Question

A study was conducted to examine the difference in house prices in A, B, C and D. Selling pricing were sampled from four houses in each city. The prices for Cities are given (in thousands of CAD) in the Excel spreadsheet. Based on these data, perform a Kruskal-Wallis test to decide if there is a difference in house prices between these cities. Be sure that you carefully define null and alternative hypotheses. Make your conclusions at the 0.10 significance level.

Data given below:

B	A	C	D
289	488	518	524
512	496	506	504
408	505	525	575
494	502	522	582

Answer 1

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

Sample	Value
1	488
1	496
1	505
1	502
2	289
2	512
2	408
2	494
3	518
3	506
3	525
3	522
4	524
4	504
4	575
4	542

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:

Sample	Value (In Asc. Order)
2	289
2	408
1	488
2	494
1	496
1	502
4	504
1	505
3	506
2	512
3	518
3	522
4	524
3	525
4	542
4	575

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:

Sample	Value (In Asc. Order)	Rank	Rank (Adjusted for ties)
2	289	1	1
2	408	2	2
1	488	3	3
2	494	4	4
1	496	5	5
1	502	6	6
4	504	7	7
1	505	8	8
3	506	9	9
2	512	10	10
3	518	11	11
3	522	12	12
4	524	13	13
3	525	14	14
4	542	15	15
4	575	16	16

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:

Sample	Value	Rank (Adjusted for ties)
1	488	3
1	496	5
1	502	6
1	505	8
2	289	1
2	408	2
2	494	4
2	512	10
3	506	9
3	518	11
3	522	12
3	525	14
4	504	7
4	524	13
4	542	15
4	575	16

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

R₁ = 3 + 5 + 6 + 8 = 22

R₂ = 1 + 2 + 4 + 10 = 17

R₃ = 9 + 11 + 12 + 14 = 46

R₄ = 7 + 13 + 15 + 16 = 51

(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

The combination of sample sizes does not allow to find a critical value for this Kruskal-Wallis test, please try increasing the sample sizes. Please consult with a Kruskal-Wallis test to check which combinations of sample sizes lead to meaningful critical values.