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%?

Explanation / Answer

a.
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=252.45
Standard deviation( sd )=74.5
Sample Size(n)=64
Confidence Interval = [ 252.45 ± t a/2 ( 74.5/ Sqrt ( 64) ) ]
= [ 252.45 - 1.998 * (9.313) , 252.45 + 1.998 * (9.313) ]
= [ 233.844,271.056 ]

b.
Confidence Interval = [ 252.45 ± t a/2 ( 74.5/ Sqrt ( 64) ) ]
= [ 252.45 - 2.656 * (9.313) , 252.45 + 2.656 * (9.313) ]
= [ 227.716,277.184 ]

c.
Increasing the confidence larger the margin of error

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