The director of student health at a large university is studyinghow many calorie

Question

The director of student health at a large university is studyinghow many calories students at her university consume in a day. Shetakes a random sample of 16 students fromthe population and records the number of calories they consume in a24 hour period. From the sample, she found that the average numberof calories consumed was 1978 and thatstandard deviation was 440. The directorwishes to construct a 80% confidenceinterval for the mean number of calories consumed by thestudents.

What value of t is needed for this confidence level?1

What is the margin of error for this confidence interval?2

Using the statistics given above, calculate the confidenceinterval the director is looking for.

Explanation / Answer

Given n=18, xbar=2106, s=440 =0.2 What value of t is needed for this confidence level? t(0.01, df=n-1=18-1=17)=2.57 (check student ttable) What is the margin of error for this confidence interval? s/n =440/sqrt(18) = 103.709 confidence interval xbar±t*s/n --> 2106 ± 2.57 * 103.709 --> (1839.468, 2372.532)

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