Question

In 1996, the General Social Survey (which uses a method similar to simple random sampling) asked, "On the whole, do you think it should be the government's responsibility to provide decent housing for those who can't afford it?" For this question, 240 people said that it definitely should out of 1572 randomly selected people. We will make a 90% confidence interval for p. What is the margin of error of the confidence interval? A. 0.01497 B. 0.0091 C. 0.000082 D. 0.017745

Answer 1

Answer: A. 0.01497

Step-by-step explanation:

Let $\hat{p}$ denotes the sample proportion.

As per given , we have

n=1572

$\hat{p}=\dfrac{240}{1572}=0.152672$

Margin of error = $z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$(1.645)\sqrt{\dfrac{0.152672(1-0.152672)}{1572}}\\\\\approx 0.01497$

Hence, the  margin of error of the confidence interval =  0.01497

Answer 2

Using confidence interval concepts, it is found that the margin of error is given by:

A. 0.01497

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 $\frac{1+\alpha}{2}$.

The margin of error is:

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

In this problem:

• 240 people out of 1572, thus $n = 1572, p = \frac{240}{1572} = 0.1527$.
• 90% confidence level, thus $\alpha = 0.9$, z has a p-value of $\frac{1 + 0.9}{2} = 0.95$, thus z = 1.645.

Then

$M = 1.645\sqrt{\frac{0.1527(0.8473)}{1572}} = 0.01497$

Option A.

A similar problem is given at https://brainly.com/question/16807970