Question

Clark Heter is an industrial engineer at Lyons Products. He would like to determine whether, on average, there are more units produced on the night shift (Population n) than on the day shift (Population d ). A sample of 52 day-shift workers showed the mean number of units produced is 355. The population standard deviation of day-shift production is 21 units. A sample of 58 night-shift workers showed that the mean number of units produced is 363. The population standard deviation of night-shift production is 29 units.

At the .05 significance level, is the average number of units produced on night shift higher than on day shift?

(1)	This is a (Click to select)twoone-tailed test.

(2)	The decision rule is to reject H0: μd≥μnH0: μd⁢≥μn if Z < . (Negative amounts should be indicated by a minus sign. Round your answer to 2 decimal places.)

(3)	The test statistic is Z = . (Negative amounts should be indicated by a minus sign. Carry at least 3 decimal places in your intermediate calculations. Round your answer to 2 decimal places.)

(4)	What is your decision regarding H0H0?
	(Click to select)Do not reject.Reject.

Answer 1

using excel>Addin>phstat>two sample test

we have

Z Test for Differences in Two Means
Data
Hypothesized Difference	0
Level of Significance	0.05
Population 1 Sample
Sample Size	52
Sample Mean	355
Population Standard Deviation	21
Population 2 Sample
Sample Size	58
Sample Mean	363
Population Standard Deviation	29
Intermediate Calculations
Difference in Sample Means	-8
Standard Error of the Difference in Means	4.793826
Z-Test Statistic	-1.66881
Lower-Tail Test
Lower Critical Value	-1.64485
p-Value	0.047577
Reject the null hypothesis

(1)	This is a one-tailed test.

(2)	The decision rule is to reject H0: μd≥μnH0: μd⁢≥μn if Z < -1.65

(3)	The test statistic is Z = . -1.67

(4)	What is your decision regarding H0H0?
	(.Reject. the null hypothesis