Problem 1) A survey is conducted by the American Automobile Association to inves

ID: 3174925 • Letter: P

Question

Problem 1) 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. Based on historical data,

we assume that the standard deviation is = $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 work! Thank you

Here, n = 64 , mean = 252.45 , std. deviation = 74.50

Z value for 95% confidence interval = 1.96

Standard error = s / sqrt(n)

= 74.50 / sqrt(64)

= 9.31

CI = mean + / - z * SE

= 252.45 + / - 1.96 * 9.31

= (234.19 , 270.70)

Lower Limit = 234.19

Upper limit = 270.70

Z value for 99% confidence interval = 2.578

Standard error = s / sqrt(n)

= 74.50 / sqrt(64)

= 9.31

CI = mean + / - z * SE

= 252.45 + / - 2.578 * 9.31

= (228.44 , 276.45)

Lower Limit = 228.44

Upper limit = 276.45

(C)

ME = z*SE

For 95% CI, ME = 1.96*9.31 = 18.2476

For 99% CI, ME = 2.578*9.31 = 24.0012

ME increases as CI increases

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