Hypothesis Testing with Two Samples
Read this chapter, which discusses how to compare data from two similar groups. This is useful when, for example, you want to analyze things like how someone's income relates to another sample that you are interested in. Make sure you read the introduction as well as sections 10.1 through 10.6. Attempt the practice problems and homework at the end of the chapter.
Two Population Means with Known Standard Deviations
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:
\(\sqrt{\dfrac{(σ_1)^2}{n_1}+\dfrac{(σ_2)^2}{n_2}}\)
The test statistic (z-score) is:
\(Z_c=\dfrac{(\overline x_1– \overline x_2)–δ_0}{\sqrt{\dfrac{(σ_1)^2}{n_1}+\dfrac{(σ_2)^2}{n_2}}}\)
Example 10.7
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 10.3.
| Wax | Sample mean number of months floor wax lasts | Population standard deviation |
|---|---|---|
| 1 | 3 | 0.33 |
| 2 | 2.9 | 0.36 |
Table 10.3
This is a test of two independent groups, two population means, population standard deviations known.
Random Variable: \(\overline X_1 – \overline X_2\) = difference in the mean number of months the competing floor waxes last.
\(H_0:μ_1≤μ_2\)
\(H_a:μ_1>μ_2\)
The words "is more effective" says that wax 1 lasts longer than wax 2, on average. "Longer" is a ">" symbol and goes into Ha. 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.
\(Z_c=\dfrac{(μ_1−μ_2)−δ_0}{\sqrt{\dfrac{σ^2_1}{n_1}+\dfrac{σ^2_2}{n_2}}}=0.1\)
The estimated difference between he two means is : \(\overline X_1 – \overline X_2= 3 – 2.9 = 0.1\)
Compare calculated value and critical value and Zα: 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 H0.
Try It 10.7
| Engine | Sample mean number of RPM | Population standard deviation |
|---|---|---|
| 1 | 1,500 | 50 |
| 2 | 1,600 | 60 |
Example 10.8
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: \(\overline X_1 – \overline X_2\)= difference in the mean age of Democratic and Republican U.S. senators.
\(H_0:μ_1≤μ_2 H_0:μ_1−μ_2≤0\)
\(H_a:μ_1 > μ_2 H_a:μ_1−μ_2 > 0\)
The words "older than" translates as a ">" symbol and goes into Ha. Therefore, this is a right-tailed test.
Figure 10.8
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.