Question #325866

According to one of the teachers in your school, the senior high school students sleep for 7 hours a day on a weekday. You wanted to prove this claim by making a survey of 50 senior high school students. You will be asking the students about how long they sleep in a regular weekday. Record the results of your survey. As a researcher, you need to compute the mean and standard deviation of your sample.




Create a 90%, 95%, and 99% confidence interval for the mean time of sleep for senior high school students. In your report, the raw data should be present, and the computation should be complete

Expert's answer

Mean(M)=(X)/50=7Mean(M)=\sum(X)/50=7

D=(XM)2n1=(X7,0)249=0.37D=\frac{\sum(X-M)^2}{n-1}=\frac{\sum(X-7,0)^2}{49}=0.37


σ=D=0.37=0.6\sigma=\sqrt{D}=\sqrt{0.37}=0.6


CI=M±zσnCI=M \plusmn z\frac{\sigma}{\sqrt{n}}

For 90 % z=1.645

CI=7±1.6450.650=7±0.14CI=7 \plusmn 1.645 \frac{0.6}{\sqrt{50}}=7 \plusmn 0.14

For 95 % z=1.96

CI=7±1.960.650=7±0.17CI=7 \plusmn 1.96 \frac{0.6}{\sqrt{50}}=7 \plusmn 0.17

For 99 % z=2.58

CI=7±2.580.650=7±0.22CI=7 \plusmn 2.58 \frac{0.6}{\sqrt{50}}=7 \plusmn 0.22






Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

LATEST TUTORIALS
APPROVED BY CLIENTS