Question

The table below shows the number of students from three different high schools participating in the mathematics and physics sections of a science fair: Number of Students High School 1 High School 2 High School 3 Mathematics 8 9 12 Physics 28 20 19 Using a .05 level of significance, test the claim that the section of participation and the high school where the students were from are independent. Group of answer choices There is evidence to reject the claim that the high school and the section of participation are independent because the test value 5.991 > 0.821 There is evidence to reject the claim that the high school and the section of participation are independent because the test value 12.592 > 0.821 There is not evidence to reject the claim that the high school and the section of participation are independent because the test value 0.821 < 5.991 There is not evidence to reject the claim that the high school and the section of participation are independent because the test value 2.161 < 5.991

Answer 1

The expected values for each of the 6 cells above are computed as:
E_i = (Sum of row i)*(Sum of column i) / Grand Total

Also the chi square test statistic contribution for i^th cell is computed here as:

$\chi^2_i = \frac{(O_i - E_i)^2}{E_i}$

The computations therefore here are obtained as:

Mathematics Physics High School-1 8 (10.88) [0.76] 28 (25.12) [0.33] High School-2 9 (8.76) [0.01] 20 (20.24) [0.00] High School-3 12 (9.36) [0.74] 19 (21.64) [0.32]

The values in the circular bracket are expected frequencies here, while the ones in the square bracket are chi square test statistic contributions here.

Therefore the chi square test statistic here is computed as:

$\chi^2 = \sum \chi^2_i = \sum \frac{(O_i - E_i)^2}{E_i} = 2.1611$

The degrees of freedom here is computed as:
Df = (num of column - 1)(num of rows - 1) = (3 - 1)(2 -1) = 2

For 2 degrees of freedom, we have from chi square distribution tables here:

$P(\chi^2_2 > 5.991) = 0.05$

As the test statistic here is less than the critical value here, therefore the test is not significant here and we cannot reject the null hypothesis here. There is not evidence to reject the claim that the high school and the section of participation are independent because the test value 2.161 < 5.991