An independent group of food service personnel conducted a survey on tipping practices in a large metropolitan area. They collected information on the percentage of the bill left as a tip for 35 randomly selected bills. The average tip was 14.2% of the bill with a standard deviation of 2.5%. Assume that the tips are approximately normally distributed. Construct an interval to estimate the true average tip (as a percent of the bill) with 99% confidence. Round the endpoints to two decimal places, if necessary.

Respuesta :

An interval to estimate the true average tip is  (14.62,13.78).

According to statement is

we know that the

confidence interval = sample mean ± standard deviation /(n)^1/2

substitute the values in it then

confidence interval = 14.2  ± 2.5/(35)^1/2

confidence interval = 14.2 ± 2.5/5.91

confidence interval = 14.2 ± 0.42

confidence interval = 14.2 + 0.42 , confidence interval = 14.2 - 0.42

confidence interval = (14.62,13.78)

So, an interval to estimate the true average tip is  (14.62,13.78).

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