Question

A college student organization wants to start a nightclub for students under the age of 21. To assess support for this proposal, they will select an SRS of students and ask each respondent if he or she would patronize this type of establishment. They expect that about 74% of the student body would respond favorably.(a) What sample size is required to obtain a 95% confidence interval with an approximate margin of error of 0.03?answer: 822(b) Suppose that 54% of the sample responds favorably. Calculate the margin of error for the 95% confidence interval.

Answer 1

Answer:

a) Sample size = 822

b) Margin of error = 0.03407          

Step-by-step explanation:

We are given the following in the question:

p = 74% = 0.74

a) Sample size is required to obtain margin of error of 0.03

Formula:

$\text{Margin of error} = z_{\text{statistic}}\times \sqrt{\dfrac{p(1-p)}{n}}}$

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

Putting values, we get,

$0.03 = 1.96\times \sqrt{\dfrac{0.74(1-0.74)}{n}}\\\\n = (\dfrac{1.96}{0.03})^2(0.74)(1-0.74)\\\\n = 821.24 \approx 822$

Thus, the sample size must be approximately 822  to obtain a 95% confidence interval with an approximate margin of error of 0.03

b) Margin of error for the 95% confidence interval

p = 54% = 0.54

Formula:

$\text{Margin of error} = z_{\text{statistic}}\times \sqrt{\dfrac{p(1-p)}{n}}}$

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

Putting values, we get,

$\text{Margin of error} = 1.96\times \sqrt{\dfrac{0.54(1-0.54)}{822}}\\\\=0.03407$

The margin of error now will be 0.03407.