The employees of a large international company would like to have vision care as part of their benefits package. A researcher needs to estimate the amount of money that employees and their beneficiaries spend on such treatment, on average. A sample of 1296 employees reveals the mean amount spent last year was $1,140.05, with a standard deviation of $108.77. We assume that the population standard deviation is close to the last number.

(a)Construct a 90% confidence interval for the population mean amount spent on vision care.

Findz2=

Round the confidence interval values to 2 decimal places.

What is the 90% confidence interval?

(b)Construct a 98% confidence interval for the population mean amount spent on vision care.

Findz2=

Round the confidence interval values to 2 decimal places.

What is the 98% confidence interval?

(c) What is not true about the confidence intervals?

• The researcher is 98% confident that the confidence interval constructed in part (b) contains the true average amount spent on vision care.
• If the confidence level increases from 90% to 98%, then the confidence interval becomes shorter.
• There is a 90% chance that the confidence interval constructed in part (a) contains the true population mean amount spent on vision care.
• If the confidence level increases from 90% to 98%, then the confidence interval becomes longer.