Question

Suppose Texas A&M University-Commerce has five scholarships available for the women's basketball team. The head coach provided the two assistant coaches with the names of 10 high school players with potential to play at the university. Each assistant coach attended three games and then ranked the 10 players with respect to potential. To explain, the first coach ranked Norma Tidwell as the best player among the 10 scouted and Jeannie Black the worst. Rank, by Assistant Coach Jean Cann John Cannelli Player Cora Jean Seiple Bette Jones Jeannie Black Norma Tidwell Kathy Marchal 7 6 4 Player Candy Jenkins Rita Rosinski Anita Lockes Brenda Towne Denise Ober Rank, by Assistant Coach Jean Cann John Cannelli 3 1 9 8 10 4 2 2 9 6 10 1 Click here for the Excel Data File (Leave no cells blank - be certain to enter "O" wherever required. Negative amounts should be indicated by a minus sign.) Player Rank by Cann Rank by Cannelli d d-square 1 5 5 7 2 7 3 6 9 4 4 6 5 8 10 6 3 1 7 9 8 8 10 4 9 2 ني 10 1 2

Answer 1

a. Spearman's coefficient of rank correlation = 0.6364 The Spearman's coefficient of rank correlation () is calculated as: p = 1- 62d2 where n(n2-1)' = 60 as shown in the table below, n = 10 (number of observations. Hence, 6260 360 360 1 1-3 1 -0.3636 = 0.6364 10(100 – 1) 10(99) 990 = p=1- 1 1 0 9 4 4 00 4 4 3 9 1 3 2 1 A B с D E Α. B с D E Rank by Rank by Player d d-square Rank by Rank by Cann Cannelli Player d Cann Cannelli d-square 2 1 5 5 0 0 2 1 5 5 =B2-C2 =D2^2 3 2 7 7 이 3 2 7 7 =B3-C3 =D3^2 4 3 6 9 -3 43 6 9 =B4-C4 =D4^2 5 6 -2 54 4 6 =B5-C5 =D5^2 6 5 8 10 -2 6 5 8 10 =B6-C6 =D6^2 7 6 1 2 7 6 3 1 =B7-C7 =D742 8 7 8 1 1 8 7 9 8 =B8-C8 =D8^2 9 8 10 4 6 36 98 10 4 =B9-09 =D9^2 10 9 2 3 -1 10 9 2 =B10-C10 =D10^2 11 10 2 -1 1 11 10 1 =B11-C11 =D11^2 The left-hand side table shows the calculated values and the right-hand side table shows the formula view of the calculations. b. Ho:P < 0, HP > 0 Reject H, if t > 1.397 (Critical value at 0.1 or 10% level of significance with n - 2 = 8 degree of freedom = 1.397, one tailed) 0.6364(80.5) t> 2.985 (The t-statistic can be calculated as t = 2.985) (1-p2.0.5 (1-0.6364230.5 Ho is rejected. There is positive association between the ratings of the two coaches. (Since calculated value 2.985 is greater than the critical value 1.397, we reject the null hypothesis and conclude that there is a significant positive correlation between the ratings of the two coaches. # P(n-2)0.5 =