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vazorg [7]
4 years ago
8

A Gallup Poll used telephone interviews to survey a sample of 1030 U.S. residents over the age of 18 regarding their use of cred

it 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
Mathematics
1 answer:
Vlada [557]4 years ago
7 0

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.

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