Question

Exercise 6.12 presents the results of a poll where 48% of 331 Americans who decide to not go to college do so because they cannot a ord it. (a) Calculate a 90% con dence interval for the proportion of Americans who decide to not go to college because they cannot a ord it, and interpret the interval in context. (b) Suppose we wanted the margin of error for the 90% con dence level to be about 1.5%. How large of a survey would you recommend?

Answer 1

Answer with explanation:

As per given , we have

$\hat{p}=0.48$   , n=331

Critical value for 90% Confidence interval : $z_{\alpha/2}=1.645$

a) Confidence interval :

$\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$0.48\pm (1.645)\sqrt{\dfrac{0.48(1-0.48)}{331}}\\\\\approx 0.48\pm0.02746\\\\=(0.48-0.02746,\ 0.48+0.02746)\\\\=(0.45254,\ 0.50746)$

Hence, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot a ord it : $(0.45254,\ 0.50746)$

b) Margin of error : E=1.5%=0.015

Formula for sample size : $n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2$

For p =0.48 , we  have

$n=0.48(1-0.48)(\dfrac{1.645}{0.015})^2=3001.88373333\approx3002$

Hence, the required sample size to survey = 3002

Answer 2

According to the data given, we have that:

a) The 90% confidence interval is (0.435, 0.525), and it means that we are 90% sure that the true population proportion is in this interval.

b) A sample of 3002 is recommended.

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $\frac{1+\alpha}{2}$.

For this problem, we have that:

$n = 331, \pi = 0.48$

90% confidence level

So $\alpha = 0.9$, z is the value of Z that has a p-value of $\frac{1+0.9}{2} = 0.95$, so $z = 1.645$.  

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.48 - 1.645\sqrt{\frac{0.48(0.52)}{331}} = 0.435$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.48 + 1.645\sqrt{\frac{0.48(0.52)}{331}} = 0.525$

The 90% confidence interval is (0.435, 0.525), and it means that we are 90% sure that the true population proportion is in this interval.

Item b:

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

In this problem, we want to find n for which M = 0.015, then:

$0.015 = 1.645\sqrt{\frac{0.48(0.52)}{n}}$

$0.015\sqrt{n} = 1.645\sqrt{0.48(0.52}}$

$\sqrt{n} = \frac{1.645\sqrt{0.48(0.52}}}{0.015}$

$(\sqrt{n})^2 = (\frac{1.645\sqrt{0.48(0.52}}}{0.015})^2$

$n = 3001.9$

Rounding up:

A sample of 3002 is recommended.

A similar problem is given at https://brainly.com/question/13625548