Even though this situation is not likely (knowing the population standard deviations is very unlikely), the following example illustrates hypothesis testing for independent means with known population standard deviations. The sampling distribution for the difference between the means is normal in accordance with the central limit theorem. The random variable is \(\overline{X_1}-\overline{X_2}\). The normal distribution has the following format:
The standard deviation is:
\[\sqrt{\dfrac{\left(\sigma_1\right)^2}{n_1}+\dfrac{\left(\sigma_2\right)^2}{n_2}}\]
The test statistic ( \(z\)-score) is:
\[Z_c=\dfrac{\left(\bar{x}_1-\bar{x}_2\right)-\delta_0}{\sqrt{\dfrac{\left(\sigma_1\right)^2}{n_1}+\dfrac{\left(\sigma_2\right)^2}{n_2}}}\]
A public health department is evaluating two competing brands of anti-fatigue standing mats used in a government-wide workplace wellness program. The goal is to determine if there is a difference in the mean "lasting time" (the time until the mat loses its specified cushioning thickness) between the two brands.
Twenty workstations are randomly assigned to test Brand A, and twenty workstations are randomly assigned to test Brand B. Based on extensive previous laboratory testing by the manufacturers, the population standard deviations for both brands are known, and both populations follow a normal distribution.
Given Parameters:
- Sample Size (nA,nB): 20 mats for Brand A, 20 mats for Brand B.
- Distribution: Both populations are normally distributed.
- Standard Deviations (σA,σB): Known (This allows for the use of the Z-test).
The data are recorded in Table \(\PageIndex{1}\).
| Brand | Sample mean number of years of daily use | Population standard deviation |
|---|---|---|
| A | 3 | 0.33 |
| B | 2.9 | 0.36 |
Problem
Does the data indicate that brand A is more effective than brand B? Test at a 5% level of significance.
- Answer
-
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 years of daily use.
\[
\begin{array}{l}
H_0: \mu_1 \leq \mu_2 \\
H_a: \mu_1>\mu_2
\end{array}
\]The words "is more effective" says that Brand A lasts longer than Brand B, 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 the formula for the test statistic we find the calculated value for the problem.
\[Z_c=\dfrac{\left(\mu_1-\mu_2\right)-\delta_0}{\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}}=0.1\]
Figure \(\PageIndex{1}\): Copy and Paste Caption here. (Copyright; author via source) The estimated difference between he two means is: \(\bar{X}_1-\bar{X}_2=3-2.9=0.1\)
Compare calculated value and critical value and \(\mathbf{Z}_\alpha\) : We mark the calculated value on the graph and find the calculated value is not in the tail therefore we cannot reject the null hypothesis.
Make a decision: the calculated value of the test statistic is not in the tail, therefore you cannot 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 brand A lasts is longer (Brand A is more effective) than the mean time Brand B lasts.
A public health researcher wanted to know if the directors of community health clinics in Region A are older than directors in Region B, on average.
The researcher randomly selected 30 directors from Region A and found their mean age was 61.704 years with a standard deviation of 9.55 years. A random sample of 30 directors from Region B showed a mean age of 61.675 years with a standard deviation of 10.17 years.
Problem
Do the data indicate that clinic directors in Region A are older than directors in Region B, on average? Test at a 5% level of significance.
-
Answer
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. Since the total sample size is large, we use the Student’s t-distribution (often approximated by the normal distribution in larger samples).
\[\begin{array}{l}
H_0: \mu_1 \leq \mu_2 H_0: \mu_1-\mu_2 \leq 0 \\
H_a: \mu_1>\mu_2 H_a: \mu_1-\mu_2>0
\end{array}\]The words "older than" translates as a ">" symbol and goes into \(H_a\). Therefore, this is a right-tailed test.
Figure \(\PageIndex{2}\) Make a decision: The p-value is larger than 5%, therefore we cannot reject the null hypothesis. By calculating the test statistic we would find that the test statistic does not fall in the tail, therefore we cannot reject the null hypothesis. We reach the same conclusion using either method of a making this statistical decision.
Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of clinic directors in Region A is greater than the mean age of clinic directors in Region B.