Question

The National Football League (NFL) holds its annual draft of the nation's best college football players in April each year. Prior to the draft, various sporting news services project the players who will be drafted along with the order in which each will be selected in what are called mock drafts. Players who are considered to have superior potential as professional football players are selected earlier in the draft. Suppose the following table shows projections by one mock draft service of what position in the first round players from the Atlantic Coast Conference, the Big Ten Conference, the PAC-12 Conference, and the Southeastern Conference will be selected.

ACC		Big Ten		PAC-12		SEC
College Attended	Projected Draft Position	College Attended	Projected Draft Position	College Attended	Projected Draft Position	College Attended	Projected Draft Position
Florida State	2	Iowa	9	USC	4	Florida	1
Clemson	5	Michigan St	10	Oregon	7	Alabama	3
Miami	6	Nebraska	11	Oregon	15	Kentucky	8
Georgia Tech	13	Minnesota	26	Washington	16	Texas A&M	12
Louisville	17	Wisconsin	28	UCLA	18	Missouri	14
Wake Forest	20			UCLA	21	Alabama	19
Florida State	22			Stanford	23	LSU	25
Virginia Tech	29			Arizona St	24	LSU	27

Use the Kruskal-Wallis test to determine if there is any difference among NFL teams for players from these four conferences. Use α = 0.05.

Find the value of the test statistic. (Round your answer to two decimal places.)

What is the p-value? (Round your answer to three decimal places.)

Answer 1

First, all the data needs to be put together in one column as shown below: Value Sample 1 5 1 1 13 1 17 | 1 20 1 22 | 1 29 1 9 2 10 2 11 | | 2 26 2 28 2 4 3 7 3 15 16 3 18 21 3 23 24 co

1 3 4 8 4 12 14 19 4 25 4 27 4. 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: Value (In Asc. Order) Sample 1 4 2 1 3 | 4 4 3 5 1 6 1 7 | 3 8 4 9 2 10 2 11 | 2 12 4 st

13 | 1 14 4 15 3 16 3 17 1 18 19 4 20 1 21 3 22 1 23 3 24 3 25 4 26 2 27 4 28 2 29 1 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:

Rank (Adjusted for ties) Rank Order) Value (In Asc. Sample 1 1 1 4 2 2 2 1 4 4 4 3 5 | 5 5 1 6 6 6 7 | 7 3 8 4 9 | 2 10 10 10 2 11 11 11 2 12 12 12 4 13 13 | 13 1 14 14 14 4 15 | 15 15 16 16 16 17 17 17 1 18 18 18 3 19 19 19 4 20 20 20 1 21 21 21 3 22 22 22 1 st LO t en

23 23 23 24 24 24 25 | 25 25 26 26 26 2 27 27 27 4 28 28 28 2 29 29 29 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: Rank (Adjusted for ties) Value Sample 2 2 5 5 1 6 1 13 13 1 17 17 1 20 20 1 22 22 | 29 29 1 2 10 10 2 11 11 2 26 26 2 28 28 2 LO LO O st

3 7 7 15 15 16 | 16 18 18 21 21 23 23 3 24 24 1 4 3 4 12 12 4 14 14 19 19 25 25 4 27 27 4 easily compute the sum of ranks for each of the samples: With the information provided we can now 114 25+6 13+17 +20 +22+29 R, R2 9+10 11+26+2884 R3 47+ 15+16+18 +21+23+24 = 128 Ri 13+8 121419 25 27 109 co t

(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 a 0.05, and the number of degrees of freedom is df 1 3. Therefore, the rejection region for this Chi-Square test is R (x2 4 x27.815 (3) Test Statistics The H statistic is computed as shown in the following formula: R3 12 H N(N 1) 3(N1) + ni 109 1142 842 1282 12 3(29 1) 0.604 29(29+1) 8 5 8 8 (4) Decision about the null hypothesis Since it is observed that y2 0.604<x 7 7.815, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p 0.8954, and since p 0.8954 0.05, 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 a 0.05 significance level. l

Based on above calculation answers are as follows:

1. Test Statistics = 0.60

2. P-value = 0.896 (Obtained using online calculator for Chi-Square = 0.604 and df = 3. Screenshot attached)

P Value from Chi-square Calculator This calculator is designed to generate a p-value from a chi- square score. If you need to derive a chi-square score from raw data, you should use our chi-square calculator (which will additionally calculate the p-value for you) The calculator below should be self-explanatory, but just in case it's not: your chi-square score goes in the chi-square score box, you stick your degrees of freedom in the DFbox (df- (NColumns 1)*(NRows-1) for chi-square test for independence), select your significance level, then press the button 0.6040 Chi-square score: 3 DF. Significance Level: 0.01 0.05 O0.10 The P-Value is .895516. The result is not significant at p.05.

3. There is not enough evidence to claim that there is difference among NFL teams for players from these four conferences