MATH 399N Statistics for Decision Making Week 6 iLab Name Statistical Concepts:
ID: 3043237 • Letter: M
Question
MATH 399N Statistics for Decision Making Week 6 iLab Name Statistical Concepts: Data Simulation Confidence Intervals Normal Probabilities Short Answer Writing Assignment All answers should be complete sentences. We need to find the confidence interval for the SLEEP variable. To do this, we need to find the mean and then find the maximum error. Then we can use a calculator to find the interval, (x-E x+ E). First, find the mean. Under that column, in cell E37, type·AVERAGE(E2:E36) Under that in cell E38, type STDEV(E2:E36) Now we can find the maximum error of the confidence interval. To find the maximum error, we use the confidence" formula. In cell E39, type .CONFIDENCE.NORM(0.05,E38,35). The O 05 is based on the confidence level of 95%, the E38 is the standard deviation, and 35 is the number in our sample. You then need to calculate the confidence interval by using a calculator to subtract the maximum error from the mean (x-E) and add it to the mean (x+E). 1. Give and interpret the 95% confidence interval for the hours of sleep a student gets. (5 points) neon 14 cell E40 and type-CONFIDENCE-NORM(0.01,E38,35) to find the Then, you can go down to mum error for a 99% confidence interval. Again, you would need to use a calculator to maxi subtract this and add this to the mean to find the actual confidence interval. 2. Give and interpret the 99% confidence interval for the hours of sleep a student gets. (5 points) G91, 857 compare the 95% and 99% confidence intervals for the hours of sleep a student gets. Explain the difference between these intervals and why this difference 3 10 points) MATH399 Week 6 Lab Template/ page 1 Version 20160523
Explanation / Answer
1)
we know that the confidence interval is given as
mean +- z*sd/sqrt(n)
where z for 95% is 1.96 and for 99% is 2.58
ao we put the values in the equation
7.714 +- 1.96*1.840/sqrt(35)
7.10 and 8.32
Interpretation : we are 95% confident that the true value of the mean would lie in the range 7.10 and 8.32
the 99% ci value is
so we put the values in the equation
7.714 +- 2.58*1.840/sqrt(35)
6.911 and 8.516
Interpretation : we are 99% confident that the true value of the mean would lie in the range
6.911 and 8.516
as the confidence interval increases from 95% to 99% the width of the interval is widened and the margin of error increases
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.