Answer:
Step-by-step explanation:
Assuming that:
the promotion will be considered to be a success if more than 10% use the coupons received.
and coupons are sent to 100 credit card customers.
Then, the null hypothesis and alternative hypothesis is:
[tex]\mathbf{H_o:p\le 0.10} \\ \\ \mathbf{H_a:p> 0.10}[/tex]
sample size n = 100
Using the eagle file data;
the no of people in the sample who use the coupon = 13
then,
[tex]\hat p = \dfrac{x}{n}[/tex]
[tex]\hat p = \dfrac{13}{100}[/tex]
[tex]\hat p = 0.13[/tex]
Test statistics can be computed as:
[tex]Z = \dfrac{\hat p - p }{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]
[tex]Z = \dfrac{0.13 - 0.10 }{\sqrt{\dfrac{0.10(1-0.10)}{100}}}[/tex]
[tex]Z = \dfrac{0.03 }{\sqrt{\dfrac{0.09}{100}}}[/tex]
[tex]Z = \dfrac{0.03 }{\sqrt{0.0009}}[/tex]
[tex]Z = \dfrac{0.03 }{0.03}[/tex]
[tex]Z = 1[/tex]
[tex]P-value = P(Z > 1) \\ \\ P-value = 1 - P(Z < 1 ) \\ \\ P-value = 1 - 0.8413 \\ \\ P-value = 0.1587 \\ \\[/tex]
At ∝ = 0.05
Since, P-value is greater than ∝, then we fail to reject [tex]\mathbf{H_o}[/tex].
Therefore, Eagle should not go with the promotion; a larger sample should be taken.