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

1
Expert's answer
2022-04-11T07:50:58-0400

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!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS