Question

Researchers planning a study of the reading ability of third-grade children want to obtain a 95% confidence interval for the population mean score on a reading test, with margin of error no greater than 3 points. They carry out a small pilot study to estimate the variability of test scores. Assume the population standard deviation to be 12 points.

a. The study budget will allow as many as 100 students. Calculate the margin of error of the 99% confidence interval for the population mean based on n = 100.
b. There are many other demands on the research budget. If all of the various demands are met, there would be funds to measure only 10 children. What is the margin of error of the confidence interval based on n = 10 measurements?
c. Find the smallest value of n that would satisfy the goal of a 95% confidence interval with margin of error 3 or less. Is this sample size within the limits of the budget?

Answer 1

A) M = 3.016

B) M = 9.79

C) 62

Explanation:

A) We are given;

Population size; n = 100

standard deviation; σ = 12

Formula for the margins of error is;

M = z × σ/√n

At 99% confidence interval, z = 2.58

Thus;

M = 2.58 × 12/√100

M = 3.016

B) n is now equal to 10, thus at same confidence interval used in option A, we have;

M = 2.58 × 12/√10

M = 9.79

C) to solve this, we will use the formula;

n ≥ (zσ/m)²

Where;

z at 95% confidence interval has a value of 1.96

m is the margin of error = 3

Thus;

n ≥ (1.96 × 12/3)²

n ≥ 61.4656

Thus,approximately n = 62