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. A sample of 60 day-shift workers showed that the mean number of units produced was 334, with a population standard deviation of 23. A sample of 68 night-shift workers showed that the mean number of units produced was 341, with a population standard deviation of 28 units.At the .10 significance level, is the number of units produced on the night shift larger?1. This is a (Click to select)twoone-tailed test.2. The decision rule is to reject H0: μd ≥ μn if z < . (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)3. The test statistic is z = . (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)4. What is your decision regarding H0?

Answer 1

Explanation:

Let the subscripts d and n represent day and night respectively

The null hypothesis is

H0 : μd ≥ μn

The alternative hypothesis is

H1 : μd < μn

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

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

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

z = (xd - xn)/√σd²/nd + σn²/nn

Where

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

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

nd and nn represents number of samples

From the information given,

xd = 334

xn = 341

σd = 23

σ2 = 28

nd = 60

nn = 68

z = (334 - 341)/√23²/60 + 28²/68

= - 7/√20.34607843138

z = - 1.55

From the normal distribution table, the probability value corresponding to the z score is 0.061

Since the level of significance, 0.1 > 0.061, we would reject H0

Therefore, there is enough evidence to conclude that there are more units produced on the night shift than on the day shift.