Respuesta :
Answer:
a) Alternative hypothesis: the use of the coupons is isgnificantly higher than 10%.
Null hypothesis: the use of the coupons is not significantly higher than 10%.
The null and alternative hypothesis can be written as:
[tex]H_0: \pi=0.1\\\\H_a:\pi>0.1[/tex]
b) Point estimate p=0.13
c) At a significance level of 0.05, there is not enough evidence to support the claim that the proportion of coupons use is significantly higher than 10%.
Eagle should not go national with the promotion as there is no evidence it has been succesful.
Step-by-step explanation:
The question is incomplete.
The sample data shows that x=13 out of n=100 use the coupons.
This is a hypothesis test for a proportion.
The claim is that the proportion of coupons use is significantly higher than 10%.
Then, the null and alternative hypothesis are:
[tex]H_0: \pi=0.1\\\\H_a:\pi>0.1[/tex]
The significance level is 0.05.
The sample has a size n=100.
The point estimate for the population proportion is the sample proportion and has a value of p=0.13.
[tex]p=X/n=13/100=0.13[/tex]
The standard error of the proportion is:
[tex]\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.1*0.9}{100}}\\\\\\ \sigma_p=\sqrt{0.0009}=0.03[/tex]
Then, we can calculate the z-statistic as:
[tex]z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.13-0.1-0.5/100}{0.03}=\dfrac{0.025}{0.03}=0.833[/tex]
This test is a right-tailed test, so the P-value for this test is calculated as:
[tex]\text{P-value}=P(z>0.833)=0.202[/tex]
As the P-value (0.202) is greater than the significance level (0.05), the effect is not significant.
The null hypothesis failed to be rejected.
At a significance level of 0.05, there is not enough evidence to support the claim that the proportion of coupons use is significantly higher than 10%.
