Question

A travel analyst claims that the standard deviation of the room rates for two adults at 3-star hotels in Denver, Colorado is more than $68. A random sample of 18 3-star hotels has a standard deviation of $40. Use a 0.10 significance level to test the analyst’s claim. Assume the population is normally distributed.

a. Hypothesis (steps 1-3):

b. Value of Test Statistic (steps 5-6):

c. P-value (step 6):

d. Decision (steps 4 and 7):

e. Conclusion (step 8):

Answer 1

Solution:

Given:

Claim: the standard deviation of the room rates for two adults at 3-star hotels in Denver, Colorado is more than $68.

Sample Size = n = 18

Sample standard deviation = 40

significance level = 0.10

Part a. Hypothesis:

Since claim is directional to right side, this is right tailed test.

Thus H0 and H1:

Part b. Value of Test Statistic

Part c. P-value

df = n - 1 = 18 -1 = 17

Look in Chi-square table for df = 17 row and find the interval in which fall , then find corresponding right tail area interval, which is p-value interval.

fall between 5.697 and 6.408 and corresponding right tail area interval is between 0.990 and 0.995

thus we get: 0.990 < p-value < 0.995

To get exact p-value use Excel command:

=CHISQ.DIST.RT(_X² , df )

=CHISQ.DIST.RT(5.882,17)

=0.9939

Thus p-value = 0.9939

We can use TI 84 plus also:

Press 2ND and VARS

select cdf

Enter numbers:

Thus p-value = 0.9939

Part d. Decision

Decision Rule:
Reject H0, if p-value < 0.10 level of significance, otherwise we fail to reject H0

Since p-value = 0.9939 > 0.10 level of significance, we fail to reject H0

Part e. Conclusion

At 0.10 level of significance, there is not sufficient to support the claim that the standard deviation of the room rates for two adults at 3-star hotels in Denver, Colorado is more than $68.