Question

AARP estimated that the average annual expenditure on restaurants and carryout food was $3500 for individuals of age 50 and over. Suppose this estimate is based on a sample of 60 persons, and that the sample standard deviation is $800. [From Chapter 8]

a) At 95% confidence, what is the margin of error?

b) What is the 95% confidence interval for the population mean amount spent on restaurants and carryout food?

a. $206.85
b. $3220.15 to $3779.85
c. $296.28
d. $3400.00 to $3600.00

Answer 1

a) At 95% confidence, the margin of error is approximately $203.05. b) The 95% confidence interval for the population mean amount spent on restaurants and carryout food is approximately $3296.95 to $3703.05.

Step-by-step explanation:

a) To find the margin of error, we can use the formula:

Margin of Error = Critical Value * Standard Error

The critical value for a 95% confidence interval can be found using the z-table or calculator. For a 95% confidence interval, the critical value is approximately 1.96.

The standard error can be calculated using the formula:

Standard Error = Sample Standard Deviation / sqrt(Sample Size)

Substituting the given values:

Standard Error = 800 / sqrt(60) ≈ 103.68

Now, we can calculate the margin of error:

Margin of Error = 1.96 * 103.68 ≈ 203.05

So, the margin of error is approximately $203.05.

b) The 95% confidence interval can be calculated using the formula:

Confidence Interval = Sample Mean ± Margin of Error

Substituting the given values:

Sample Mean = $3500

Margin of Error = $203.05

Confidence Interval = $3500 ± $203.05 = $3296.95 to $3703.05

Therefore, the 95% confidence interval for the population mean amount spent on restaurants and carryout food is approximately $3296.95 to $3703.05.