P \left(\frac{\overline{X}-\mu}{\sigma / \sqrt{n}}> c+\frac{\mu_0-\mu}{\sigma / \sqrt{n}} \; | \; \mu\right)\\ &\leq P \left(\frac{\overline{X}-\mu}{\sigma / \sqrt{n}}> c \; | \; \mu\right) \quad (\textrm{ since }\mu \leq \mu_0)\\ &=1-\Phi(c) \quad \big(\textrm{ since given }\mu, \frac{\overline{X}-\mu}{\sigma / \sqrt{n}} \sim N(0,1) \big). \end{align} Thus, we can choose $\alpha=1-\Phi(c)$, which results in \begin{align} c=z_{\alpha}. \end{align} Therefore, we accept $H_0$ if \begin{align} \frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}} \leq z_{\alpha}, \end{align} and reject it otherwise.

$\quad$ $H_0$: $\mu \geq \mu_0$, $\quad$ $H_1$: $\mu \lt \mu_0$,

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9.3 Statistical Inference for Two Population Means with Unknown Population Standard Deviations

Learning objectives.

Construct and interpret a confidence interval for two population means with unknown population standard deviations.
Conduct and interpret hypothesis tests for two population means with unknown population standard deviations.

The comparison of two population means is very common. Often, we want to find out if the two populations under study have the same mean or if there is some difference in the two population means. The approach we take when studying two population means depends on whether the samples are independent or matched . In the case the samples are independent, we also have to contend with whether or not we know the population standard deviations.

Two populations are independent if the sample taken from population 1 is not related in anyway to the sample taken from population 2. In this situation, any relationship between the samples or populations is entirely coincidental.

Throughout this section, we will use subscripts to identify the values for the means, sample sizes, and standard deviations for the two populations:

In order to construct a confidence interval or conduct a hypothesis test on the difference in two population means ([latex]\mu_1-\mu_2[/latex]), we need to use the distribution of the difference in the sample means [latex]\overline{x}_1-\overline{x}_2[/latex]:

The mean of the distribution of the difference in the sample means is [latex]\displaystyle{\mu_{\overline{x}_1-\overline{x}_2}}=\mu_1-\mu_2[/latex].
The standard deviation of the distribution of the difference in the sample means is [latex]\displaystyle{\sigma_{\overline{x}_1-\overline{x}_2}=\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}[/latex].
Both populations are normally distributed.
The sample sizes are large enough ([latex]n_1 \geq 30[/latex] and [latex]n_2 \geq 30[/latex]).

[latex]\displaystyle{z=\frac{(\overline{x}_1-\overline{x}_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}}[/latex]

As we have seen previously when working with confidence intervals and hypothesis testing for a single population, when the population standard deviation is unknown and we must use the sample standard deviation as an estimate for the population standard deviation, we use a [latex]t[/latex]-distribution. We do the same thing when working with the two population means. When the population standard deviations are unknown, we use the sample standard deviations as estimates for the population standard deviations [latex]\sigma_1[/latex] and [latex]\sigma_2[/latex]. In this situation, we use a [latex]t[/latex]-distribution for the distribution of the difference in the sample means. So, when the population standard deviations are unknown for a confidence interval or hypothesis test on the difference in two population means, we will use a [latex]t[/latex]-distribution. The [latex]t[/latex]-score and the degrees of freedom are:

[latex]\begin{eqnarray*} t & = & \frac{(\overline{x}_1-\overline{x}_2)-(\mu_1-\mu_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \\ \\ df & = & \frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{1}{n_1-1} \times \left(\frac{s_1^2}{n_1}\right)^2+\frac{1}{n_2-1} \times \left(\frac{s_2^2}{n_2}\right)^2} \end{eqnarray*}[/latex]

Obviously, the degrees of freedom formula is somewhat complicated. But a computer makes the calculation a bit more manageable. The output from the degrees of freedom formula is rarely a whole number. After calculating the value of [latex]df[/latex] using the above formula, round the output from this formula down to the next whole number to get the degrees of freedom for the [latex]t[/latex]-distribution.

Constructing a Confidence Interval for the Difference in Two Population Means with Unknown Population Standard Deviations

Suppose a sample of size [latex]n_1[/latex] with sample mean [latex]\overline{x}_1[/latex] and standard deviation [latex]s_1[/latex] is taken from population 1 and a sample of size [latex]n_2[/latex] with sample mean [latex]\overline{x}_2[/latex] and standard deviation [latex]s_2[/latex] is taken from population 2 where the populations are independent and the population standard deviations are unknown . The limits for the confidence interval with confidence level [latex]C[/latex] for the difference in the population means [latex]\displaystyle{\mu_1-\mu_2}[/latex] are:

[latex]\begin{eqnarray*} \\ \mbox{Lower Limit} & = & \overline{x}_1-\overline{x}_2-t \times \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}} \\ \\ \mbox{Upper Limit} & = & \overline{x}_1-\overline{x}_2+t \times \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}} \\ \\\end{eqnarray*}[/latex]

where [latex]t[/latex] is the positive [latex]t[/latex]-score of the [latex]t[/latex]-distribution with [latex]\displaystyle{df = \frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{1}{n_1-1} \times \left(\frac{s_1^2}{n_1}\right)^2+\frac{1}{n_2-1} \times \left(\frac{s_2^2}{n_2}\right)^2}}[/latex] so that the area under the curve in between [latex]-t[/latex] and [latex]t[/latex] is [latex]C\%[/latex].

In order to construct the confidence interval for the difference in two population means with independent samples, we need to check that the distribution of the difference in the sample means follows a normal distribution. This means that we need to check that either the populations are normal or that the sample sizes are large enough (greater than or equal to 30).
When the population standard deviations are unknown, we must use a [latex]t[/latex]-distribution in the construction of the confidence interval.
The value of degrees of freedom must be a whole number. After using the formula, remember to round the value down to the next whole number to get the required degrees of freedom for the [latex]t[/latex]-distribution.

CALCULATING THE [latex]\textcolor{white}t[/latex]-SCORE FOR A CONFIDENCE INTERVAL IN EXCEL

To find the [latex]t[/latex]-score to construct a confidence interval with confidence level [latex]C[/latex], use the t.inv.2t(area in the tails, degrees of freedom) function.

For area in the tails , enter the sum of the area in the tails of the [latex]t[/latex]-distribution. For a confidence interval, the area in the tails is [latex]1-C[/latex].
For degrees of freedom , enter the degrees of freedom calculated using [latex]\displaystyle{df = \frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{1}{n_1-1} \times \left(\frac{s_1^2}{n_1}\right)^2+\frac{1}{n_2-1} \times \left(\frac{s_2^2}{n_2}\right)^2}}[/latex].

The output from the t . inv.2t function is the value of [latex]t[/latex]-score needed to construct the confidence interval.

The t.inv.2t function requires that we enter the sum of the area in both tails. The area in the middle of the distribution is the confidence level [latex]C[/latex], so the sum of the area in both tails is the leftover area [latex]1-C[/latex].
The degrees of freedom for a [latex]t[/latex]-distribution must be a whole number . The output from the degrees of freedom formula [latex]\displaystyle{df = \frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{1}{n_1-1} \times \left(\frac{s_1^2}{n_1}\right)^2+\frac{1}{n_2-1} \times \left(\frac{s_2^2}{n_2}\right)^2}}[/latex] is almost never a whole number. After calculating the value of [latex]df[/latex] using the formula, round the value down to the next whole number. Remember to entered the rounded down value of [latex]df[/latex] for the degrees of freedom in the t.inv.2t function.

A company that manufacturers and services photocopiers wants to study the difference in the average repair time for the two different models of photocopiers they make. In a sample of 60 repairs of photocopier A, the mean repair time was 84.2 minutes with a standard deviation of 19.4 minutes. In a sample of 70 repairs of photocopier B, the mean repair time was 91.6 minutes with a standard deviation of 18.8 minutes.

Construct a 95% confidence interval for the difference in the mean repair time for the two photocopiers.
Interpret the confidence interval found in part 1.
Is there evidence to suggest that the mean repair times for the photocopiers is the same? Explain.

To find the confidence interval, we need to find the [latex]t[/latex]-score for the 95% confidence interval. This means that we need to find the [latex]t[/latex]-score so that the area in the tails is [latex]1-0.95=0.05[/latex].

[latex]\begin{eqnarray*} \\ df & = & \frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{1}{n_1-1} \times \left(\frac{s_1^2}{n_1}\right)^2+\frac{1}{n_2-1} \times \left(\frac{s_2^2}{n_2}\right)^2} \\ & = & \frac{\left(\frac{19.4^2}{60}+\frac{18.8^2}{70}\right)^2}{\frac{1}{60-1} \times \left(\frac{19.4^2}{60}\right)^2+\frac{1}{70-1} \times \left(\frac{18.8^2}{70}\right)^2} \\ & = & 123.68.... \\ & \Rightarrow & 123 \\ \\ \end{eqnarray*}[/latex]

Graph of a t-distribution curve. Along the horizontal axis the point t is labeled. There is a vertical line from t to the normal distribution curve. The area under the curve in the middle of the distribution is labeled 95%. The area in the left tail is labeled 2.5%. The area in the right tail is labeled 2.5%.

So [latex]t=1.9794...[/latex]. The 95% confidence interval is

[latex]\begin{eqnarray*} \\ \mbox{Lower Limit} & = & \overline{x}_1-\overline{x}_2-t \times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ & = & 84.2-91.6-1.9794... \times \sqrt{\frac{19.4^2}{60}+\frac{18.8^2}{70}} \\ & = & -14.06 \\ \\ \mbox{Upper Limit} & = & \overline{x}_1-\overline{x}_2+t \times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ & = & 84.2-91.6+1.9794... \times \sqrt{\frac{19.4^2}{60}+\frac{18.8^2}{70}} \\ & = & -0.74 \\ \\ \end{eqnarray*}[/latex]

We are 95% confident that the difference in the mean repair time for the two photocopiers is between -14.06 minutes and -0.74 minutes.
Because 0 is outside the confidence interval and both limits are negative, it suggests that the difference in the means [latex]\displaystyle{\mu_1-\mu_2}[/latex] is less than 0. That is, [latex]\displaystyle{\mu_1-\mu_2 \lt 0}[/latex] ([latex]\mu_1 \lt \mu_2[/latex]). This suggests that the mean for population 1 (photocopier A) is less than the mean for population 2 (photocopier B). So the mean repair time for photocopier A is less than the mean repair time for photocopier B.
When calculating the limits for the confidence interval keep all of the decimals in the [latex]t[/latex]-score and other values throughout the calculation. This will ensure that there is no round-off error in the answers. You can use Excel to do the calculation of the limits, clicking on the cells containing the [latex]t[/latex]-score and any other values, to ensure that all of the decimal places are used in the calculation.
When writing down the interpretation of the confidence interval, make sure to include the confidence level, the actual difference in the population means captured by the confidence interval (i.e. be specific to the context of the question), and appropriate units for the limits.
The value of the degrees of freedom must be a whole number. After using the formula, remember to round the value down to the next whole number to get the required degrees of freedom for the [latex]t[/latex]-distribution.

