Question

Clark Heter is an industrial engineer at Lyons Products. He would like to determine whether there are more units produced on the night shift than on the day shift. The mean number of units produced by a sample of 56 day-shift workers was 336. The mean number of units produced by a sample of 61 night-shift workers was 341. Assume the population standard deviation of the number of units produced on the day shift is 19 and 25 on the night shift. At the 0.02 significance level, is the number of units produced on the night shift larger?
A. Is this a one-tailed or a two-tailed test?
B. State the decision rule.
C. Compute the value of the test statistic.
D. What is your decision regarding H0?
1. Reject H0.
2. Do not reject H0.

Answer 1

Answer:

Step-by-step explanation:

Let the subscripts n and d represent day and night shifts respectively.

The null hypothesis is

H0 : μn = μd, μn - μd = 0

The alternative hypothesis is

H1 : μn > μd, μn - μd > 0

A) it is a one-tailed and also a right tailed test because of the greater than symbol in the alternative hypothesis.

B)The decision rule is to reject H0: μd ≥ μn If 0.10 > p value

C) Since the population standard deviations are known, we would use the formula to determine the test statistic(z score)

z = (xn - xd)/√σn²/nn + σd²/nd

Where

xn and xd represents sample means for night and day respectively.

σn and σd represents population standard deviations for night and day respectively.

nn and nd represents number of samples

From the information given,

xn = 341

xd = 336

σn = 25

σd = 19

nn = 61

nd = 56

z = (341 - 336)/√25²/61 + 19²/56

z = 1.22

Test statistic = 1.22

D) From the normal distribution table, the probability value corresponding to the area above the z score is 1 - 0.8888 = 0.1112

Since the level of significance, 0.02 < 0.1112, do not reject H0.