Question

At a branch office of the company, 350 people are polled and 125 indicate they are in favor of using a corporate American Express card instead of the lots of individual cards which require extensive amounts of paperwork for reimbursement purposes. A recent survey at the main corporate headquarters revealed that the 400 employees who responded to the survey indicated that 350 wanted to stay with the American Express Card. At the .05 level of significance, is the percentage of employees at the headquarters who want the American Express Card greater than the percentage of employees at the branch office?
Please select the correct P value
a. 0.013
b. 0.857
c. 0.777
d. 0.000

Answer 1

Answer:

The P-value is Less than 0.0005%.

Step-by-step explanation:

We are given that at a branch office of the company, 350 people are polled and 125 indicate they are in favor of using a corporate American Express card.

A recent survey at the main corporate headquarters revealed that the 400 employees who responded to the survey indicated that 350 wanted to stay with the American Express Card.

Let $p_1$ = proportion of employees at the headquarters who want the American Express Card.

$p_2$ = proportion of employees at the branch office who want the American Express Card.

So, Null Hypothesis, $H_0$ : $p_1\leq p_2$      {means that the percentage of employees at the headquarters who want the American Express Card is smaller or equal to the percentage of employees at the branch office}

Alternate Hypothesis, $H_A$ : $p_1>p_2$      {means that the percentage of employees at the headquarters who want the American Express Card is greater than the percentage of employees at the branch office}

The test statistics that would be used here Two-sample z proportion statistics;

                      T.S. =  $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$  ~ N(0,1)

where, $\hat p_1$ = sample proportion of employees at the headquarters wanted to use the American Express Card = $\frac{350}{400}$ = 0.875

$\hat p_2$ = sample proportion of employees at the branch office wanted to use the American Express Card = $\frac{125}{350}$ = 0.357

$n_1$ = sample of people at the main corporate headquarters = 400

$n_2$ = sample of people at a branch office = 350

So, the test statistics  =  $\frac{(0.875-0.357)-(0)}{\sqrt{\frac{0.875(1-0.875)}{400}+\frac{0.357(1-0.357)}{350} } }$

                                     =  16.99

The value of z test statistics is 16.99.

Now, P-value of the test statistics is given by;

            P(Z > 16.99) = Less than 0.0005%