Question

A Gallup Poll used telephone interviews to survey a sample of 1030 U.S. residents over the age of 18 regarding their use of credit cards. The poll reported that 72% of Americans said that they had at least one credit card. Give the 95% margin of error for this estimate. m

Answer 1

Answer:

The 95% margin of error for this estimate is 0.0274 = 2.74 percentage points.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

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

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

The margin of error is:

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

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

72% of Americans said that they had at least one credit card

This means that $p = 0.72$

Give the 95% margin of error for this estimate.

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

$M = 1.96\sqrt{\frac{0.72*0.28)}{1030}} = 0.0274$

The 95% margin of error for this estimate is 0.0274 = 2.74 percentage points.