Steps to Conduct a Hypothesis Test for the Difference in Two Independent Population Means with Unknown Population Standard Deviations

[latex]\begin{eqnarray*} \\ H_0: & & \mu_1-\mu_2=0 \\ \end{eqnarray*}[/latex]

[latex]\begin{eqnarray*} \\ H_a: \mu_1-\mu_2 0 & & (\mu_1 \gt \mu_2) \\ H_a: \mu_1-\mu_2 \neq 0 & & (\mu_1 \neq \mu_2) \\ \\ \end{eqnarray*}[/latex]

Use the form of the alternative hypothesis to determine if the test is left-tailed, right-tailed, or two-tailed.
Collect the sample information for the test and identify the significance level.

The results of the sample data are significant. There is sufficient evidence to conclude that the null hypothesis [latex]H_0[/latex] is an incorrect belief and that the alternative hypothesis [latex]H_a[/latex] is most likely correct.
The results of the sample data are not significant. There is not sufficient evidence to conclude that the alternative hypothesis [latex]H_a[/latex] may be correct.
Write down a concluding sentence specific to the context of the question.

USING EXCEL TO CALCULE THE P -VALUE FOR A HYPOTHESIS TEST ON TWO INDEPENDENT POPULATION MEANS WITH UNKNOWN POPULATION STANDARD DEVIATIONS

Assuming that the population standard deviations are unknown, the p -value for a hypothesis test on the difference in two independent population means is the area in the tail(s) of the [latex]t[/latex]-distribution.

If the p -value is the area in the left tail:

For t-score , enter the value of [latex]t[/latex] calculated from [latex]\displaystyle{t = \frac{(\overline{x}_1-\overline{x}_2)-(\mu_1-\mu_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}}[/latex].
For the logic operator , enter true . Note: Because we are calculating the area under the curve, we always enter true for the logic operator.

If the p -value is the area in the right tail:

If the p -value is the sum of the area in the two tails:

For t-score , enter the absolute value of [latex]t[/latex] calculated from [latex]\displaystyle{t = \frac{(\overline{x}_1-\overline{x}_2)-(\mu_1-\mu_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}}[/latex]. Note: In the t.dist.2t function, the value of the [latex]t[/latex]-score must be a positive number. If the [latex]t[/latex]-score is negative, enter the absolute value of the [latex]t[/latex]-score into the t.dist.2t function.

The degrees of freedom for a [latex]t[/latex]-distribution must be a whole number . The output from the degrees of freedom formula [latex]\displaystyle{df = \frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{1}{n_1-1} \times \left(\frac{s_1^2}{n_1}\right)^2+\frac{1}{n_2-1} \times \left(\frac{s_2^2}{n_2}\right)^2}}[/latex] is almost never a whole number. After calculating the value of [latex]df[/latex] using the formula, round the value down to the next whole number. Remember to entered the rounded down value of [latex]df[/latex] for the degrees of freedom in the t.dist functions.

A researcher wants to study the difference between the average amount of time boys and girls aged seven to eleven spend playing sports each day. In a sample of 9 girls, the average number of hours spent playing sports per day is 2 hours with a standard deviation of 0.866 hours. In a sample of 16 boys, the average number of hours spent playing sports per day is 3.2 hours with a standard deviation of 1 hours. Both populations have a normal distribution. At the 5% significance level, is there a difference in the mean amount of time boys and girls aged seven to eleven play sports each day?

Let girls be population 1 and boys be population 2. These populations are independent because there is no relationship between the two groups. From the questions, we have the following information:

Hypotheses:

[latex]\begin{eqnarray*} H_0: & & \mu_1-\mu_2=0 \\ H_a: & & \mu_1-\mu_2 \neq 0 \end{eqnarray*}[/latex]

This is a test on a the difference in two population means where the population standard deviation are unknown. So we use a [latex]t[/latex]-distribution to calculate the p -value. Because the alternative hypothesis is a [latex]\neq[/latex], the p -value is the sum of areas in the tails of the distribution.

This is a t distribution curve. The peak of the curve is at 0 on the horizontal axis. The point -t and t are also labeled. A vertical line extends from point t to the curve with the area to the right of this vertical line shaded with the shaded area labeled half of the p-value. A vertical line extends from -t to the curve with the area to the left of this vertical line shaded with the shaded area labeled half of the p-value. The p-value equals the area of these two shaded regions.

To use the t.dist.2t function, we need to calculate out the [latex]t[/latex]-score and the degrees of freedom:

[latex]\begin{eqnarray*} t & = & \frac{(\overline{x}_1-\overline{x}_2)-(\mu_1-\mu_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \\ & = & \frac{(2-3.2)-0}{\sqrt{\frac{0.866^2}{9}+\frac{1^2}{16}}} \\ & = & -3.1423...\\ \\ df & = & \frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{1}{n_1-1} \times \left(\frac{s_1^2}{n_1}\right)^2+\frac{1}{n_2-1} \times \left(\frac{s_2^2}{n_2}\right)^2} \\ & = & \frac{\left(\frac{0.866^2}{9}+\frac{1^2}{16}\right)^2}{\frac{1}{9-1} \times \left(\frac{0.866^2}{9}\right)^2+\frac{1}{16-1} \times \left(\frac{1^2}{16}\right)^2} \\ & = & 18.846... \\ & \Rightarrow & 18 \end{eqnarray*}[/latex]

So the p -value[latex]=0.0056[/latex].

Conclusion:

Because p -value[latex]=0.0056 \lt 0.05=\alpha[/latex], we reject the null hypothesis in favour of the alternative hypothesis. At the 5% significance level there is enough evidence to suggest that there is a difference in the mean amount of time boys and girls aged seven to eleven play sports each day.

The null hypothesis [latex]\mu_1-\mu_2=0[/latex] is the claim that there is no difference in the mean amount of time boys and girls spend playing sports each day. That is, the two populations have the same mean.
The alternative hypothesis [latex]\mu_1 -\mu_2 \neq 0[/latex] is the claim that there is a difference in the mean amount of time boys and girls spend playing sports each day ([latex]\mu_1 \neq \mu_2[/latex]). That is, the two populations have different means.
Keep all of the decimals throughout the calculation (i.e. in the [latex]t[/latex]-score, etc.) to avoid any round-off error in the calculation of the p -value. This ensures that we get the most accurate value for the p -value. Use Excel to do the calculations, and then click on the cells in subsequent calculations.
The t.dist.2t function requires that the value entered for the [latex]t[/latex]-score is positive . A negative [latex]t[/latex]-score entered into the t.dist.2t function generates an error in Excel. In this case, the value of the [latex]t[/latex]-score is negative, so we must enter the absolute value of this [latex]t[/latex]-score into field 1.
The p -value of 0.0056 is a small probability compared to the significance level, and so is unlikely to happen assuming the null hypothesis is true. This suggests that the assumption that the null hypothesis is true is most likely incorrect, and so the conclusion of the test is to reject the null hypothesis in favour of the alternative hypothesis. In other words, there is a difference in the mean amount of time boys and girls spend playing sports each day.

A town has two colleges. A local community group believes that students who graduate from College A have taken more math classes than the students who graduate from College B. In a sample of 11 graduates from College A, the average is 4 math classes per graduate with a standard deviation of 1.5 math classes. In a sample of 9 graduates from College B, the average is 3.5 math classes per graduate with a standard deviation of 1 math class. Both populations have a normal distribution. At the 1% significance level, test the community groups claim that graduates from College A have taken more math classes than graduates from College B.

Let College A be population 1 and College B be population 2. These populations are independent because there is no relationship between the two groups. From the questions, we have the following information:

[latex]\begin{eqnarray*} H_0: & & \mu_1-\mu_2=0 \\ H_a: & & \mu_1-\mu_2 \gt 0 \end{eqnarray*}[/latex]

This is a test on a the difference in two population means where the population standard deviation are unknown. So we use a [latex]t[/latex]-distribution to calculate the p -value. Because the alternative hypothesis is a [latex]\gt[/latex], the p -value is the area in the right tail of the distribution.

This is a t-distribution curve. The peak of the curve is at 0 on the horizontal axis. The point t is also labeled. A vertical line extends from point t to the curve with the area to the right of this vertical line shaded. The p-value equals the area of this shaded region.

To use the t.dist.rt function, we need to calculate out the [latex]t[/latex]-score and the degrees of freedom:

[latex]\begin{eqnarray*} t & = & \frac{(\overline{x}_1-\overline{x}_2)-(\mu_1-\mu_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \\ & = & \frac{(4-3.5)-0}{\sqrt{\frac{1.5^2}{11}+\frac{1^2}{9}}} \\ & = & 0.8899...\\ \\ df & = & \frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{1}{n_1-1} \times \left(\frac{s_1^2}{n_1}\right)^2+\frac{1}{n_2-1} \times \left(\frac{s_2^2}{n_2}\right)^2} \\ & = & \frac{\left(\frac{1.5^2}{11}+\frac{1^2}{9}\right)^2}{\frac{1}{11-1} \times \left(\frac{1.5^2}{11}\right)^2+\frac{1}{9-1} \times \left(\frac{1^2}{9}\right)^2} \\ & = & 17.397... \\ & \Rightarrow & 17 \end{eqnarray*}[/latex]

So the p -value[latex]=0.1930[/latex].

Because p -value[latex]=0.1930 \gt 0.01=\alpha[/latex], we do not reject the null hypothesis. At the 1% significance level there is not enough evidence to suggest that, on average, graduates of College A take more math classes than graduates of College B.

The null hypothesis [latex]\mu_1-\mu_2=0[/latex] is the claim that the average number of math classes taken by graduates of College A equals the average number of math classes taken by graduates of College B. That is, the two populations have the same mean.
The alternative hypothesis [latex]\mu_1 -\mu_2 \gt 0[/latex] is the claim that, on average, graduates of College A taken more math classes than graduates of College B ([latex]\mu_1 \gt \mu_2[/latex]).
The p -value of 0.1930 is a large probability compared to the significance level, and so is likely to happen assuming the null hypothesis is true. This suggests that the assumption that the null hypothesis is true is most likely correct, and so the conclusion of the test is to not reject the null hypothesis. In other words, graduates from the two colleges take, on average, the same number of math classes.

A professor at a large community college taught both an online section and a face-to-face section of his statistics course. The professor wants to study the difference in the average score on the final exam, believing that the mean score for the online section would be lower than the face-to-face section. The professor randomly selected 30 final exam scores from each section and recorded the scores in the tables below.

Online Section:

Face-to-Face Section:

At the 5% significance level, is the mean of the final exam score for the online section lower than the mean of the final exam score for the face-to-face section?

Let the online section be population 1 and the face-to-face section be population 2. These populations are independent because there is no relationship between the two groups. From the questions, we have the following information:

[latex]\begin{eqnarray*} H_0: & & \mu_1-\mu_2=0 \\ H_a: & & \mu_1-\mu_2 \lt 0 \end{eqnarray*}[/latex]

This is a test on a the difference in two population means where the population standard deviation are unknown. So we use a [latex]t[/latex]-distribution to calculate the p -value. Because the alternative hypothesis is a [latex]\lt[/latex], the p -value is the area in the left tail of the distribution.

his is a t-distribution curve. The peak of the curve is at 0 on the horizontal axis. The point t is also labeled. A vertical line extends from point t to the curve with the area to the left of this vertical line shaded. The p-value equals the area of this shaded region.

