Question

Suppose the U.S. president wants to estimate the proportion of the population that supports his current policy toward revisions in the health care system. The president wants the margin of error to be 0.05. Assume a 98% level of confidence. The president’s political advisors found a similar survey from two years ago that reported that 46% of people supported health care revisions. (Use t Distribution Table & z Distribution Table.)
Required:

a. How large of a sample is required? (Round up your final answer to the next whole number.)

b. How large of a sample would be necessary if no estimate were available for the proportion supporting current policy? (Round up your final answer to the next whole number.)

Answer 1

The sample size for part (a) is 540, and the sample size for part (b) is 543 if the margin of error to be 0.05.

What is the margin of error(MOE)?

It is defined as an error that provides an estimate of the percentage of errors in real statistical data.

The formula for finding the MOE:

$\rm MOE = Z\times \dfrac{s}{\sqrt{n}}$

Where   Z is the z-score at the confidence interval

            s is the standard deviation

            n is the number of samples.

We know:

$\rm MOE = Z \times \sqrt{\dfrac{pq}{n}}$

We have MOE = 0.05, p = 46% = 0.46,

q = 1-p = 1-0.46 = 0.54

Z at 0.02(because 98% level of confidence) = 2.33

Plug this value in the formula we get:

$\rm 0.05 = 2.33 \times \sqrt{\dfrac{0.46\times0.54}{n}}$

After solving:

n = 539.41 ≈ 540

If no estimate is given:

p = 0.5 and q = 0.5

$\rm 0.05 = 2.33 \times \sqrt{\dfrac{0.5\times0.5}{n}}$

n = 542.89 ≈ 543

Thus, the sample size for part (a) is 540, and the sample size for part (b) is 543 if the margin of error to be 0.05.

Learn more about the Margin of error here:

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