Question

In: Statistics and Probability

6. As a class, we own and operate four car washes in Tucson. We notice that...

6. As a class, we own and operate four car washes in Tucson. We notice that fewer people are out driving right now, so we might have to make the difficult decision to down-size our operation by closing a location. We gather the following information about each location.

Location Washes
Oracle 213

Tangerine 170

Kolb 201

Speedway 216

At the 0.05 significance level, is there a difference in the number of full-service washes purchased at each location?

(A) Write out the null and alternate hypotheses for this problem. (8)

(B) State the decision rule at the 0.05 level of significance. (8)

(C) Compute the chi-square value. (14)

(D) Make a decision about the hypotheses. Be sure to clearly articulate what you would do about our car wash locations. (10)

Expert Solution

Chi-Square Goodness of Fit test

(1) Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
H0: p1 =0.25, p2 =0.25, p3=0.25, p4=0.25
Ha: Some of the population proportions differ from the values stated in the null hypothesis

This corresponds to a Chi-Square test for Goodness of Fit.

(2)Critical Value and Rejection Region
Based on the information provided, the significance level is α=0.05, the number of degrees of freedom is df=n-1=4-1=3, so the critical value becomes 7.8147.

Decision rule
Then the rejection region i.e. reject H0 for this test if R={χ2:χ2>7.8147}.

(3) Degrees of Freedom
The number of degrees of freedom is df=n-1=4-1=3

Location	Washes	Expected Percent	Expected Frequency	Σ(O-E)^2/E
Oracle	213	1/4	200.0000	0.8450
Tangerine	170	1/4	200.0000	4.5000
Kolb	201	1/4	200.0000	0.0050
Speedway	216	1/4	200.0000	1.2800
Total	800	1.0000	800	6.6300

Test Statistics
The Chi-Squared statistic is computed as follows:

P-value
The P-value is the probability that a chi-square statistic having 3 degrees of freedom is more extreme than 7.8147.
The p-value is p=Pr(χ2>6.63)=0.0847

(4) The decision about the null hypothesis
Since it is observed that χ2=6.63≤χc2=7.8147, it is then concluded that the null hypothesis is not rejected.

Conclusion
It is concluded that the null hypothesis Ho is not rejected. Therefore, there is NOT enough evidence to claim that some of the population proportions differ from those stated in the null hypothesis, at the α=0.05 significance level. Since there is no significant difference between the four locations, therefore it would be more prudent not to close down any or choose one randomly to close down, most likely Tangerine since it contributes maximum to the chi-square statistic.

orchestra answered 2 years ago

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