Two Population Means with Known Standard Deviations Quiz

PDF page 553: 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 . The normal distribution has the following format: Normal distribution is

PDF page 553: 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.

PDF page 555: 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.

PDF page 553: The test statistic (z-score) is

PDF page 553: 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.

PDF page 554: 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: = difference in the mean number of months the competing floor waxes last.

PDF page 554: 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, the distribution is

PDF page 554: Since μ1 ≤ μ2, then μ1 – μ2 ≤ 0 and the mean for the normal distribution is zero.

PDF page 554: Calculate the p value using the normal distribution: p value = 0.1799 Graph:

PDF page 554: 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.

PDF page 554: 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.

PDF page 555: 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.

PDF page 555: 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.