Question

Use graphing calculator to explain answer please. Round results to the nearest 4 decimal places

(3 points) A news report states that the 99% confidence interval for the mean number of daily calories consumed by participants in a medical study is (1870, 2120). Assume the population distribution for daily calories consumed is normally distributed and that the confidence interval was based on a simple random sample of 15 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? Help Entering Answers Preview My Answers Submit Answers

Answer 1

Answer:

Given,

99% confidence interval for population mean = (xbar - s/sqrt(n)*t (0.005 , n-1) , xbar + s/sqrt(n)*t (0.005 , n-1)

n = 15

xbar - s/sqrt(15)*t (0.005 , 14) = 1870

xbar + s/sqrt(15)*t (0.005 , 14) = 2120

So,

a)

Sample mean = (1870 + 2120)/2

= 1995

b)

Margin of error = (2120 - 1870)/2

= 125

c)

So consider,

Margin of error = s/sqrt(15)*t(0.005,14) = 125

s/sqrt(15)*2.977 = 125

s = 125*sqrt(15) / 2.977

s = 162.62