A survey is conducted by the American Automobile Association to investigate the

Question

A survey is conducted by the American Automobile Association to investigate the

daily expense of a family of four while on vacation. Suppose that a sample of 64 families of four

vacationing at Niagara Falls resulted in sample mean of $252.45 per day and a sample standard

deviation of $74.50.

a. Develop a 95% confidence interval estimate of the mean amount spent per day by a family of

four visiting Niagara Falls.

b. What is the difference between problem 1 and problem 2?

Please show work! Thank you

Explanation / Answer

Solution:-

a) We have, s = $74.50, X-bar = $252.45, = 1-0.95 = 0.05
Z (/2) = Z (0.05/2)
= Z (0.025)
= 1.96
95% confidence interval is given by:-
X-bar ± Z (/2)*/n
=>252.45 ± 1.96*74.50/65
=>252.45 ± 1.96*9.24
=>252.45 ± 18.11
=>234.34, 270.56

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