To use the t.dist function, we need to calculate out the [latex]t[/latex]-score and the degrees of freedom:

[latex]\begin{eqnarray*} t & = & \frac{(\overline{x}_1-\overline{x}_2)-(\mu_1-\mu_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \\ & = & \frac{(72.85-84.98)-0}{\sqrt{\frac{16.918...^2}{30}+\frac{11.714...^2}{30}}} \\ & = & -3.228...\\ \\ df & = & \frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{1}{n_1-1} \times \left(\frac{s_1^2}{n_1}\right)^2+\frac{1}{n_2-1} \times \left(\frac{s_2^2}{n_2}\right)^2} \\ & = & \frac{\left(\frac{16.918...^2}{30}+\frac{11.714...^2}{30}\right)^2}{\frac{1}{30-1} \times \left(\frac{16.918...^2}{30}\right)^2+\frac{1}{30-1} \times \left(\frac{11.714...^2}{30}\right)^2} \\ & = & 51.608... \\ & \Rightarrow & 51 \end{eqnarray*}[/latex]

So the p -value[latex]=0.0011[/latex].

Because p -value[latex]=0.0011 \lt 0.05=\alpha[/latex], we do reject the null hypothesis in favour of the alternative hypothesis. At the 5% significance level there is enough evidence to suggest that the mean final exam score for the online section is lower than the face-to-face section.

The null hypothesis [latex]\mu_1-\mu_2=0[/latex] is the claim that the average final exam score is the same for both sections. That is, the two populations have the same mean.
The alternative hypothesis [latex]\mu_1 -\mu_2 \lt 0[/latex] is the claim that average final exam score for the online section is lower than the face-to-face section ([latex]\mu_1 \lt \mu_2[/latex]).
Keep all of the decimals throughout the calculation (i.e. in the sample means, sample standard deviations, in the [latex]t[/latex]-score, etc.) to avoid any round-off error in the calculation of the p -value. This ensures that we get the most accurate value for the p -value. Use Excel to do the calculations, and then click on the cells in subsequent calculations.
The p -value of 0.0011 is a small probability compared to the significance level, and so is unlikely to happen assuming the null hypothesis is true. This suggests that the assumption that the null hypothesis is true is most likely incorrect, and so the conclusion of the test is to reject the null hypothesis in favour of the alternative hypothesis. In other words, the average final exam score for the online section is lower than for the face-to-face section.

A study is done to determine if Company A retains its workers longer than Company B. Company A samples 15 workers, and their average time with the company is 5 years with a standard deviation of 1.2 years. Company B samples 20 workers, and their average time with the company is 4.5 years with a standard deviation of 0.8 years. The populations are normally distributed. At the 5% significance level, on average, do workers at Company A stay longer than workers at Company B?

Let Company A be population 1 and Company B be population 2.

[latex]\begin{eqnarray*} t & = & \frac{(\overline{x}_1-\overline{x}_2)-(\mu_1-\mu_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \\ & = & \frac{(5-4.5)-0}{\sqrt{\frac{1.2^2}{15}+\frac{0.8^2}{20}}} \\ & = & 1.3975... \\ \\ df & = & \frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{1}{n_1-1} \times \left(\frac{s_1^2}{n_1}\right)^2+\frac{1}{n_2-1} \times \left(\frac{s_2^2}{n_2}\right)^2} \\ & = & \frac{\left(\frac{1.2^2}{15}+\frac{0.8^2}{20}\right)^2}{\frac{1}{15-1} \times \left(\frac{1.2^2}{15}\right)^2+\frac{1}{20-1} \times \left(\frac{0.8^2}{20}\right)^2} \\ & = & 23.005... \\ & \Rightarrow & 23 \end{eqnarray*}[/latex]

Because p -value[latex]=0.0878 \gt 0.05=\alpha[/latex], we do not reject the null hypothesis. At the 5% significance level there is not enough evidence to suggest that, on average, workers at Company A stay longer than workers at Company B.

Watch this video: Confidence Intervals for Two Population Means, Sigma Unknown by ExcelIsFun [16:11]

Watch this video: Hypothesis Testing for Two Population Means, Sigma Unknown by ExcelIsFun [17:29]

Concept Review

The general form of a confidence interval for the difference in two independent population means with unknown population standard deviations is

where [latex]t[/latex] is the positive [latex]t[/latex]-score of the [latex]t[/latex]-distribution with [latex]\displaystyle{df = \frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{1}{n_1-1} \times \left(\frac{s_1^2}{n_1}\right)^2+\frac{1}{n_2-1} \times \left(\frac{s_2^2}{n_2}\right)^2}}[/latex] so that the area under the [latex]t[/latex]-distribution in between [latex]-t[/latex] and [latex]t[/latex] is [latex]C[/latex].

The hypothesis test for the difference in two independent population means with unknown population standard deviations is a well established process:

Write down the null and alternative hypotheses in terms of the differences in the population means [latex]\mu_1-\mu_2[/latex].

Compare the p -value to the significance level and state the outcome of the test.

Attribution

“ 10.1 Two Population Means with Unknown Standard Deviations “ and “ 10.2 Two Population Means with Known Standard Deviations “ in Introductory Statistics by OpenStax is licensed under a Creative Commons Attribution 4.0 International License.

Module 10: Hypothesis Testing With Two Samples

Two population means with unknown standard deviations, learning outcomes.

Classify hypothesis tests by type
Conduct and interpret hypothesis tests for two population means, population standard deviations unknown
The two independent samples are simple random samples from two distinct populations.
if the sample sizes are small, the distributions are important (should be normal)
if the sample sizes are large, the distributions are not important (need not be normal)

Note: The test comparing two independent population means with unknown and possibly unequal population standard deviations is called the Aspin-Welch t-test. The degrees of freedom formula was developed by Aspin-Welch.

The comparison of two population means is very common. A difference between the two samples depends on both the means and the standard deviations. Very different means can occur by chance if there is great variation among the individual samples. In order to account for the variation, we take the difference of the sample means, [latex]\displaystyle\overline{{X}}_{{1}}-\overline{{X}}_{{2}} [/latex], and divide by the standard error in order to standardize the difference. The result is a t-score test statistic.

Because we do not know the population standard deviations, we estimate them using the two sample standard deviations from our independent samples. For the hypothesis test, we calculate the estimated standard deviation, or standard error , of the difference in sample means, [latex]\displaystyle\overline{{X}}_{{1}}-\overline{{X}}_{{2}} [/latex].

The standard error is: [latex]\displaystyle\sqrt{\frac{(s_1)^2}{n_1}+\frac{(s_2)^2}{n_2}} [/latex]

The test statistic ( t -score) is calculated as follows: [latex]\frac{(\overline{x}_1-\overline{x}_2)-(\overline{\mu}_1-\overline{\mu}_2)}{\displaystyle\sqrt{\frac{(s_1)^2}{n_1}+\frac{(s_2)^2}{n_2}}} [/latex]

Where: s 1 and s 2 , the sample standard deviations, are estimates of σ 1 and σ 2 , respectively. σ 1 and σ 1 are the unknown population standard deviations. [latex]\displaystyle\overline{{x}}_{{1}} [/latex] and [latex]\overline{{x}}_{{2}} [/latex] are the population means.

The number of degrees of freedom ( df ) requires a somewhat complicated calculation. However, a computer or calculator calculates it easily. The df are not always a whole number. The test statistic calculated previously is approximated by the Student’s t -distribution with df as follows:

[latex]\displaystyle{df}=\frac{((\frac{(s_1)^2}{n_1})+(\frac{(s_2)^2}{n_2}))^2}{(\frac{1}{n_1-1})(\frac{(s_1)^2}{n_1})^2+(\frac{1}{n_2-1})(\frac{(s_2)^2}{n_2})^2} [/latex]

When both sample sizes n 1 and n 2 are five or larger, the Student’s t approximation is very good. Notice that the sample variances ( s 1 ) 2 and ( s 2 ) 2 are not pooled. (If the question comes up, do not pool the variances.)

Note: It is not necessary to compute this by hand. A calculator or computer easily computes it.

The average amount of time boys and girls aged seven to 11 spend playing sports each day is believed to be the same. A study is done and data are collected, resulting in the data in the table below. Each populations has a normal distribution.

Is there a difference in the mean amount of time boys and girls aged seven to 11 play sports each day? Test at the 5% level of significance.

The population standard deviations are not known. Let g be the subscript for girls and b be the subscript for boys. Then, μ g is the population mean for girls and μ b is the population mean for boys. This is a test of two independent groups , two population means.

Random variable: [latex]\displaystyle\overline{{X}}_{{{g}}}-\overline{{X}}_{{b}} [/latex] = difference in the sample mean amount of time girls and boys play sports each day.

[latex]H_0:\mu_g=\mu_b [/latex]; [latex]H_0:\mu_g-\mu_b=0 [/latex]

[latex]H_a:\mu_g\neq\mu_b [/latex]; [latex]H_a:\mu_g-\mu_b\neq{0} [/latex]

The words “ the same ” tell you H 0 has an equal sign. Since there are no other words to indicate H a , assume it says “ is different .” This is a two-tailed test.

Distribution for the test: Use t df where df is calculated using the df formula for independent groups, two population means. Using a calculator, df is approximately 18.8462. Do not pool the variances.

Calculate the p -value using a Student’s t -distribution: p -value = 0.0054

This is a normal distribution curve representing the difference in the average amount of time girls and boys play sports all day. The mean is equal to zero, and the values -1.2, 0, and 1.2 are labeled on the horizontal axis. Two vertical lines extend from -1.2 and 1.2 to the curve. The region to the left of x = -1.2 and the region to the right of x = 1.2 are shaded to represent the p-value. The area of each region is 0.0028.

So, [latex]\displaystyle\overline{x}_g-\overline{x}_b=2-3.2=-1.2 [/latex]

Half the p -value is below –1.2 and half is above 1.2.

Make a decision: Since α > p -value, reject H0 . This means you reject μg = μb . The means are different.

Press STAT .
Arrow over to TESTS and press 4:2-SampTTest .
Arrow over to Stats and press ENTER .
Arrow down and enter 2 for the first sample mean, [latex]0.866[/latex] for Sx1, 9 for n1, 3.2 for the second sample mean, 1 for Sx2, and 16 for n2.
Arrow down to μ1: and arrow to does not equal μ2.
Press ENTER .
Arrow down to Pooled: and No .
Arrow down to Calculate and press ENTER .

The p -value is p = 0.0054, the dfs are approximately 18.8462, and the test statistic is –3.14.

Do the procedure again but instead of Calculate do Draw.

Conclusion: At the 5% level of significance, the sample data show there is sufficient evidence to conclude that the mean number of hours that girls and boys aged seven to 11 play sports per day is different (mean number of hours boys aged seven to 11 play sports per day is greater than the mean number of hours played by girls OR the mean number of hours girls aged seven to 11 play sports per day is greater than the mean number of hours played by boys).

Two samples are shown in the table. Both have normal distributions. The means for the two populations are thought to be the same. Is there a difference in the means? Test at the 5% level of significance.

The p -value is 0.4125, which is much higher than 0.05, so we decline to reject the null hypothesis. There is not sufficient evidence to conclude that the means of the two populations are not the same.

Note: When the sum of the sample sizes is larger than 30 ( n 1 + n 2 > 30) you can use the normal distribution to approximate the Student’s t .

A study is done by a community group in two neighboring colleges to determine which one graduates students with more math classes. College A samples 11 graduates. Their average is four math classes with a standard deviation of 1.5 math classes. College B samples nine graduates. Their average is 3.5 math classes with a standard deviation of one math class. The community group believes that a student who graduates from college A has taken more math classes , on the average. Both populations have a normal distribution. Test at a 1% significance level. Answer the following questions.

