I have completed the problem. But I am knowing how to check the normality assumptions of both population. I used minitab for output.
Lesson 10, Practice Problems
1. Problem 6.8: First check assumptions by verification that there are both (1) independent samples from the two populations and (2) that datain each sample are about normal or large samples. In this problem, both samples are independent because these samples are selected from populations which have no relationship to each other. One population is Type I emission control devices and the other is Type II emission control devices. However these are not large samples because both samples consist of ten measurements (n = 10 and n ≤ 30). Thus, we need to check the normality assumptions from both populations.
Two different emission-control devices were tested to determine the average amount of nitric oxide emitted by an automobile over a 1-hour period of time. Twenty cars of the same model and year were selected for the study. Ten cars were randomly selected and equipped with a Type I emission-control device, and the remaining cars were equipped with Type II devices. Each of the 20 cars was then monitored for a 1-hour period to determine the amount of nitric oxide emitted.
Use the following data to test the research hypothesis that the meanlevel of emission for Type I devices (µ1) is greater than the mean emission level for Type II devices (µ2). Use sigma = 0.01.
Type I Device Type II Device
1.35 1.28 1.01 0.96
1.16 1.21 0.98 0.99
1.23 1.25 0.95 0.98
1.20 1.17 1.02 1.01
1.32 1.19 1.05 1.02
Two-sample T for C1 vs C2
N Mean StDev SE Mean
C1 10 1.2360 0.0636 0.020
C2 10 0.9970 0.0306 0.0097
Difference = mu (C1) – mu (C2)
Estimate for difference: 0.239000
99% lower bound for difference: 0.182022
T-Test of difference = 0 (vs >): T-Value = 10.71 P-Value = 0.000 DF = 18
Both use Pooled StDev = 0.0499
Since p-value is less than 1% we reject Ho and conclude that the mean level of emission for Type I devices (µ1) is greater than the mean emission level for Type II devices (µ2).
This solution gives answers for checking the normality assumptions of populations.