Question

A survey reported that 70% of those responding to a national survey of college freshmen were interested in taking summer courses. Using the sample size of 49, calculate a 95% confidence interval for the proportion of college students who are interested in taking summer courses.

Answer 1

Answer:

95% confidence interval: (0.5717,0.8283)          

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 49

p = 70% = 0.7

Alpha, α = 0.05

The confidence interval can be calculated as:

$p \pm z_{critical}\sqrt{\dfrac{p(1-p)}{n}}$

Putting the values, we get,

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

$0.7 \pm 1.96(\sqrt{\dfrac{0.7(1-0.7)}{49}} ) = 0.7 \pm 0.1283 =(0.5717,0.8283)$

95% confidence interval: (0.5717,0.8283)