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

15-04-2020
Mathematics

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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.

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Answer:

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.