Question

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 $74.50.

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?
Select Yes / No

Answer 1

A	B	C	D	E	F	G	H	I	J	K
2
3		a)
4		Number of families in sample (N)	64
5		Sample Mean (M)	$252.45
6		Sample Standard Deviation (S)	$74.50
7
8		Standard error of estimate (σM)	=S/sqrt(N)
9			$1.16
10
11		Z Value for 95% confidence	1.96
12
13		Lower confidence interval	=M-Z.95*σM
14			$250.17
15
16		Upper confidence interval	=M+Z.95*σM
17			$254.73
18		Therefore,
19		The confidence interval is	=($250.17, $254.73)
20
21		b)
22		Since the population standard deviation is not known, therefore t-statistics is to be used
23		to test that the sample mean is different from the population mean or not.
24
25		Sample Mean	$252.45
26		Sample St. Deviation	$74.50
27
28		Inputs--
29		Hypothesized Population mean (µ):	215.60
30		Sample Standard Deviation (s):	74.50
31		Sample Size (n):	64
32		Sample mean (X-bar):	252.45
33		Alpha	0.05	(for 95% Confidence)
34		Hypothesis formulation:
35		Null hyposthesis is an assertion that is true unless there is sufficient statistical evidance to conclude otherwise.
36		Alternation hypothesis is the negation of the null hypothesis.
37
38		Null Hypothesis: µ = 215.60
39		Alternate Hypothesis: µ ≠ 215.60
40
41		Intermediate calculation--
42		Standard Error of the Estimate(S.E.):	9.31	=s/sqrt(n)		=D30/SQRT(D31)
43		Test Statistic(t):	3.96	=(X-bar - µ)/S.E.		=(D32-D29)/D42
44		Degree of freedom (d.f.):	63	=n-1		=D31-1
45
46		Results:
47		Two tailed, H0:µ = 215.60, p=	0.000	=1-T.DIST(D43,D44,TRUE)
48
49		For Alpha level given, H0 should be:
50		Alpha	0.05
51		Null hypothesis should be	Rejected	(As p-value is lower than alpha)
52		Thus it can be concluded that for a significance level of 0.05, there is sufficient evidence to support
53		that the population mean amount for families visiting the Nigeria Falls differs from the mean
54		as reported by the American automobile association.
55
56		Hence the answer is Yes.
57