Respuesta :
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 [tex]p_1[/tex] = proportion of employees at the headquarters who want the American Express Card.
[tex]p_2[/tex] = proportion of employees at the branch office who want the American Express Card.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]p_1\leq p_2[/tex] {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, [tex]H_A[/tex] : [tex]p_1>p_2[/tex] {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. = [tex]\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} } }[/tex] ~ N(0,1)
where, [tex]\hat p_1[/tex] = sample proportion of employees at the headquarters wanted to use the American Express Card = [tex]\frac{350}{400}[/tex] = 0.875
[tex]\hat p_2[/tex] = sample proportion of employees at the branch office wanted to use the American Express Card = [tex]\frac{125}{350}[/tex] = 0.357
[tex]n_1[/tex] = sample of people at the main corporate headquarters = 400
[tex]n_2[/tex] = sample of people at a branch office = 350
So, the test statistics = [tex]\frac{(0.875-0.357)-(0)}{\sqrt{\frac{0.875(1-0.875)}{400}+\frac{0.357(1-0.357)}{350} } }[/tex]
= 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%
