Question

A travel association says the daily lodging cost for a family in the United States is $152. You work for a tourist publication and want to test this claim. You randomly select 10 U.S. families and find out they spent on lodging for one overnight trip as below. At α = 0.02, can you reject the travel association’s claim?

164, 137, 142, 155, 119, 104, 74, 204, 148, 181

Answer #1

Here, we have to use one sample t test for the population mean.

The null and alternative hypotheses are given as below:

Null hypothesis: H_{0}: the daily lodging cost for a
family in the United States is $152.

Alternative hypothesis: H_{a}: the daily lodging cost
for a family in the United States is different from $152.

H_{0}: µ = 152 versus H_{a}: µ ≠ 152

This is a two tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 152

Xbar = 142.8

S = 37.51977257

n = 10

df = n – 1 = 9

α = 0.02

Critical value = - 2.8214 and 2.8214

(by using t-table or excel)

t = (Xbar - µ)/[S/sqrt(n)]

t = (142.8 - 152)/[ 37.51977257/sqrt(10)]

t = -0.7754

P-value = 0.4580

(by using t-table)

P-value > α = 0.02

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that the daily lodging cost for a family in the United States is $152.

Math 15Assessment # 41. A travel association says that the daily
lodging cost for a family traveling in California is the same as in
Florida. The daily lodging cost for 35 families traveling in
California is $136 and the standard deviation is $25. The daily
lodging cost for 35 families traveling in Florida is $140 and the
standard deviation is $30. At α = 0.10 is there enough evidence to
reject the travel association’s claim?
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