Is this a test of two means or two proportions?
Are the populations standard deviations known or unknown?
Which distribution do you use to perform the test?
What is the random variable?
What are the null and alternate hypotheses?
Is this test right-, left-, or two-tailed?
What is the p -value?
Do you reject or not reject the null hypothesis?
Student’s t
[latex]\displaystyle\overline{{X}}_{{{A}}}-\overline{{X}}_{{B}} [/latex]
H 0 : μ A ≤ μ B H a : μ A > μ B

This is a normal distribution curve with mean equal to 0. A vertical line near the tail of the curve to the right of zero extends from the axis to the curve. The region under the curve to the right of the line is shaded.

Do not reject.

Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a student who graduates from college A has taken more math classes, on the average, than a student who graduates from college B.

A study is done to determine if Company A retains its workers longer than Company B. Company A samples 15 workers, and their average time with the company is five years with a standard deviation of 1.2. Company B samples 20 workers, and their average time with the company is 4.5 years with a standard deviation of 0.8. The populations are normally distributed.

Are the population standard deviations known?
Conduct an appropriate hypothesis test. At the 5% significance level, what is your conclusion?
They are unknown.
The p -value = 0.0878. At the 5% level of significance, there is insufficient evidence to conclude that the workers of Company A stay longer with the company.

A professor at a large community college wanted to determine whether there is a difference in the means of final exam scores between students who took his statistics course online and the students who took his face-to-face statistics class. He believed that the mean of the final exam scores for the online class would be lower than that of the face-to-face class. Was the professor correct? The randomly selected 30 final exam scores from each group are listed in the two tables below:

Online Class:

Face-to-face Class:

Is the mean of the Final Exam scores of the online class lower than the mean of the Final Exam scores of the face-to-face class? Test at a 5% significance level. Answer the following questions:

Are the population standard deviations known or unknown?
What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols.
Is this test right, left, or two tailed?
At the ___ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______.

(Review the conclusion in Example 2, and write yours in a similar fashion)

Be careful not to mix up the information for Group 1 and Group 2!

First put the data for each group into two lists (such as L1 and L2). Press STAT. Arrow over to TESTS and press 4:2SampTTest. Make sure Data is highlighted and press ENTER. Arrow down and enter L1 for the first list and L2 for the second list. Arrow down to μ 1 : and arrow to ≠ μ 2 (does not equal). Press ENTER. Arrow down to Pooled: No. Press ENTER. Arrow down to Calculate and press ENTER.

H 0 : μ 1 = μ 2 Null hypothesis: the means of the final exam scores are equal for the online and face-to-face statistics classes.
H a : μ 1 < μ 2 Alternative hypothesis: the mean of the final exam scores of the online class is less than the mean of the final exam scores of the face-to-face class.
left-tailed

This is a normal distribution curve with mean equal to zero. A vertical line near the tail of the curve to the left of zero extends from the axis to the curve. The region under the curve to the left of the line is shaded representing p-value = 0.0011.

Reject the null hypothesis
The professor was correct. The evidence shows that the mean of the final exam scores for the online class is lower than that of the face-to-face class.At the 5% level of significance, from the sample data, there is (is/is not) sufficient evidence to conclude that the mean of the final exam scores for the online class is less than the mean of final exam scores of the face-to-face class.

Cohen’s Standards for Small, Medium, and Large Effect Sizes

Cohen’s d is a measure of effect size based on the differences between two means. Cohen’s d , named for United States statistician Jacob Cohen, measures the relative strength of the differences between the means of two populations based on sample data. The calculated value of effect size is then compared to Cohen’s standards of small, medium, and large effect sizes.

Cohen’s Standard Effect Sizes

Cohen’s d is the measure of the difference between two means divided by the pooled standard deviation:

[latex]\displaystyle{d}=\frac{{\overline{{x}}_{{1}}-\overline{{x}}_{{2}}}}{{{s}_{{\text{pooled}}}}} \text{ where } {s}_{{\text{pooled}}}=\sqrt{{\frac{{{({n}_{{1}}-{1})}{{s}_{{1}}^{{2}}}+{({n}_{{2}}-{1})}{{s}_{{2}}^{{2}}}}}{{{n}_{{1}}+{n}_{{2}}-{2}}}}} [/latex]

Calculate Cohen’s d for Example 2. Is the size of the effect small, medium, or large? Explain what the size of the effect means for this problem.

μ 1 = 4 s 1 = 1.5 n 1 = 11

μ 2 = 3.5 s 2 = 1 n 2 = 9

The effect is small because 0.384 is between Cohen’s value of 0.2 for small effect size and 0.5 for medium effect size. The size of the differences of the means for the two colleges is small indicating that there is not a significant difference between them.

Calculate Cohen’s d for Example 3. Is the size of the effect small, medium or large? Explain what the size of the effect means for this problem.

d = 0.834; Large, because 0.834 is greater than Cohen’s 0.8 for a large effect size. The size of the differences between the means of the Final Exam scores of online students and students in a face-to-face class is large indicating a significant difference.

Weighted alpha is a measure of risk-adjusted performance of stocks over a period of a year. A high positive weighted alpha signifies a stock whose price has risen while a small positive weighted alpha indicates an unchanged stock price during the time period. Weighted alpha is used to identify companies with strong upward or downward trends. The weighted alpha for the top 30 stocks of banks in the northeast and in the west as identified by Nasdaq on May 24, 2013 are listed in the two tables below.

Is there a difference in the weighted alpha of the top 30 stocks of banks in the northeast and in the west? Test at a 5% significance level. Answer the following questions:

Calculate Cohen’s d and interpret it.
Student’s-t
H 0 : μ 1 = μ 2 Null hypothesis: the means of the weighted alphas are equal.
H a : μ 1 ≠ μ 2 Alternative hypothesis : the means of the weighted alphas are not equal.
p -value = 0.8787
Do not reject the null hypothesis

This is a normal distribution curve with mean equal to zero. Both the right and left tails of the curve are shaded. Each tail represents 1/2(p-value) = 0.4394.

d = 0.040, Very small, because 0.040 is less than Cohen’s value of 0.2 for small effect size. The size of the difference of the means of the weighted alphas for the two regions of banks is small indicating that there is not a significant difference between their trends in stocks.

Concept Review

Two population means from independent samples where the population standard deviations are not known

Random Variable: [latex]\displaystyle\overline{{X}}_{{1}}-\overline{{X}}_{{2}} [/latex] = the difference of the sampling means
Distribution: Student’s t -distribution with degrees of freedom (variances not pooled)

Formula Review

Standard error: [latex]\displaystyle{S}{E}=\sqrt{{\frac{{({s}_{{1}})}^{{2}}}{{n}_{{1}}}+\frac{{({s}_{{2}})}^{{2}}}{{n}_{{2}}}}} [/latex]

Test statistic ( t -score): [latex]\displaystyle{t}=\frac{{{(\overline{{x}}_{{1}}-\overline{{x}}_{{2}})}-{(\mu_{{1}}-\mu_{{2}})}}}{\sqrt{{\frac{{({s}_{{1}})}^{{2}}}{{n}_{{1}}}+\frac{{({s}_{{2}})}^{{2}}}{{n}_{{2}}}}}} [/latex]

Degrees of freedom:

[latex]\displaystyle{d}{f}=\frac{{{(\frac{{({s}_{{1}})}^{{2}}}{{n}_{{1}}}+\frac{{({s}{2})}^{{2}}}{{n}_{{2}}})}^{{2}}}}{{{(\frac{{1}}{{{n}_{{1}}-{1}}})}{(\frac{{({s}_{{1}})}^{{2}}}{{n}_{{1}}})}^{{2}}+{(\frac{{1}}{{{n}_{{2}}-{1}}})}{(\frac{{({s}_{{2}})}^{{2}}}{{n}^{{2}}})}^{{2}}}} [/latex] where: s 1 and s 2 are the sample standard deviations, and n 1 and n 2 are the sample sizes.

[latex]\displaystyle\overline{{x}}_{{1}}{\quad\text{and}\quad}\overline{{x}}_{{2}} [/latex] are the sample means.

Cohen’s d is the measure of effect size:

Two Population Means with Unknown Standard Deviations. Provided by : OpenStax. Located at : . License : CC BY: Attribution
Introductory Statistics . Authored by : Barbara Illowski, Susan Dean. Provided by : Open Stax. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]
One-tailed and two-tailed tests | Inferential statistics | Probability and Statistics | Khan Academy. Authored by : Khan Academy. Located at : https://www.youtube.com/embed/mvye6X_0upA . License : All Rights Reserved . License Terms : Standard YouTube License

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8.2 A Single Population Mean (Unknown σ)

To find the standard error, we need the population standard deviation, [latex]\sigma[/latex]. Unfortunately, this value isn’t generally unknown. In this situation, the next best thing we can do is to use the sample standard deviation, [latex]s[/latex], as a substitute for [latex]\sigma[/latex]. This substitution works well for large samples but this estimate is off for small sample sizes. In the case of small sample sizes taken from an underlying normal distribution, a different kind of distribution, called a t -distribution gives better results.

Choosing appropriate distribution: What do I use: z or t distribution? Do we know population standard deviation, σ?

If YES, use normal distribution ( z -distribution)
If NO, then use t -distribution. Note : You may use normal distribution if sample size is at least 30 ( n ≥ 30) even if σ is unknown. For n ≥ 30, you can use sample standard deviation ( s ) in place of population standard deviation ([latex]\sigma[/latex]). Note that since the t and z -distributions look similar ( see Desmos demo ) for larger sample sizes, probability calculations from these distributions will lead to similar results. For this reason, when dealing with large sample sizes from populations with unknown standard deviations (unknown [latex]\sigma[/latex]), many simply prefer to use t -distribution instead of z .

Finding Critical Values

Use Desmos or StatKey. (Note: If you need to use more than 3 decimal places for the critical value, you may want to skip StatKey)

Critical t -value using Desmos | Critical t -value using StatKey (no audio)

Please complete the following practice exercise: Finding the critical value t for a desired confidence level

EXAMPLE: MARGIN OF ERROR

A survey of 46 people showed that the respondents spent an average of $31 on their child’s last birthday gift with a standard deviation of $9. Find the critical value for a 95% confidence level, the standard error, and the margin of error. Assume that the population is normally distributed.

The mean and the standard deviations given here are about a sample , as it says in the question — a sample of size 46 with a mean of $31 and a standard deviation of $9 .

Given facts are:

[latex]n=46[/latex]

[latex]\bar x = $31[/latex]

[latex]s = $9[/latex]. This is not σ (The notation σ represents the population standard deviation. What does s represent?)

Since the population standard deviation is unknown, we use t -distribution. If the sample size is at least 30, the result from using normal distribution is approximately equal to the one from that of using a t -distribution. Therefore, using normal distribution ( z -distribution) wouldn’t be too far off, but t -distribution is there, so why not use it.

Critical value: Use a t -distribution to find the critical value. What would be the degrees of freedom for the t -distribution here?

Standard Error: Standard Error is given by [latex]\sigma_{\bar x} = \dfrac{s}{\sqrt n}[/latex]

Now that we have the pieces sorted out, let’s use the EBM (or Margin or Error, ME) formula to find the margin of error.

Margin of Error = (Critical Value) • (Standard Error)

EXAMPLE: MEAN WITH STATISTICS

The effectiveness of a blood-pressure drug is being investigated. An experimenter finds that, on average, the reduction in systolic blood pressure is 40.9 for a sample of size 20 and standard deviation 11.7.

Estimate how much the drug will lower a typical patient’s systolic blood pressure (using a 98% confidence level).

