Question

Do people think that diversity in a large population is healthy or damaging? At one college, responses of a randomly selected group of students found that 10 students majoring in behavioral sciences thought diversity healthy and 3 thought it damaging. In the natural sciences, 4 thought diversity healthy and 8 thought it damaging. In the arts, 18 thought diversity healthy and 7 thought it damaging. In the languages, 6 thought diversity healthy and 7 thought it damaging. Finally, in the field of history, 4 students thought diversity healthy and 9 thought it damaging. Do these results suggest that opinions about diversity vary over different majors? (Use the .05 significance level.)

Choose the appropriate test and use the five steps of hypothesis testing!   

Answer 1

Step 1:

H0: Null Hypothesis: Opinions about diversity do not vary over different majors

HA: Alternative Hypothesis: Opinions about diversity vary over different majors (Claim)

Step 2:
Observed Frequencies:

	Healthy	Damaging	Total
behavioral sciences	10	3	13
the natural sciences	4	8	12
arts	18	7	25
languages	6	7	13
history	4	9	13
Total	42	34	76

Expected Frequencies:

	Healthy	Damaging	Total
behavioral sciences	42X13/76=7.184	34X13/76=5.816	13
the natural sciences	42X12/76=6.632	34X12/76=5.368	12
arts	42X25/76=13.816	34X25/76=11.184	25
languages	42X13/76=7.184	34X13/76=5.816	13
history	42X13/76=7.184	34X13/76=5.816	13
Total	42	34	76

Test Statistic (Chi Square) is got as follows:

Observed (O)	Expected (E)	(O - E)2/E
10	7.184	1.104
3	5.816	1.36
4	6.632	1.044
8	5.368	1.29
18	13.816	1.267
7	11.184	1.565
6	7.184	0.195
7	5.816	0.241
4	7.184	1.411
9	5.816	1.743
Total = Test Statistic (Chi Square) =		11.225

So,

Test Statistic (Chi Square) = 11.225

Step 3:

df = (5 - 1) X (2 - 1) = 4

alpha = 0.05

From Table,

critical value of chi square = 9.88

Step 4:
Since calculated value of chi square = 11.225 is greater than critical value of chi square = 9.88, the difference is significant. Reject null hypothesis.

Conclusion:
The data support th claim that opinions about diversity vary over different majors.

Step 5:

By Technology,

p - value = 0.0242

Since p - value = 0.0242 is less than alpha = 0.05, the difference is significant. Reject null hypothesis.

Conclusion:
The data support th claim that opinions about diversity vary over different majors.