Question

(3 points) A news report states that the 90% confidence interval for the mean number of daily calories consumed by participants in a medical study is (1750, 1980). Assume the population distribution for daily calories consumed is normally distributed and that the confidence interval was based on a simple random sample of 18 observations. Calculate the sample mean, the margin of error, and the sample standard deviation based on the stated confidence interval and the given sample size. Use the t distribution in any calculations and round non-integer results to 4 decimal places. 1. What is the sample mean? 2. What is the margin of error? 3. What is the sample standard deviation?

Answer 1

1) sample mean = (1980 + 1750) / 2 = 1865

2) margin of error = (1980 - 1750) / 2 = 115

3) t score for 90% confidence interval = t_0.05,17 = 1.740

Margin of error = t_0.05,17 * s / sqrt(n)

or, 115 = 1.74 * s / sqrt(18)

or, s = 280.4044