Question

The mean and standard deviation of a random sample of n measurements are equal to 33.5 and 3.2 , respectively. a. Find a 99% confidence interval for μ if n=49. b. Find a 99% confidence interval for μ if n=196. c. Find the widths of the confidence intervals found in parts a and b. What is the effect on the width of a confidence interval of quadrupling the sample size while holding the confidence coefficient fixed?

Answer 1

a. For n=49: CI = 33.5 ± 1.262 (99% confidence interval: 32.238, 34.762). b. For n=196: CI = 33.5 ± 0.679 (99% confidence interval: 32.821, 34.179). c. The width of the confidence interval is reduced by half when quadrupling the sample size while holding the confidence coefficient fixed.

Step-by-step explanation:

To find the confidence interval for μ, we can use the formula: CI = x ± (z * σ/√n), where x is the sample mean, z is the z-score corresponding to the desired confidence level, σ is the population standard deviation, and n is the sample size.

a. For n=49, the z-score for a 99% confidence level is approximately 2.576. Using the given values of x=33.5, σ=3.2, and n=49, we can calculate the confidence interval:

CI = 33.5 ± (2.576 * 3.2/√49) = 33.5 ± 1.262.

Therefore, the 99% confidence interval for μ when n=49 is (32.238, 34.762).

b. For n=196, the z-score for a 99% confidence level is still approximately 2.576. Using the given values of x=33.5, σ=3.2, and n=196, we can calculate the confidence interval:

CI = 33.5 ± (2.576 * 3.2/√196) = 33.5 ± 0.679.

Therefore, the 99% confidence interval for μ when n=196 is (32.821, 34.179).

c. The width of a confidence interval is determined by the formula: Width = 2 * (z * σ/√n). When quadrupling the sample size while holding the confidence coefficient fixed, the z-score and σ remain the same, but the √n term becomes double. Therefore, the width of the confidence interval is reduced by half.