Barbara Illowsky; Susan Dean

Even though this situation is not likely (knowing the population standard deviations), 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} ~ N [μ_{1} - μ_{2}, \sqrt{\frac{{(σ_{1})}^{2}}{n_{1}} + \frac{{(σ_{2})}^{2}}{n_{2}}}] .

The standard deviation is

\sqrt{\frac{{(σ_{1})}^{2}}{n_{1}} + \frac{{(σ_{2})}^{2}}{n_{2}}} .

10.2

The test statistic (z-score) is

z = \frac{({\bar{x}}_{1} - {\bar{x}}_{2}) - (μ_{1} - μ_{2})}{\sqrt{\frac{{(σ_{1})}^{2}}{n_{1}} + \frac{{(σ_{2})}^{2}}{n_{2}}}} .

Example 10.6

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 distribution. The data are recorded in Table 10.8.

Wax	Sample Mean Number of Months Floor Wax Lasts	Population Standard Deviation
1	3	0.33
2	2.9	0.36

Table 10.8

Problem

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

Solution

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.

H₀: μ₁ ≤ μ₂

H_a: μ₁ > μ₂

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 the formula, the distribution is

${\bar{X}}_{1} - {\bar{X}}_{2} ~ N (0, \sqrt{\frac{{0.33}^{2}}{20} + \frac{{0.36}^{2}}{20}}) .$

Since μ₁ ≤ μ₂, then μ₁ – μ₂ ≤ 0 and the mean for the normal distribution is zero.

Calculate the p value using the normal distribution: p value = 0.1799

Graph:

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.

Figure 10.5

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

Compare α and the p value: α = 0.05 and p value = 0.1799. Therefore, α < p value.

Make a decision: Since α < p value, do not reject H₀.

Conclusion: At the 5 percent 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.

Using the TI-83, 83+, 84, 84+ Calculator

Press STAT. Arrow over to TESTS and press 3:2-SampZTest. Arrow over to Stats and press ENTER. Arrow down and enter .33 for sigma1, .36 for sigma2, 3 for the first sample mean, 20 for n1, 2.9 for the second sample mean, and 20 for n2. Arrow down to μ1: and arrow to > μ₂. Press ENTER. Arrow down to Calculate and press ENTER. The p value is p = 0.1799, and the test statistic is 0.9157. Do the procedure again, but instead of Calculate do Draw.

Try It 10.6

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 10.9 shows the result. Do the data indicate that Engine 2 has higher RPM than Engine 1? Test at a 5 percent level of significance.

Engine	Sample Mean Number of RPM	Population Standard Deviation
1	1,500	50
2	1,600	60

Table 10.9

Example 10.7

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 (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 (61.704 years) with a standard deviation of 9.55 years.

Problem

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

Solution

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.

H₀: µ₁ ≤ µ₂ H₀: µ₁ – µ₂ ≤ 0

H_a: µ₁ > µ₂ H_a: µ₁ – µ₂ > 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 [0, \sqrt{\frac{{(9.55)}^{2}}{30} + \frac{{(10.17)}^{2}}{30}}]

Since µ₁ ≤ µ₂, µ₁ – µ₂ ≤ 0 and the mean for the normal distribution is zero.

Calculating the p value using the normal distribution gives p value = 0.4040.

Graph:

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.

Figure 10.6

Compare α and the p value: α = 0.05 and p value = 0.4040. Therefore, α < p value.

Make a decision: Since α < p value, do not reject H₀.

Conclusion: At the 5 percent 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.

10.2 Two Population Means with Known Standard Deviations

Problem

Solution

Problem

Solution