Answer to Question #325866 in Statistics and Probability for joy

Question

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

Answer 1

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

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

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

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

For 90 % z=1.645

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

For 95 % z=1.96

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

For 99 % z=2.58

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