Question

Question 1: A survey conducted by the AAA 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 standard deviation of $74.50.

1. Construct the 90% confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls and interpret your interval
2. Do we have to assume spending is normally distributed and please explain why or why not
3. If we increase the confidence from 90% to 95% then what should happen to the interval in part a) and please explain your answer
4. If we increase the sample size from 64 to 300 families then what should happen to the interval in part a) and please explain your answer
5. Based on the CI 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 AAA and please explain your answer

Answer 1

Part a

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± t*S/sqrt(n)

We are given

Xbar = 252.45

S = 74.5

n = 64

df = n – 1 = 64 – 1 = 63

Confidence level = 90%

Critical t value = 1.6694

(by using t-table or excel)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 252.45 ± 1.6694*74.5/sqrt(64)

Confidence interval = 252.45 ± 1.6694*9.3125

Confidence interval = 252.45 ± 15.5463

Lower limit = 252.45 - 15.5463 = 236.90

Upper limit = 252.45 + 15.5463 = 268.00

WE are 90% confident that the family of four average spends will lies within $236.90 and $268.00.

Part b

Yes, we have to assume spending is normally distributed because the use of t or z distribution is depends on the nature of population distribution.

Part c

If we increase the confidence from 90% to 95%, the confidence interval in part a will becomes as below:

Confidence interval = Xbar ± t*S/sqrt(n)

We are given

Xbar = 252.45

S = 74.5

n = 64

df = n – 1 = 64 – 1 = 63

Confidence level = 95%

Critical t value = 1.9983

(by using t-table or excel)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 252.45 ± 1.9983*74.5/sqrt(64)

Confidence interval = 252.45 ± 1.9983*9.3125

Confidence interval = 252.45 ± 18.6095

Lower limit = 252.45 - 18.6095 = 233.84

Upper limit = 252.45 + 18.6095 = 271.06

This means, if we increase the confidence level, the width of the confidence interval also increases.

Part d

Here, we have to use n = 300 instead of 64 for the part a.

Confidence interval = Xbar ± t*S/sqrt(n)

We are given

Xbar = 252.45

S = 74.5

n = 300

df = n – 1 = 300 – 1 = 299

Confidence level = 90%

Critical t value = 1.65

(by using t-table or excel)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 252.45 ± 1.65*74.5/sqrt(300)

Confidence interval = 252.45 ± 1.65* 4.301259505

Confidence interval = 252.45 ± 7.0969

Lower limit = 252.45 - 7.0969= 245.35

Upper limit = 252.45 + 7.0969= 259.55

This means, as we increase the sample size, the width of the confidence interval decreases.

Part e

From part a, we are 90% confident that the family of four average spends will lies within $236.90 and $268.00. The value $215.60 is not lies within this interval. So, it does not appear that the population mean amount spent per day by families visiting Niagara Falls differs from the mean reported by the AAA.