Question

In a survey of 280 adults over 50, 75% said they were taking vitamin supplements. Find the margin of error for this survey if we want a 99% confidence in our estimate of the percent of adults over 50 who take vitamin supplements. (Round to the nearest thousandths.)

Answer 1

The margin of error for this survey if we want a 99% confidence in our estimate of the percent is 0.067

The formula for calculating the margin of error is expressed as:

$e= z \cdot \sqrt{\frac{p(1-p)}{n} }$

p is the proportion

n is the sample size

Given the following parameters;

p = 75% = 0.75

n = 280

z = 2.576

Substitute into the formula

$e=\sqrt{\frac{0.75(1-0.75)}{280} }\\e=\sqrt{\frac{0.1875}{280} } \\e=\sqrt{0.0006964} \\e=0.0258 \times 2.576\\e \approx 0.067$

Hence the margin of error for this survey if we want a 99% confidence in our estimate of the percent is 0.067

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