The mean amount of money spent on lunch per week for a sample of 592592 students is $15$ 15. If the margin of error for the population mean with a 98%98% confidence interval is 1.701.70, construct a 98%98% confidence interval for the mean amount of money spent on lunch per week for all students.

The 98% confidence interval for the mean amount of money spent on lunch per week for all students is between $13.3 and $16.7.

Step-by-step explanation:

A confidence interval has the following format.

In which is the mean of the sample and M is the margin of error.

In this problem, we have that:

The 98% confidence interval for the mean amount of money spent on lunch per week for all students is between $13.3 and $16.7.

4sqrt5

step-by-step explanation:

the base has dimensions of 8 x 4, so the diagonal is 64 + 16 square rooted or 4sqrt5.