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 = <em>proportion of employees at the headquarters who want the American Express Card.</em>
= <em>proportion of employees at the branch office who want the American Express Card.</em>
So, Null Hypothesis, : {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, : {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 <u>Two-sample z proportion</u> <u>statistics</u>;
T.S. = ~ N(0,1)
where, = sample proportion of employees at the headquarters wanted to use the American Express Card = = 0.875
= sample proportion of employees at the branch office wanted to use the American Express Card = = 0.357
= sample of people at the main corporate headquarters = 400
= sample of people at a branch office = 350
So, <u><em>the test statistics</em></u> =
= 16.99
The value of z test statistics is 16.99.
<u></u>
<u>Now, P-value of the test statistics is given by;</u>
P(Z > 16.99) = Less than 0.0005%