A sample of n = 25 n=25 diners at a local restaurant had a mean lunch bill of $16 with a standard deviation of σ = $ 4 σ=$4. We obtain a 95% confidence interval as (14.43, 17.57) (14.43,17.57). Which choice correctly interprets this interval? None of the answer options are correct. 95% of all lunches will cost between $14.43 and $17.57. 95% of the time, the average price for a lunch will be between $14.43 and $17.57. 95% of all samples of size n = 25 n=25 will have an average lunch price between $14.43 and $17.57.

Question

Mathematics

Dottie

12 July, 01:55

A sample of n = 25 n=25 diners at a local restaurant had a mean lunch bill of $16 with a standard deviation of σ = $ 4 σ=$4. We obtain a 95% confidence interval as (14.43, 17.57) (14.43,17.57). Which choice correctly interprets this interval? None of the answer options are correct. 95% of all lunches will cost between $14.43 and $17.57. 95% of the time, the average price for a lunch will be between $14.43 and $17.57. 95% of all samples of size n = 25 n=25 will have an average lunch price between $14.43 and $17.57.

+5

Answers (1)

Know the Answer?

Remington Wright · Answer 1

None of the answer options are correct.

Step-by-step explanation:

A confidence interval of the mean is a range within is expected to find the population mean, with a certain degree of confidence.

In this case, the population mean is the average price for a lunch for a n=25 sample.

The interpretation of a 95% confidence interval of (14.43,17.57) should be that there is a 95% of confidence that the population mean (average price for a lunch out of a sample of 25 lunches) is within this interval.