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. Develop a 99% confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls.

c. What happens to the margin of error as the confidence level is increased from 95% to 99%?

PLEASE SHOW ALL MATH

Explanation / Answer

sample:

mean = $252.45

standard deviation= $74.50

n=64

a) Confidence interval is given by
sample mean +/- (confidence coefficient)*(standard error of mean)

For 95% CI   confidence coefficient=1.96
252.45 +/- 1.96*74.50/sqrt{64}
252.45 +/- 18.25
loweer boundary is 252.45 - 18.25 = 234.20

(b)

For 99% CI   confidence coefficient=2.58
252.45 +/- 2.58*74.50/sqrt{64}
252.45 +/- 24.03
loweer boundary is 228.42
upper boundary is 276.48
The CI is ($228.42, $276.48)

(c)

Margin of error = (confidence coefficient)*(standard error of mean)

Margin of error increases as confidence level increases

Increased from 18.25 to 24.03

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