1. A simple random sample of 60 items resulted in a sample mean of 68. The popul

Question

1. A simple random sample of 60 items resulted in a sample mean of 68. The population standard deviation is 14.

a. Compute the 95% confidence interval for the population mean (to 1 decimal).

b. Assume that the same sample mean was obtained from a sample of 120 items. Provide a 95% confidence interval for the population mean (to 2 decimals).

2. A survey conducted by the American Automobile Association showed that a family of four spends an average of $215.60 per day while on vacation. Suppose a sample of 64 families of four vacationing at Niagara Falls resulted in a sample mean of $252.45 per day and a sample standard deviation of $71.50.

a. Develop a 95% confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls (to 2 decimals).

b. Based on the confidence interval from part (a), does it appear that the population mean amount spent per day by families visiting Niagara Falls differs from the mean reported by the American Automobile Association?

(YES/NO)

Explanation / Answer

Q1.
a.
Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=68
Standard deviation( sd )=14
Sample Size(n)=60
Confidence Interval = [ 68 ± Z a/2 ( 14/ Sqrt ( 60) ) ]
= [ 68 - 1.96 * (1.81) , 68 + 1.96 * (1.81) ]
= [ 64.46,71.54 ]

b.
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=68
Standard deviation( sd )=14
Sample Size(n)=120
Confidence Interval = [ 68 ± Z a/2 ( 14/ Sqrt ( 120) ) ]
= [ 68 - 1.96 * (1.28) , 68 + 1.96 * (1.28) ]
= [ 65.5,70.5 ]

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