Question

Hypothesis Testing for Two Populations Scenario Assignment Scenario Caption: Labor negotiations have reared their ugly head again! Princess Foods Corporation's Human Resources department is concerned about the perception of the hours worked per week at the Baltimore facility compared to hours worked at the Burlington facility. Parthika, the HR Manager has asked for your help. Parthika: The Baltimore managers claim they are working more hours than the Burlington managers, for the same compensation We need to know if this is real, or if this is just a rumor that has taken on a life of its own Caption: Morale at the Baltimore facility is down and the labor situation there has grown increasingly contentious Bonnie: Let's tackle this head on. Let's take a look at the hours worked per person at each facility. Can you analyze this data? Assignment Assume a normal distribution. From the Baltimore facility, 175 samples were obtained and compared with a sample of 168 from the Burlington facility. The mean hours per week and standard deviations for middle management who had been with the organizations for approximately the same length of time were: Mean 62.5 39 deviation 23.7 8.9 Using this data, complete the following 1. State your hypothesis and conduct a hypothesis testing using a significance of 0.10 2. What is your conclusion? And why? Interpret it.

Answer 1

1:

Let sample 1: Baltimore

Sample 2: Burlington

Hypotheses are:

$H_{0}:\mu_{1}=\mu_{2}$

$H_{a}:\mu_{1}\neq\mu_{2}$

Here we have

$n_{1}=175,n_{2}=168, \bar{x}_{1}=62.5,s_{1}=23.7, \bar{x}_{2}=39.7,s_{2}=8.9,$

Since it is not given that variances are equal so degree of freedom of the test is

$df=\frac{\left ( \frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}} \right )^{2}}{\frac{\left ( s_{1}^{2}/n_{1} \right )^{2}}{n_{1}-1}+\frac{\left ( s_{2}^{2}/n_{2} \right )^{2}}{n_{2}-1}}=223$

So df is 223.

And test statistics will be

$t=\frac{\bar{x}_{1}-\bar{x}_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}=11.88$

Test is two tailed so p-value of the test is: 0.0000

2:

Since p-value is less than 0.10 so we reject the null hypothesis.

So we cannot conclude that the population means are equal.