Respuesta :
Answer:
a) p = 0.625
b) (0.563, 0.687)
c) This means that we are 99% sure that the true proportion of all viewers who would prefer the comedy show over the crime investigation show is between (0.563, 0.687).
Step-by-step explanation:
a. Estimate the value of the population proportion.
Sample: 250 viewers would prefer the comedy show over the crime investigation show, out of 400 viewers sampled. So
[tex]p = \frac{250}{400} = 0.625[/tex]
b. Develop a 99 percent confidence interval for the population proportion.
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
[tex]n = 400, \pi = 0.625[/tex]
99% confidence level
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.625 - 2.575\sqrt{\frac{0.625*0.375}{400}} = 0.563[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.625 + 2.575\sqrt{\frac{0.625*0.375}{400}} = 0.687[/tex]
c. Interpret your findings.
This means that we are 99% sure that the true proportion of all viewers who would prefer the comedy show over the crime investigation show is between (0.563, 0.687).
