Answer:
Step-by-step 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.
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