A survey was recently conducted in which random samples of car owners of Chrysle

Question

A survey was recently conducted in which random samples of car owners of Chrysler, GM, and Ford cars were surveyed to determine their satisfaction. Each owner was asked to rate overall satisfaction on a scale of 1 (poor) to 1000 (excellent). The following data were recorded: If the analysts are not willing to assume that the population ratings are normally distributed and will use the Kruskal-Wallis test to determine if the three companies have different median ratings, what is correct test statistic for these data?

Explanation / Answer

Solution:

First, all the data needs to be put together in one column as shown below:

Now, the data needs to be organized in ascending order by value (keeping track of what sample the values belongs to). The results are shown below:

Now, we need to assign ranks to the values that are already organized in ascending order. Make sure that take the average of ranks in case of rank ties (Ex. If two values shared the first place in the list, instead of assigning rank 1 and rank 2 to them, assign rank 1.5 to both) The following ranks are obtained:

In order to compute the sum of rank for each sample, it is easier to organize the above table by samples. The following is obtained:

With the information provided we can now easily compute the sum of ranks for each of the samples:

R1 = 3.5 + 6.5 + 9 + 11.5 + 14 + 17 = 61.5

R2 = 1.5 + 1.5 + 3.5 + 5 + 14 + 17 = 42.5

R3 = 6.5 + 9 + 9 + 11.5 + 14 + 17 = 67

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: The samples come from populations with equal medians

Ha: The samples come from populations with medians that are not all equal

The above hypotheses will be tested using the Kruskal-Wallis test.

(2) Rejection Region

Based on the information provided, the significance level is lpha = 0.05=0.05, and the number of degrees of freedom is df = 3 - 1 = 2df=31=2. Therefore, the rejection region for this Chi-Square test is R = {chi^2: chi^2 > 5.991}R={2:2>5.991}.

(3) Test Statistics

The H statistic is computed as shown in the following formula:

H = rac{12}{N(N+1)}left( rac{R_1^2}{n_1}+rac{R_2^2}{n_2}+ cdots + rac{R_k^2}{n_k} ight) - 3(N+1)

H=[12/18(18+1)}(61.5^2/6+42.5^2/6+67^2/6)3(18+1)=1.933

4) Decision about the null hypothesis

Since it is observed that chi^2 = 1.933 leq chi_U^2 = 5.9912=1.933U2=5.991, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.3805p=0.3805, and since p = 0.3805 geq 0.05p=0.38050.05, it is concluded that the null hypothesis is not rejected.

(5) 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 medians are unequal, at the =0.05 significance level.

Sample Value 1 650 1 700 1 500 1 800 1 900 1 750 2 400 2 800 2 500 2 400 2 600 2 900 3 700 3 750 3 650 3 800 3 900 3 700

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