Question

A researcher wants to construct a 98% confidence interval for the proportion of elementary school students in Seward County who receive free or reduced-price school lunches. A state-wide survey indicates that the proportion is 0.60. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.07?

Question 1 options:

8

265

189

19

Answer 1

Answer: option B.

Explanation:

We can use the formula for the margin of error of a confidence interval for a proportion:

Margin of error = zsqrt(p(1-p)/n)

where z is the critical value from the standard normal distribution for the desired confidence level (98% in this case), p is the estimated proportion (0.60), and n is the sample size.

We are given that the margin of error should be 0.07. Setting this equal to the above expression, we have:

0.07 = zsqrt(0.60(1-0.60)/n)

We need to solve for n. To do this, we first need to find the appropriate value of z for a 98% confidence level. Using a standard normal distribution table or calculator, we can find that z = 2.33.

Substituting this into the above equation and solving for n, we have:

0.07 = 2.33sqrt(0.60(1-0.60)/n)

Squaring both sides and solving for n, we get:

n = (2.33^2)(0.60(1-0.60))/(0.07^2) ≈ 265

Therefore, the sample size needed to construct a 98% confidence interval with a margin of error of 0.07 is approximately 265.

So the correct answer is option B.