Question

Question #1 A researcher wants to know if the average time in jail for robbery has increased from what it was when the average sentence was 7 years. He obtains data on 400 more-recent robberies and finds an average time served of 7.5 years. Find the 95% confidence interval for the average time served is (assume jail time follows a Normal distribution with standard deviation 3 years). Interpret your interval.
Question #2 The manager at a movie theater would like to estimate the true mean amount of money spent by customers on popcorn only. He selects a simple random sample of 26 receipts and calculates a 92% confidence interval for true mean to be ($12.45, $23.32). The confidence interval can be interpreted to mean that, in the long run:a.92% of similarly constructed intervals would contain the population mean. b. 92% of similarly constructed intervals would contain the sample mean.c.92% of all customers who buy popcorn spend between $12.45 and $23.22.

Answer 1

Question #1.

$95\%CI=(\bar x-1.96\frac{\sigma}{\sqrt{n}},\;\bar x+1.96\frac{\sigma}{\sqrt{n}})=\\ =(7.5-1.96\frac{3}{\sqrt{400}},\;7.5+1.96\frac{3}{\sqrt{400}})=(7.206, 7.794).$

We are 95% confident that the average time in jail for robbery lies between 7.206 years and 7.794 years.

Question #2.

a.92% of similarly constructed intervals would contain the population mean.