Assume the data is from a normally distributed population . Round your answers to 3 decimals .

Confidence level, c = 0.98 (for a 98% confidence interval)

Sample info (These include sample statistics):

Sample Mean, [latex]\bar x = 40.9[/latex] Sample Standard Deviation, [latex]s = 11.7[/latex] Sample Size, [latex]n = 20[/latex]

Population standard deviation is not known. Sample size is fewer than 30 and the population is normally distributed. Therefore, use a t-distribution. Note that confidence interval is:

Point Estimate ± Margin of Error and Margin of Error = Critical Value • Standard Error , where t c is the critical value and the standard error of the mean = [latex]\dfrac{s}{\sqrt n}[/latex].

Margin of Error = [latex]t_c \cdot \dfrac{s}{\sqrt n}[/latex]

So, confidence interval is:

Point Estimate ± Critical Value • Standard Error Point Estimate [latex]\pm \:{\color{#e03e2d} {t_c}} \cdot \dfrac{s}{\sqrt n}[/latex]. First, recognize that the sample mean is our point estimate. x̅ = 40.9 . And our sample standard deviation is given as well: s = 11.7 , while the sample size n = 20 .

Let’s update our CI:

Point Estimate ± Margin of Error = Point Estimate [latex]\pm \:{\color{#e03e2d}{ t_c}} \cdot \dfrac{s}{\sqrt {n}}[/latex] = [latex]40.9 \pm \:{\color{#e03e2d} {t_c}} \cdot \dfrac{11.7}{\sqrt {20}}[/latex]

Now, we just need the critical value. With [latex]n = 20[/latex], degrees of freedom, [latex]d.f. = n - 1 = .........[/latex]

Use Desmos or StatKey or TiCalculator to find the critical value, [latex]t_c[/latex], for a 98% confidence level. After you have computed the value, click on Show More below to show critical value and more:

[latex]t_c= 2.53948319062[/latex]

Let’s plug this [latex]t_c[/latex] into the formula above to find the confidence interval: [latex]40.9 \pm 2.53948319062[/latex] • [latex]\dfrac{11.7}{\sqrt {20}}[/latex] = [latex]40.9 \pm 6.64379473907[/latex]

This is our confidence interval in ± notation.

Three Ways to Write Confidence Intervals

1) So, the confidence interval in interval notation: (34.25620530117019, 47.54379469882981) → Round to 3 decimals: (34.256, 47.544)

2) Confidence interval in tri-inequality notation: 34.25620530117019 < [latex]\mu[/latex] < 47.54379469882981 34.256 < [latex]\mu[/latex] < 47.544

3) Confidence interval in plus-minus notation: Margin of Error, ME or EBM = 47.54379469882981 − 40.9 = 6.64379469883 ≈ 6.644 Confidence interval: 40.9 ± 6.644

The SUBEDI calculator gives answers in ± notation, whereas the LibreText calculator ‘s results are in interval notation. Be sure to convert CI in one notation to another. (See Three Ways to Write a Confidence Interval for additional details on notations).

ONLINE CALCULATOR Approach

Go to Confidence Interval for a Mean calculator @ rsubedi.com

Confidence Level (in decimal), [latex]c: \fbox{$\mathstrut \;0.98\;$}[/latex]

Number of Samples

Distribution Type to Use?

Sample Size, [latex]n: \fbox{$\mathstrut \;20\;$}[/latex]

Sample Mean, [latex]\bar x: \fbox{$\mathstrut \;40.9\;$}[/latex]

Sample Standard Deviation, [latex]n: \fbox{$\mathstrut \;11.7\;$}[/latex]

CALCULATE Results show in a panel to the right. CI is displayed in [latex]\pm[/latex] notation.

Go to: Confidence Interval for a Mean With Statistics from the list of online calculators.

Enter the following values and press Calculate .

Results displayed are:

EXAMPLE: MEAN WITH DATA

You are interested in finding a 90% confidence interval for the mean number of visits for physical therapy patients. The data below show the number of visits for 11 randomly selected physical therapy patients.

Confidence level, c = 0.90 (for a 90% confidence interval)

Population standard deviation is not known. Sample size is fewer than 30 and the population is normally distributed. Therefore, use a t -distribution.

Confidence Level (in decimal), [latex]c: \fbox{$\mathstrut \;0.90\;$}[/latex]

Enter your data in the spreadsheet column shown for data entry.

Select the above data to copy. Once copied, on SUBEDI Calc click on the first cell of the spreadsheet and paste the data (Control+V or Command + V).

Go to: Confidence Interval Calculator with Data from the list of online calculators Enter the following values and press Calculate .

Data: (Separate each value with a comma)

[latex]\fbox{$\mathstrut \quad14, 12, 6, 27, 12, 13, 21, 20, 20, 13, 19\quad$}[/latex]

[latex]\text{CL}: \fbox{$\mathstrut \;0.90\;$}[/latex]

CALCULATE Results displayed are:

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9.3 Probability Distribution Needed for Hypothesis Testing

Earlier in the course, we discussed sampling distributions. Particular distributions are associated with various types of hypothesis testing.

The following table summarizes various hypothesis tests and corresponding probability distributions that will be used to conduct the test (based on the assumptions shown below):

Assumptions

When you perform a hypothesis test of a single population mean μ using a normal distribution (often called a z-test), you take a simple random sample from the population. The population you are testing is normally distributed , or your sample size is sufficiently large. You know the value of the population standard deviation , which, in reality, is rarely known.

When you perform a hypothesis test of a single population mean μ using a Student's t-distribution (often called a t -test), there are fundamental assumptions that need to be met in order for the test to work properly. Your data should be a simple random sample that comes from a population that is approximately normally distributed. You use the sample standard deviation to approximate the population standard deviation. (Note that if the sample size is sufficiently large, a t -test will work even if the population is not approximately normally distributed).

When you perform a hypothesis test of a single population proportion p , you take a simple random sample from the population. You must meet the conditions for a binomial distribution : there are a certain number n of independent trials, the outcomes of any trial are success or failure, and each trial has the same probability of a success p . The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np and nq must both be greater than five ( n p > 5 n p > 5 and n q > 5 n q > 5 ). Then the binomial distribution of a sample (estimated) proportion can be approximated by the normal distribution with μ = p μ = p and σ = p q n σ = p q n . Remember that q = 1 - p q q = 1 - p q .

Hypothesis Test for the Mean

Going back to the standardizing formula we can derive the test statistic for testing hypotheses concerning means.

The standardizing formula cannot be solved as it is because we do not have μ, the population mean. However, if we substitute in the hypothesized value of the mean, μ 0 in the formula as above, we can compute a Z value. This is the test statistic for a test of hypothesis for a mean and is presented in Figure 9.3 . We interpret this Z value as the associated probability that a sample with a sample mean of X ¯ X ¯ could have come from a distribution with a population mean of H 0 and we call this Z value Z c for “calculated”. Figure 9.3 and Figure 9.4 show this process.

In Figure 9.3 two of the three possible outcomes are presented. X ¯ 1 X ¯ 1 and X ¯ 3 X ¯ 3 are in the tails of the hypothesized distribution of H 0 . Notice that the horizontal axis in the top panel is labeled X ¯ X ¯ 's. This is the same theoretical distribution of X ¯ X ¯ 's, the sampling distribution, that the Central Limit Theorem tells us is normally distributed. This is why we can draw it with this shape. The horizontal axis of the bottom panel is labeled Z and is the standard normal distribution. Z α 2 Z α 2 and -Z α 2 -Z α 2 , called the critical values , are marked on the bottom panel as the Z values associated with the probability the analyst has set as the level of significance in the test, (α). The probabilities in the tails of both panels are, therefore, the same.

Notice that for each X ¯ X ¯ there is an associated Z c , called the calculated Z, that comes from solving the equation above. This calculated Z is nothing more than the number of standard deviations that the hypothesized mean is from the sample mean. If the sample mean falls "too many" standard deviations from the hypothesized mean we conclude that the sample mean could not have come from the distribution with the hypothesized mean, given our pre-set required level of significance. It could have come from H 0 , but it is deemed just too unlikely. In Figure 9.3 both X ¯ 1 X ¯ 1 and X ¯ 3 X ¯ 3 are in the tails of the distribution. They are deemed "too far" from the hypothesized value of the mean given the chosen level of alpha. If in fact this sample mean it did come from H 0 , but from in the tail, we have made a Type I error: we have rejected a good null. Our only real comfort is that we know the probability of making such an error, α, and we can control the size of α.

Figure 9.4 shows the third possibility for the location of the sample mean, x _ x _ . Here the sample mean is within the two critical values. That is, within the probability of (1-α) and we cannot reject the null hypothesis.

This gives us the decision rule for testing a hypothesis for a two-tailed test:

This rule will always be the same no matter what hypothesis we are testing or what formulas we are using to make the test. The only change will be to change the Z c to the appropriate symbol for the test statistic for the parameter being tested. Stating the decision rule another way: if the sample mean is unlikely to have come from the distribution with the hypothesized mean we cannot accept the null hypothesis. Here we define "unlikely" as having a probability less than alpha of occurring.

P-Value Approach

An alternative decision rule can be developed by calculating the probability that a sample mean could be found that would give a test statistic larger than the test statistic found from the current sample data assuming that the null hypothesis is true. Here the notion of "likely" and "unlikely" is defined by the probability of drawing a sample with a mean from a population with the hypothesized mean that is either larger or smaller than that found in the sample data. Simply stated, the p-value approach compares the desired significance level, α, to the p-value which is the probability of drawing a sample mean further from the hypothesized value than the actual sample mean. A large p -value calculated from the data indicates that we should not reject the null hypothesis . The smaller the p -value, the more unlikely the outcome, and the stronger the evidence is against the null hypothesis. We would reject the null hypothesis if the evidence is strongly against it. The relationship between the decision rule of comparing the calculated test statistics, Z c , and the Critical Value, Z α , and using the p -value can be seen in Figure 9.5 .

The calculated value of the test statistic is Z c in this example and is marked on the bottom graph of the standard normal distribution because it is a Z value. In this case the calculated value is in the tail and thus we cannot accept the null hypothesis, the associated X ¯ X ¯ is just too unusually large to believe that it came from the distribution with a mean of µ 0 with a significance level of α.

If we use the p -value decision rule we need one more step. We need to find in the standard normal table the probability associated with the calculated test statistic, Z c . We then compare that to the α associated with our selected level of confidence. In Figure 9.5 we see that the p -value is less than α and therefore we cannot accept the null. We know that the p -value is less than α because the area under the p-value is smaller than α/2. It is important to note that two researchers drawing randomly from the same population may find two different P-values from their samples. This occurs because the P-value is calculated as the probability in the tail beyond the sample mean assuming that the null hypothesis is correct. Because the sample means will in all likelihood be different this will create two different P-values. Nevertheless, the conclusions as to the null hypothesis should be different with only the level of probability of α.

Here is a systematic way to make a decision of whether you cannot accept or cannot reject a null hypothesis if using the p -value and a preset or preconceived α (the " significance level "). A preset α is the probability of a Type I error (rejecting the null hypothesis when the null hypothesis is true). It may or may not be given to you at the beginning of the problem. In any case, the value of α is the decision of the analyst. When you make a decision to reject or not reject H 0 , do as follows:

If α > p -value, cannot accept H 0 . The results of the sample data are significant. There is sufficient evidence to conclude that H 0 is an incorrect belief and that the alternative hypothesis , H a , may be correct.
If α ≤ p -value, cannot reject H 0 . The results of the sample data are not significant. There is not sufficient evidence to conclude that the alternative hypothesis, H a , may be correct. In this case the status quo stands.
When you "cannot reject H 0 ", it does not mean that you should believe that H 0 is true. It simply means that the sample data have failed to provide sufficient evidence to cast serious doubt about the truthfulness of H 0 . Remember that the null is the status quo and it takes high probability to overthrow the status quo. This bias in favor of the null hypothesis is what gives rise to the statement "tyranny of the status quo" when discussing hypothesis testing and the scientific method.

