Question

The mean amount of money spent on lunch per week for a sample of 458458 college students is $43$ 43. If the margin of error for the population mean with a 99%99% confidence interval is 1.601.60, construct a 99%99% confidence interval for the mean amount of money spent on lunch per week for all college students.

Answer 1

Answer:

The 99% confidence interval for the mean amount of money spent on lunch per week for all college students is between $41.40 and $44.60.

Step-by-step explanation:

A confidence interval has two bounds, a lower bounds and an upper bound.

These bounds depend on the sample mean and the margin of error.

The lower bound is the sample mean subtracted by the margin of error.

The upper bound is the margin of error added to the sample mean.

In this problem, we have that:

The sample mean is $43.

The margin of error is $1.60 for a 99% confidence interval.

Lower bound: 43 - 1.60 = $41.40

Upper bound: 43 + 1.60 = $44.60

The 99% confidence interval for the mean amount of money spent on lunch per week for all college students is between $41.40 and $44.60.