Question

The CEO of a large manufacturing company is curious if there is a difference in productivity level of her warehouse employees based on the region of the country the warehouse is located. She randomly selects 35 employees who work in warehouses on the East Coast (Group 1) and 35 employees who work in warehouses in the Midwest (Group 2) and records the number of parts shipped out from each for a week. She finds that East Coast group ships an average of 1287 parts and knows the population standard deviation to be 348. The Midwest group ships an average of 1449 parts and knows the population standard deviation to be 298. Using a 0.01 level of significance, test if there is a difference in productivity level. What are the correct hypotheses for this problem

Answer 1

Answer:

Step-by-step explanation:

Let μ1 be the population mean for group 1 and and μ2 be the population mean for group 2. The random variable is μ1 - μ2 = difference in the population mean for group 1 and the population mean for group 2.

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

H1 : μ1 ≠ μ2 H1 : μ1 - μ2 ≠ 0

This is a 2 tailed test because of the 'unequal to' symbol in the alternative hypothesis.

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

z = (x1 - x2)/√σ1²/n1 + σ2²/n2

Where

x1 and x2 represents sample means for group 1 and group 2 respectively.

σd and σn represents population standard deviations for group 1 and group 2 respectively.

n1 and n2 represents number of samples

From the information given,

x1 = 1287

x2 = 1449

σ1 = 348

σ2 = 298

n1 = 35

n2 = 35

z = (1287 - 1449)/√348²/35 + 298²/35

= - 2.09

test statistic = - 2.09

Since it is a two tailed test, we would determine the probability for the area in the left tail and double it to account for the area on the right scale.

From the normal distribution table, the probability value corresponding to the z score is 0.0183

P value = 0.0183 × 2 = 0.0366

Since the level of significance, 0.01 < the p value, 0.0366, we would fail to reject the null hypothesis.

Therefore, at 1% level of significance, there is not enough evidence to conclude that there is a difference in productivity level.