Both decision rules will result in the same decision and it is a matter of preference which one is used.

One and Two-tailed Tests

The discussion of Figure 9.3 - Figure 9.5 was based on the null and alternative hypothesis presented in Figure 9.3 . This was called a two-tailed test because the alternative hypothesis allowed that the mean could have come from a population which was either larger or smaller than the hypothesized mean in the null hypothesis. This could be seen by the statement of the alternative hypothesis as μ ≠ 100, in this example.

It may be that the analyst has no concern about the value being "too" high or "too" low from the hypothesized value. If this is the case, it becomes a one-tailed test and all of the alpha probability is placed in just one tail and not split into α/2 as in the above case of a two-tailed test. Any test of a claim will be a one-tailed test. For example, a car manufacturer claims that their Model 17B provides gas mileage of greater than 25 miles per gallon. The null and alternative hypothesis would be:

H 0 : µ ≤ 25
H a : µ > 25

The claim would be in the alternative hypothesis. The burden of proof in hypothesis testing is carried in the alternative. This is because failing to reject the null, the status quo, must be accomplished with 90 or 95 percent confidence that it cannot be maintained. Said another way, we want to have only a 5 or 10 percent probability of making a Type I error, rejecting a good null; overthrowing the status quo.

This is a one-tailed test and all of the alpha probability is placed in just one tail and not split into α/2 as in the above case of a two-tailed test.

Figure 9.6 shows the two possible cases and the form of the null and alternative hypothesis that give rise to them.

where μ 0 is the hypothesized value of the population mean.

Effects of Sample Size on Test Statistic

In developing the confidence intervals for the mean from a sample, we found that most often we would not have the population standard deviation, σ. If the sample size were less than 30, we could simply substitute the point estimate for σ, the sample standard deviation, s, and use the student's t -distribution to correct for this lack of information.

When testing hypotheses we are faced with this same problem and the solution is exactly the same. Namely: If the population standard deviation is unknown, and the sample size is less than 30, substitute s, the point estimate for the population standard deviation, σ, in the formula for the test statistic and use the student's t -distribution. All the formulas and figures above are unchanged except for this substitution and changing the Z distribution to the student's t -distribution on the graph. Remember that the student's t -distribution can only be computed knowing the proper degrees of freedom for the problem. In this case, the degrees of freedom is computed as before with confidence intervals: df = (n-1). The calculated t-value is compared to the t-value associated with the pre-set level of confidence required in the test, t α , df found in the student's t tables. If we do not know σ, but the sample size is 30 or more, we simply substitute s for σ and use the normal distribution.

Table 9.5 summarizes these rules.

A Systematic Approach for Testing a Hypothesis

A systematic approach to hypothesis testing follows the following steps and in this order. This template will work for all hypotheses that you will ever test.

Set up the null and alternative hypothesis. This is typically the hardest part of the process. Here the question being asked is reviewed. What parameter is being tested, a mean, a proportion, differences in means, etc. Is this a one-tailed test or two-tailed test?

Decide the level of significance required for this particular case and determine the critical value. These can be found in the appropriate statistical table. The levels of confidence typical for businesses are 80, 90, 95, 98, and 99. However, the level of significance is a policy decision and should be based upon the risk of making a Type I error, rejecting a good null. Consider the consequences of making a Type I error.

Next, on the basis of the hypotheses and sample size, select the appropriate test statistic and find the relevant critical value: Z α , t α , etc. Drawing the relevant probability distribution and marking the critical value is always big help. Be sure to match the graph with the hypothesis, especially if it is a one-tailed test.

Take a sample(s) and calculate the relevant parameters: sample mean, standard deviation, or proportion. Using the formula for the test statistic from above in step 2, now calculate the test statistic for this particular case using the parameters you have just calculated.
The test statistic is in the tail: Cannot Accept the null, the probability that this sample mean (proportion) came from the hypothesized distribution is too small to believe that it is the real home of these sample data.
The test statistic is not in the tail: Cannot Reject the null, the sample data are compatible with the hypothesized population parameter.
Reach a conclusion. It is best to articulate the conclusion two different ways. First a formal statistical conclusion such as “With a 5 % level of significance we cannot accept the null hypotheses that the population mean is equal to XX (units of measurement)”. The second statement of the conclusion is less formal and states the action, or lack of action, required. If the formal conclusion was that above, then the informal one might be, “The machine is broken and we need to shut it down and call for repairs”.

All hypotheses tested will go through this same process. The only changes are the relevant formulas and those are determined by the hypothesis required to answer the original question.

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/introductory-business-statistics-2e/pages/1-introduction

Authors: Alexander Holmes, Barbara Illowsky, Susan Dean
Publisher/website: OpenStax
Book title: Introductory Business Statistics 2e
Publication date: Dec 13, 2023
Location: Houston, Texas
Book URL: https://openstax.org/books/introductory-business-statistics-2e/pages/1-introduction
Section URL: https://openstax.org/books/introductory-business-statistics-2e/pages/9-3-probability-distribution-needed-for-hypothesis-testing

© Dec 6, 2023 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

Interpret all statistics for 1 Variance

In this topic, null hypothesis and alternative hypothesis, confidence interval (ci) and bounds, test statistic, interpretation.

In the output, the null and alternative hypotheses help you to verify that you entered the correct value for the hypothesized standard deviation or hypothesized variance.

The sample size (N) is the total number of observations in the sample.

The sample size affects the confidence interval and the power of the test.

Usually, a larger sample size results in a narrower confidence interval. A larger sample size also gives the test more power to detect a difference. For more information, go to What is power? .

The standard deviation is the most common measure of dispersion, or how spread out the data are about the mean. The symbol σ (sigma) is often used to represent the standard deviation of a population, while s is used to represent the standard deviation of a sample. Variation that is random or natural to a process is often referred to as noise.

The standard deviation uses the same units as the data.

The standard deviation of the sample data is an estimate of the population standard deviation.

Because the standard deviation is based on sample data and not on the entire population, it is unlikely that the sample standard deviation equals the population standard deviation. To better estimate the population standard deviation, use the confidence interval.

hypothesis testing for mean standard deviation unknown

Hospital discharge times

Administrators track the discharge time for patients who are treated in the emergency departments of two hospitals. Although the average discharge times are about the same (35 minutes), the standard deviations are significantly different. The standard deviation for hospital 1 is about 6. On average, a patient's discharge time deviates from the mean (dashed line) by about 6 minutes. The standard deviation for hospital 2 is about 20. On average, a patient's discharge time deviates from the mean (dashed line) by about 20 minutes.

The variance measures how spread out the data are about their mean. The variance is equal to the standard deviation squared.

The variance of the sample data is an estimate of the population variance.

Because the variance is based on sample data and not on the entire population, it is unlikely that the sample variance equals the population variance. To better estimate the population variance, use the confidence interval.

The confidence interval provides a range of likely values for the population standard deviation or the population variance. Because samples are random, two samples from a population are unlikely to yield identical confidence intervals. But, if you repeated your sample many times, a certain percentage of the resulting confidence intervals or bounds would contain the unknown population standard deviation or population variance. The percentage of these confidence intervals or bounds that contain the standard deviation or the variance is the confidence level of the interval. For example, a 95% confidence level indicates that if you take 100 random samples from the population, you could expect approximately 95 of the samples to produce intervals that contain the population standard deviation or the population variance.

An upper bound defines a value that the population standard deviation or population variance is likely to be less than. A lower bound defines a value that the population standard deviation or population variance is likely to be greater than.

The confidence interval helps you assess the practical significance of your results. Use your specialized knowledge to determine whether the confidence interval includes values that have practical significance for your situation. If the interval is too wide to be useful, consider increasing your sample size. For more information, go to Ways to get a more precise confidence interval .

When you enter a column of data, Minitab only calculates a confidence interval for the standard deviation.

Minitab cannot calculate the Bonett method with summarized data.

Descriptive Statistics

In these results, the estimate of the population standard deviation for the length of beams is 0.871, and the estimate of the population variance is 0.759. Because the data did not pass a normality test, use the Bonett method. You can be 95% confident that the population standard deviation is between 0.704 and 1.121.

The test statistic is a statistic for chi-square tests that measures the ratio between an observed variance and its hypothesized variance.

You can compare the test statistic to critical values of the chi-square distribution to determine whether to reject the null hypothesis. However, using the p-value of the test to make the same determination is usually more practical and convenient. The p-value has the same meaning for any size test, but the same chi-square statistic can indicate opposite conclusions depending on the sample size.

The test statistic is used to calculate the p-value. Because there is no test statistic for the Bonnet method, Minitab uses the rejection regions that are defined by the confidence limits to calculate a p-value.

The degrees of freedom (DF) indicate the amount of information that is available in your data to estimate the values of the unknown parameters, and to calculate the variability of these estimates. For a 1 variance test, the degrees of freedom are determined by the number of observations in your sample.

Minitab uses the degrees of freedom to determine the test statistic. The degrees of freedom are determined by the sample size. Increasing your sample size provides more information about the population, which increases the degrees of freedom.

The p-value is a probability that measures the evidence against the null hypothesis. A smaller p-value provides stronger evidence against the null hypothesis.

Use the p-value to determine whether the population variance or standard deviation is statistically different from the hypothesized variance or standard deviation.

When you have summarized data, Minitab cannot calculate a p-value for the Bonett method.

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11.3: Two Population Means with Known Standard Deviations

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Even though this situation is not likely (knowing the population standard deviations is not likely), the following example illustrates hypothesis testing for independent means, known population standard deviations. The sampling distribution for the difference between the means is normal and both populations must be normal. The random variable is $\bar{X}_{1} - \bar{X}_{2}$. The normal distribution has the following format:

Normal distribution is:

\[\bar{X}_{1} - \bar{X}_{2} \sim{N}\left[\mu_{1} - \mu_{2}, \sqrt{\dfrac{(\sigma_{1})^{2}}{n_{1}} + \dfrac{(\sigma_{2})^{2}}{n_{2}}}\right] \label{eq1}\]

The standard deviation is:

\[\sqrt{\dfrac{(\sigma_{1})^{2}}{n_{1}} + \dfrac{(\sigma_{2})^{2}}{n_{2}}}\label{eq2}\]

The test statistic ( z -score) is:

\[z = \dfrac{(\bar{x}_{1} - \bar{x}_{2}) - (\mu_{1} - \mu_{2})}{\sqrt{\dfrac{(\sigma_{1})^{2}}{n_{1}} + \dfrac{(\sigma_{2})^{2}}{n_{2}}}} \label{eq3}\]

Example $\PageIndex{1}$

Independent groups, population standard deviations known: The mean lasting time of two competing floor waxes is to be compared. Twenty floors are randomly assigned to test each wax. Both populations have a normal distributions. The data are recorded in Table.

Does the data indicate that wax 1 is more effective than wax 2 ? Test at a 5% level of significance.

This is a test of two independent groups, two population means, population standard deviations known.

Random Variable: $\bar{X}_{1} - \bar{X}_{2} =$ difference in the mean number of months the competing floor waxes last.

