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

Answer:

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.

$\mu_{x} \pm M$

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.