Question

In an effort to estimate the mean amount spent per customer for dinner at a major Atlanta restaurant, data were collected for a sample of 49 customers over a three-week period. If the sample provided a mean of $24.80 and the population standard deviation was known to be $5.
a) What is the 95% confidence interval for the population mean
b. Develop a 95% confidence interval estimate of the mean amount spent for dinner.Enter your answer using parentheses and a comma, in the form (n1,n2). Do not use dollar sign or spaces or include a comma in a number.

Answer 1

Answer:

95% Confidence interval = (23.4,26.2)

Step-by-step explanation:

In this problem we have to develop a 95% CI for the mean.

The sample size is n=49, the mean of the sample is M=24.8 and the standard deviation of the population is σ=5.

We know that for a 95% CI, the z-value is 1.96.

The CI is

$M-z*\sigma/\sqrt{n}\leq\mu\leq M+z*\sigma/\sqrt{n}\\\\24.8-1.96*5/\sqrt{49}\leq\mu\leq 24.8+1.96*5/\sqrt{49}\\\\ 24.8-1.4\leq\mu\leq 24.8+1.4\\\\23.4\leq\mu\leq 26.2$