$H_{0}: \mu_{1} \leq \mu_{2}$
$H_{a}: \mu_{1} > \mu_{2}$

The words "is more effective" says that wax 1 lasts longer than wax 2 , on average. "Longer" is a “>” symbol and goes into $H_{a}$. Therefore, this is a right-tailed test.

Distribution for the test: The population standard deviations are known so the distribution is normal. Using Equation \ref{eq1}, the distribution is:

\[\bar{X}_{1} - \bar{X}_{2} \sim{N} \left(0, \sqrt{\dfrac{0.33^{2}}{20} + \dfrac{0.36^{2}}{20}}\right)\]

Since $\mu_{1} \leq \mu_{2}$ then $\mu_{1} - \mu_{2} \leq 0$ and the mean for the normal distribution is zero.

Calculate the $p\text{-value}$ using the normal distribution: $p\text{-value} = 0.1799$

This is a normal distribution curve with mean equal to zero. The values 0 and 0.1 are labeled on the horiztonal axis. A vertical line extends from 0.1 to the curve. The region under the curve to the right of the line is shaded to represent p-value = 0.1799.

$\bar{X}_{1} - \bar{X}_{2} = 3 – 2.9 = 0.1$

Compare $\alpha$ and the $p\text{-value}$ : $\alpha = 0.05$ and $p\text{-value} = 0.1799$. Therefore, $\alpha < p\text{-value}$.

Make a decision: Since $\alpha < p\text{-value}$, do not reject $H_{0}$.

Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean time wax 1 lasts is longer (wax 1 is more effective) than the mean time wax 2 lasts.

The Two Independent Samples With Statistics and Known Population Standard Deviations calculator is much more direct. Just enter in:

First Sample Sample Size = 20, First Sample Sample Size = 3, First Sample Population Standard Deviation = 0.33

Second Sample Sample Size = 20, Second Sample Sample Size = 2.9, Second Sample Population Standard Deviation = 0.36

Check ">" and click on Calculate. The $p\text{-value}$ is $p = 0.1799$ and the test statistic is 0.9157.

Two Independent Samples with statistics, Population Standard Deviation known Calculator

Enter in the statistics, the tail type and the confidence level and hit Calculate and the test statistic, t, the p-value, p, the confidence interval's lower bound, LB, and the upper bound, UB will be shown. Be sure to enter the confidence level as a decimal, e.g., 95% has a CL of 0.95.

Exercise $\PageIndex{1}$

The means of the number of revolutions per minute of two competing engines are to be compared. Thirty engines are randomly assigned to be tested. Both populations have normal distributions. Table shows the result. Do the data indicate that Engine 2 has higher RPM than Engine 1? Test at a 5% level of significance.

The $p\text{-value}$ is almost 0, so we reject the null hypothesis. There is sufficient evidence to conclude that Engine 2 runs at a higher RPM than Engine 1.

Example $\PageIndex{2}$: Age of Senators

An interested citizen wanted to know if Democratic U. S. senators are older than Republican U.S. senators, on average. On May 26 2013, the mean age of 30 randomly selected Republican Senators was 61 years 247 days old (61.675 years) with a standard deviation of 10.17 years. The mean age of 30 randomly selected Democratic senators was 61 years 257 days old (61.704 years) with a standard deviation of 9.55 years.

Do the data indicate that Democratic senators are older than Republican senators, on average? Test at a 5% level of significance.

This is a test of two independent groups, two population means. The population standard deviations are unknown, but the sum of the sample sizes is 30 + 30 = 60, which is greater than 30, so we can use the normal approximation to the Student’s-t distribution. Subscripts: 1: Democratic senators 2: Republican senators

Random variable: $\bar{X}_{1} - \bar{X}_{2} =$ difference in the mean age of Democratic and Republican U.S. senators.

$H_{0}: \mu_{1} \leq \mu_{2} H_{0}: \mu_{1} - \mu_{2} \leq 0$
$H_{a}: \mu_{1} > \mu_{2} H_{a}: \mu_{1} - \mu_{2} > 0$

The words "older than" translates as a “>” symbol and goes into $H_{a}$. Therefore, this is a right-tailed test.

Distribution for the test: The distribution is the normal approximation to the Student’s t for means, independent groups. Using the formula, the distribution is: \[\bar{X}_{1} - \bar{X}_{2} \sim N\left[0, \sqrt{\dfrac{(9.55)^{2}}{30} + \dfrac{(10.17)^{2}}{30}}\right]\]

Since $\mu_{1} \leq \mu_{2}, \mu_{1} - \mu_{2} \leq 0$ and the mean for the normal distribution is zero.

(Calculating the p -value using the normal distribution gives $p\text{-value} = 0.4040$)

This is a normal distribution curve with mean equal to zero. A vertical line to the right of zero extends from the axis to the curve. The region under the curve to the right of the line is shaded representing p-value = 0.4955.

Compare $\alpha$ and the $p\text{-value}$ : $\alpha = 0.05$ and $p\text{-value} = 0.4040$. Therefore, $\alpha < p\text{-value}$.

Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of Democratic senators is greater than the mean age of the Republican senators.

Data from the United States Census Bureau. Available online at http://www.census.gov/prod/cen2010/b...c2010br-02.pdf
Hinduja, Sameer. “Sexting Research and Gender Differences.” Cyberbulling Research Center, 2013. Available online at http://cyberbullying.us/blog/sexting...r-differences/ (accessed June 17, 2013).
“Smart Phone Users, By the Numbers.” Visually, 2013. Available online at http://visual.ly/smart-phone-users-numbers (accessed June 17, 2013).
Smith, Aaron. “35% of American adults own a Smartphone.” Pew Internet, 2013. Available online at http://www.pewinternet.org/~/media/F...martphones.pdf (accessed June 17, 2013).
“State-Specific Prevalence of Obesity AmongAduls—Unites States, 2007.” MMWR, CDC. Available online at http://www.cdc.gov/mmwr/preview/mmwrhtml/mm5728a1.htm (accessed June 17, 2013).
“Texas Crime Rates 1960–1012.” FBI, Uniform Crime Reports, 2013. Available online at: http://www.disastercenter.com/crime/txcrime.htm (accessed June 17, 2013).

Chapter Review

A hypothesis test of two population means from independent samples where the population standard deviations are known (typically approximated with the sample standard deviations), will have these characteristics:
Random variable: $\bar{X}_{1} - \bar{X}_{2} =$ the difference of the means
Distribution: normal distribution

Formula Review

Normal Distribution:

\[\bar{X}_{1} - \bar{X}_{2} \sim N\left[\mu_{1} - \mu_{2}, \sqrt{\dfrac{(\sigma_{1})^{2}}{n_{1}} + \dfrac{(\sigma_{2})^{2}}{n_{2}}}\right]\]

Generally $\mu_{1} - \mu_{2} = 0$.

Test Statistic ( z -score):

\[z = \dfrac{(\bar{x}_{1} - \bar{x}_{2}) - (\mu_{1} - \mu_{2})}{\sqrt{\dfrac{(\sigma_{1})^{2}}{n_{1}} + \dfrac{(\sigma_{2})^{2}}{n_{2}}}}\]

$\sigma_{1}$ and $\sigma_{2}$ are the known population standard deviations. $n_{1}$ and $n_{1}$ are the sample sizes. $\bar{x}_{1}$ and $\bar{x}_{2}$ are the sample means. $\mu_{1}$ and $\mu_{2}$ are the population means

Contributors and Attributions

Barbara Illowsky and Susan Dean (De Anza College) with many other contributing authors. Content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] .

T-test for two Means – Unknown Population Standard Deviations

Instructions : Use this T-Test Calculator for two Independent Means calculator to conduct a t-test for two population means ($\mu_1$ and $\mu_2$), with unknown population standard deviations. This test apply when you have two-independent samples, and the population standard deviations $\sigma_1$ and $\sigma_2$ and not known. Please select the null and alternative hypotheses, type the significance level, the sample means, the sample standard deviations, the sample sizes, and the results of the t-test for two independent samples will be displayed for you:

The T-test for Two Independent Samples

More about the t-test for two means so you can better interpret the output presented above: A t-test for two means with unknown population variances and two independent samples is a hypothesis test that attempts to make a claim about the population means ($\mu_1$ and $\mu_2$).

More specifically, a t-test uses sample information to assess how plausible it is for the population means $\mu_1$ and $\mu_2$ to be equal. The test has two non-overlapping hypotheses, the null and the alternative hypothesis.

The null hypothesis is a statement about the population means, specifically the assumption of no effect, and the alternative hypothesis is the complementary hypothesis to the null hypothesis.

Properties of the two sample t-test

The main properties of a two sample t-test for two population means are:

Depending on our knowledge about the "no effect" situation, the t-test can be two-tailed, left-tailed or right-tailed
The main principle of hypothesis testing is that the null hypothesis is rejected if the test statistic obtained is sufficiently unlikely under the assumption that the null hypothesis is true
The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true
In a hypothesis tests there are two types of errors. Type I error occurs when we reject a true null hypothesis, and the Type II error occurs when we fail to reject a false null hypothesis

How do you compute the t-statistic for the t test for two independent samples?

The formula for a t-statistic for two population means (with two independent samples), with unknown population variances shows us how to calculate t-test with mean and standard deviation and it depends on whether the population variances are assumed to be equal or not. If the population variances are assumed to be unequal, then the formula is:

On the other hand, if the population variances are assumed to be equal, then the formula is:

Normally, the way of knowing whether the population variances must be assumed to be equal or unequal is by using an F-test for equality of variances.

With the above t-statistic, we can compute the corresponding p-value, which allows us to assess whether or not there is a statistically significant difference between two means.

Why is it called t-test for independent samples?

This is because the samples are not related with each other, in a way that the outcomes from one sample are unrelated from the other sample. If the samples are related (for example, you are comparing the answers of husbands and wives, or identical twins), you should use a t-test for paired samples instead .

What if the population standard deviations are known?

The main purpose of this calculator is for comparing two population mean when sigma is unknown for both populations. In case that the population standard deviations are known, then you should use instead this z-test for two means .

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8.4: Hypothesis Test on a Single Standard Deviation

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A test of a single standard deviation assumes that the underlying distribution is normal . The null and alternative hypotheses are stated in terms of the population standard deviation (or population variance). The test statistic is:

\[\chi^{2} = \frac{(n-1)s^{2}}{\sigma^{2}} \label{test}\]

$n$ is the the total number of data
$s^{2}$ is the sample variance
$\sigma^{2}$ is the population variance

The requirements to be able to perform a hypothesis test on a population standard deviation are:

the sample must be obtained from a simple random sample or from a randomized experiment
the population has a normal distribution

You may think of $s$ as the random variable in this test. The number of degrees of freedom is $df = n - 1$. A test of a single standard deviation may be right-tailed, left-tailed, or two-tailed. The next example will show you how to set up the null and alternative hypotheses. The null and alternative hypotheses contain statements about the population variance.

Example 8.4.1

Math instructors are not only interested in how their students do on exams, on average, but how the exam scores vary. To many instructors, the standard deviation may be more important than the average.

Suppose a math instructor believes that the standard deviation for his final exam is five points. One of his best students thinks otherwise. The student claims that the standard deviation is more than five points. If the student were to conduct a hypothesis test, what would the null and alternative hypotheses be?

