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Problem Value: 3 point(s). Problem Score: 0 %. Attempts Remaining: 3 attempts. (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 (1800, 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 21 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

Given, 99 % cofiden ce inkeaval for mean number ot daily calories consumed by paaticipants in study a medical is (1800, 1980) je where E: Margin ot esovr X: sample mean XIe=(8oo (980 adding equation 4e get 2X= 3780 X1890 sample mean subtiacting equatien faom D (we get 2 28 =180 Magin E = go Heve, Masgin 1 * Jo = 2.84534 S 21 n= 21 given go 21 (e 2.84534 S= 144.9499 225 Sample standaod deviation S144 9499