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 1299 parts and knows the population standard deviation to be 350. The Midwest group ships an average of 1456 parts and knows the population standard deviation to be 297.Using a 0.01 level of significance, test if there is a difference in productivity level. What is the p-value? (Round to four decimal places) p-value =

The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.

While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.

For this question, we want to test if there is a difference in productivity level of the two warehouses (East Coast and the Midwest Coast).

Hence, the null hypothesis would be that there isn't significant evidence to suggest that there is a difference in productivity level of two warehouses (East Coast and the Midwest Coast). That is, there is no difference in the productivity level of two warehouses (East Coast and the Midwest Coast).

The alternative hypothesis is that there is significant evidence to suggest that there is a difference in productivity level of two warehouses (East Coast and the Midwest Coast).

Mathematically, if the average productivity level of the East Coast group is μ₁, the average productivity level of the Midwest group is μ₂ and the difference in productivity level is μ = μ₂ - μ₁

The null hypothesis is represented as

H₀: μ = 0 or μ₂ = μ₁

The alternative hypothesis is represented as

Hₐ: μ ≠ 0 or μ₂ ≠ μ₁

So, to perform this test, we need to compute the test statistic

Test statistic for 2 sample mean data is given as

Test statistic = (μ₂ - μ₁)/σ

σ = √[(s₂²/n₂) + (s₁²/n₁)]

μ₁ = average productivity level of the East Coast group = 1299 parts shipped

n₁ = sample size of East Coast group surveyed = 35

s₁ = standard deviation of the East Coast group sampled = 350

μ₂ = average productivity level of the Midwest group = 1456 parts shipped

n₂ = sample size of Midwest group surveyed = 35

s₂ = standard deviation of the Midwest group sampled = 297

σ = √[(297²/35) + (350²/35)] = 77.5903160379 = 77.59

We will use the t-distribution as no information on population standard deviation is provided

t = (1456 - 1299) ÷ 77.59

= 2.02

checking the tables for the p-value of this t-statistic

Degree of freedom = df = n₁ + n₂ - 2 = 35 + 35 - 2 = 68

Significance level = 0.01

The hypothesis test uses a two-tailed condition because we're testing in both directions.

p-value (for t = 2.02, at 0.01 significance level, df = 68, with a two tailed condition) = 0.047326

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 0.01

p-value = 0.047326

0.047326 > 0.01

Hence,

p-value > significance level

This means that we fail to reject the null hypothesis & say that there isn't enough evidence to suggest that there is a difference in productivity level of two warehouses (East Coast and the Midwest Coast).

Hope this Helps!!!