$H_{0}: \sigma = 5$
$H_{a}: \sigma > 5$

Exercise 8.4.2

A SCUBA instructor wants to record the collective depths each of his students dives during their checkout. He is interested in how the depths vary, even though everyone should have been at the same depth. He believes the standard deviation is three feet. His assistant thinks the standard deviation is less than three feet. Suppose the instructor finds a random sample of 25 SCUBA students and found that the sample standard deviation is 2.8 feet.

With a significance level of 5%, test the claim that the diving depths is less than 3 feet.

$H_{0}: \sigma = 3$

$H_{a}: \sigma < 3$

The word "less" tells you this is a left-tailed test.

Distribution for the test: $\chi^{2}_{24}$, where $n = \text{the number of customers sampled}$ $df = n - 1 = 25 - 1 = 24$

Calculate the test statistic (Equation \ref{test}):

\[\chi^{2} = \frac{(n-1)s^{2}}{\sigma^{2}} = \frac{(25-1)(2.8)^{2}}{3^{2}} = 20.91 \nonumber\]

where $n = 25$, $s = 2.8$, and $\sigma = 3$.

A7D7D06C-938F-400F-99F0-F697F318FB16 copy.jpeg

Probability statement: $p\text{-value} = P(\chi^{2} < 20.91) = 0.356$

In 2nd DISTR , use 7:χ2cdf . The syntax is (lower, upper, df) for the parameter list. For Example , χ2cdf(-1E99,20.91,24) . The $p\text{-value} = 0.356$.

Compare $\alpha$ and the $p\text{-value}$ :

Given $\alpha = 0.05, p\text{-value} = 0.356 \}, \text{ then }\alpha < p\text{-value} \nonumber$

Make a decision: Since $\alpha < p\text{-value}$, fail to reject $H_{0}$. This means that you are not reject $\sigma = 3$. In other words, you do think the standard deviation of diving depth less than 3 feet.

Conclusion: At a 5% level of significance, from the data, there is not sufficient evidence to conclude that a SCUBA students' collective depth is less than 3 feet.

Example \(\{2}

“AppleInsider Price Guides.” Apple Insider, 2013. Available online at http://appleinsider.com/mac_price_guide (accessed May 14, 2013).
Data from the World Bank, June 5, 2012.

To test variability, use the chi-square test of a single variance. The test may be left-, right-, or two-tailed, and its hypotheses are always expressed in terms of the variance (or standard deviation).

Formula Review

$\chi^{2} = \frac{(n-1) \cdot s^{2}}{\sigma^{2}}$ Test of a single variance statistic where:

$n: \text{sample size}$

$s: \text{sample standard deviation}$

$\sigma: \text{population standard deviation}$

$df = n – 1 \text{Degrees of freedom}$

Test of a Single Standard Deviation

Use the test to determine standard deviation.
The degrees of freedom is the $\text{number of samples} - 1$.
The test statistic is $\frac{(n-1) \cdot s^{2}}{\sigma^{2}}$, where $n = \text{the total number of data}$, $s^{2} = \text{sample variance}$, and $\sigma^{2} = \text{population variance}$.
The test may be left-, right-, or two-tailed.

IMAGES

Hypothesis Test, Two Variances (Standard Deviations)
Hypothesis Testing for Mean (Unknown Population Standard Deviation
Hypothesis Testing for Means with Unknown Standard Deviation
Determine the p value for a hypothesis test for the mean population
How to Calculate a Sample Standard Deviation
Chapter 7 Hypothesis Testing with One Sample Larson Farber

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Two-Sample Hypothesis Testing
What is Null Hypothesis Testing Mean? || Academic Research || Ettienne-Murphy
MATH 1342
MATH 1342
Session 10: Part 3
Basic Statistical Literacies 25 Aug 2022

COMMENTS

8.3: Hypothesis Test Examples for Means with Unknown Standard Deviation
Full Hypothesis Test Examples. Example 8.3.6 8.3. 6. Statistics students believe that the mean score on the first statistics test is 65. A statistics instructor thinks the mean score is higher than 65. He samples ten statistics students and obtains the scores 65 65 70 67 66 63 63 68 72 71.
8.7 Hypothesis Tests for a Population Mean with Unknown Population
The p-value for a hypothesis test on a population mean is the area in the tail(s) of the distribution of the sample mean. When the population standard deviation is unknown, use the [latex]t[/latex]-distribution to find the p-value.. If the p-value is the area in the left-tail: Use the t.dist function to find the p-value. In the t.dist(t-score, degrees of freedom, logic operator) function:
Hypothesis Testing for Means with Unknown Standard Deviation ...
Statistics tutorial video that explains the steps for performing a hypothesis test for a population mean with unknown population standard deviation (sigma) u...
3.3: Hypothesis Test about the Population Mean when the Population
Hypothesis Test about the Population Mean (μ) when the Population Standard Deviation (σ) is Unknown. Frequently, the population standard deviation (σ) is not known. We can estimate the population standard deviation (σ) with the sample standard deviation (s). However, the test statistic will no longer follow the standard normal distribution.
10.2: Two Population Means with Unknown Standard Deviations
The test comparing two independent population means with unknown and possibly unequal population standard deviations is called the Aspin-Welch t t -test. The degrees of freedom formula was developed by Aspin-Welch. The comparison of two population means is very common.
Hypothesis Testing: 1 Mean, Sigma Unknown
Steps to conduct a Test for 1 Mean, σ Unknown: Identify all the symbols listed above (all the stuff that will go into the formulas). This includes n n, df d f, μ μ, ¯x x ¯, s s, and α α. Identify the null and alternative hypotheses. Calculate the test statistic, t = ¯x −μ s √n t = x ¯ − μ s n. Find the critical value (s) OR the ...
Hypothesis Testing for the Mean
In this case, the null hypothesis is a simple hypothesis and the alternative hypothesis is a two-sided hypothesis (i.e., it includes both $\mu \lt \mu_0$ and $\mu>\mu_0$). We call this hypothesis test a two-sided test.
PDF Hypothesis Testing for population mean
Hypothesis Testing for Population Mean with Known and Unknown Population Standard Deviation Hypothesis tests are used to make decisions or judgments about the value of a parameter, such as the population mean. There are two approaches for conducting a hypothesis test; the critical value approach and the P-value approach.
Ch. 9 Introduction
Conduct and interpret hypothesis tests for a single population mean, population standard deviation unknown. Conduct and interpret hypothesis tests for a single population proportion. One job of a statistician is to make statistical inferences about populations based on samples taken from the population.
Hypothesis Testing
Statistics Calculators: https://www.statssolver.com/Two-tailed tests in the sigma unknown case are similar to the ones in the sigma known case, except for a ...
9.3 Statistical Inference for Two Population Means with Unknown
The hypothesis test for the difference in two independent population means with unknown population standard deviations is a well established process: Write down the null and alternative hypotheses in terms of the differences in the population means [latex]\mu_1-\mu_2[/latex].
Hypothesis Testing: 1 Mean, Sigma Known
Steps to conduct a Test for 1 Mean, σ Known: Identify all the symbols listed above (all the stuff that will go into the formulas). This includes n n, μ μ, ¯x x ¯, σ σ, and α α. Identify the null and alternative hypotheses. Calculate the test statistic, z = ¯x − μ σ √n z = x ¯ − μ σ n. Find the critical value (s) OR the p ...
Two Population Means with Unknown Standard Deviations
Then, μg is the population mean for girls and μb is the population mean for boys. This is a test of two independent groups, two population means. Random variable: ¯¯¯¯¯Xg −¯¯¯¯¯Xb X ¯ g − X ¯ b = difference in the sample mean amount of time girls and boys play sports each day. H0: μg = μb H 0: μ g = μ b; H0:μg −μb =0 H ...
Hypothesis Testing, Standard Deviation Unknown
Professor Hildebrandt works through another example from Ch 10: Hypothesis Testing in Business Statistics. In this example, a one-tailed test where we do not...
8.3: Hypothesis Testing of Single Mean
Thus the test statistic is. T = x¯ −μ0 s/ n−−√ T = x ¯ − μ 0 s / n. and has the Student t t -distribution with n − 1 = 5 − 1 = 4 n − 1 = 5 − 1 = 4 degrees of freedom. Step 3. From the data we compute x¯ = 169 x ¯ = 169 and s = 10.39 s = 10.39. Inserting these values into the formula for the test statistic gives.
8.2 A Single Population Mean (Unknown σ)
The mean and the standard deviations given here are about a sample, as it says in the question — a sample of size 46 with a mean of $31 and a standard deviation of $9. Given facts are: [latex]n=46[/latex] [latex]\bar x = $31[/latex] [latex]s = $9[/latex]. This is not σ (The notation σ represents the population standard deviation.
Hypothesis Testing for Means with Unknown Standard Deviation
Statistics tutorial that explains the steps of performing a hypothesis test for a population mean with an unknown population standard deviation using the rej...
9.3 Probability Distribution Needed for Hypothesis Testing
When testing hypotheses we are faced with this same problem and the solution is exactly the same. Namely: If the population standard deviation is unknown, and the sample size is less than 30, substitute s, the point estimate for the population standard deviation, σ, in the formula for the test statistic and use the student's t-distribution.
Interpret all statistics for 1 Variance
A hypothesis test uses sample data to determine whether to reject the null hypothesis. Null hypothesis The null hypothesis states that a population parameter (such as the mean, the standard deviation, and so on) is equal to a hypothesized value. ... a certain percentage of the resulting confidence intervals or bounds would contain the unknown ...
10.2: Two Population Means with Unknown Standard Deviations
The comparison of two population means is very common. A difference between the two samples depends on both the means and the standard deviations. Very different means can occur by chance if there is great variation among the individual samples. In order to account for the variation, we take the difference of the sample means, X¯1 −X¯2 X ...
11.3: Two Population Means with Known Standard Deviations
Therefore, this is a right-tailed test. Distribution for the test: The population standard deviations are known so the distribution is normal. Using Equation 11.3.1, the distribution is: ˉX1 − ˉX2 ∼ N(0, √0.332 20 + 0.362 20) Since μ1 ≤ μ2 then μ1 − μ2 ≤ 0 and the mean for the normal distribution is zero.
T-test for two Means
Instructions : Use this T-Test Calculator for two Independent Means calculator to conduct a t-test for two population means ( \mu_1 μ1 and \mu_2 μ2 ), with unknown population standard deviations. This test apply when you have two-independent samples, and the population standard deviations \sigma_1 σ1 and \sigma_2 σ2 and not known.
One Sample Test of Hypothesis Module (pdf)
Single Sample: Tests Concerning a Single Mean (Variance Known) Example 2] A manufacturer of sports equipment has developed a new synthetic fishing line that he claims has a mean breaking strength of 8 kg with a standard deviation of 0.5 kg. Test the hypothesis that μ = 8 kg against the alternative hypothesis that μ ≠ 8 kg.
8.1.3: Distribution Needed for Hypothesis Testing
Formula Review. If there is no given preconceived $\alpha$, then use $\alpha = 0.05$. Types of Hypothesis Tests. Single population mean, known population variance (or standard deviation): Normal test. Single population mean, unknown population variance (or standard deviation): Student's $t$-test. Single population proportion: Normal test. For a single population mean, we may use a normal ...
8.4: Hypothesis Test on a Single Standard Deviation
A test of a single standard deviation assumes that the underlying distribution is normal. The null and alternative hypotheses are stated in terms of the population standard deviation (or population variance). The test statistic is: χ2 = (n − 1)s2 σ2 (8.4.1) (8.4.1) χ 2 = ( n − 1) s 2 σ 2. where: