Of the 23 first year male students at State U. admitted from Jim Thorpe High Sch

ID: 3320587 • Letter: O

Question

Of the 23 first year male students at State U. admitted from Jim Thorpe High School, 8 were offered baseball scholarships and 7 were offered football scholarships. The University admissions committee looked at the students' composite ACT scores (shown in the table), wondering if the University was lowering their standards for athletes. Assuming that this group of students is representative of all admitted students, what do you think? Composite ACT Score Baseball Non-athletes Foothall 21 27 29 26 30 27 26 23 19 25 24 25 24 23 21 24 27 19 23 17 ANOVA table Sums of Mean Source df F-ratio P-value Squares Squares Team 71.00 2 ? Error 155.6120? Total 226.6122 Means and SDs 0.023

Explanation / Answer

Solution:

We have to complete ANOVA table.

Total number of observations = 23

Total degrees of freedom = 23 – 1 = 22

Total number of treatments = 3

Degrees of freedom for treatment = 3 – 1 = 2

Error degrees of freedom = total df – treatment df = 22 – 2 = 20

MST = SST / df for treatment

MST = 71/2 = 35.5

MSE = SSE / df for error

MSE = 155.61 / 20 = 7.7805

Test statistic = F-ratio = MST / MSE = 35.5/7.7805 = 4.562689

df1 = 2, df2 = 20

P-value = 0.023312

(By using F-table or excel)

Required ANOVA table is given as below:

Source of variation

Sum of squares

Mean Squares

F - ratio

P-value

Team

35.5

4.562689

0.023312

Error

155.61

7.7805

Total

226.61

At 95% confidence, the sports teams mean ACT scores are different because p-value is given as 0.02 which is less than 0.05 level of significance.