Question

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.

Answer 1

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.

Explanation:

A confidence interval has the following format.

`\mu_(x) \pm M`

In which
`\mu_(x)` is the mean of the sample and M is the margin of error.

In this problem, we have that:

`\mu_(x) = 15, M = 1.7`

`\mu_(x) - M = 15 - 1.7 = 13.3`

`\mu_(x) + M = 15 + 1.7 = 16.7`

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.