Question

The mean amount of money spent on lunch per week for a sample of 100 students is $21. If the margin of error for the population mean with a 99% confidence interval is 1.50, construct a 99% confidence interval for the mean amount of money spent on lunch per week for all students.

Answer 1

From the given mean and margin of error, the 99% confidence interval for the mean amount of money spent on lunch per week for all students is:

[$19.5, $22.5].

How to calculate a confidence interval given the sample mean and the margin of error?

The confidence interval is given by the sample mean plus/minus the margin of error, hence:

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

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

For this problem, we have that:

• The sample mean is of $21.

• The margin of error is of $1.50.

Hence the bounds are given as follows:

• Lower bound: 21 - 1.50 = $19.50.

• Upper bound: 21 + 1.50 = $22.50.

Hence the interval is [$19.50, $22.50].

More can be learned about confidence intervals at brainly.com/question/25890103

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