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.

distance covered by pamela in 9 l of fuel = 99km

let us apply unitary method here.

distance covered by pamela in 1 l of fuel =99/9 =11 km

now let us find distance covered by pamela if she has 12 l of fuel.

distance covered in 12 l = distance covered in 1 l of fuel * 12

we know that distance covered by here in 1 l is 11 km.

so distance covered in 12 l of fuel = 11*12 = 132 km.

c

step-by-step explanation: