Question

First Generation College Attendance	Men	Women
Firsts	817.14	1118.86
Nonfirsts	1132.86	1551.14

Calculate Chi-squared and degree of freedom. Is there a significant difference at the p<0.05 level? Explain.

Answer 1

SOLUTION:

From given data,

First Generation College Attendance	Men	Women
Firsts	817.14	1118.86
Nonfirsts	1132.86	1551.14

Add up rows and columns:

First Generation College Attendance(O)	Men	Women	Total
Firsts	817.14	1118.86	1936
Nonfirsts	1132.86	1551.14	2684
Total	1950	2670	4620

Calculate "Expected Value" for each entry:

Multiply each row total by each column total and divide by the overall total:

E = (Row total)(Column total) / Grand Total

E	Men	Women
Firsts	1936*1950 / 4620 =817.142857	1936*2670/4620 = 1118.857143
Nonfirsts	2684*1950/4620 = 1132.857143	2684*2670/4620 = 1551.1428571

Subtract expected from actual, square it, then divide by expected:

$\chi ^{2}$ = $\Sigma (O-E)^{2} /E$

	Men	Women
Firsts	(817.14-817.142857)2 / 817.142857=0.0000000099	(1118.86-1118.857143)2/1118.857143 =0.0000000073
Nonfirsts	(1132.86-1132.857143)2/1132.857143 = 0.0000000072	(1551.14-1551.1428571)2/1551.1428571 = 0.0000000053

Now add up those values:

$\chi ^{2}$ =  $\Sigma (O-E)^{2} /E$ = 0.0000000099+ 0.0000000073+0.0000000072+0.0000000053 = 0.0000000297

Chi-Square = $\chi ^{2}$ = 0.0000000297

Calculate Degrees of Freedom

= (rows − 1) $\times$ (columns − 1)

DF = (2 − 1)(2 − 1) = 1×1 = 1

Critical value :

Critical value at  $\alpha$ = 0.05 with   DF = 1

$\chi _{0.05}^{2}$ = 3.841

P- value

P- value at  $\chi ^{2}$ = 0.0000000297 with   DF = 1

P- value = 0.999

Since test statistic is less than critical value, we fail to reject null hypothesis. At 5% level , i.e., there is no significant difference at the p<0.05 level.

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