Question

Question 3

A brain specialist performed an experiment to see if sensory deprivation over an extended period of time reduces the alpha-wave frequencies produced by the brain. To determine this, 10 subjects, inmates of a Correctional Centre, were randomly selected.

First, the members of the group were allowed to remain in their cells. Seven days later, alpha-wave frequencies were measured in Hertz for all subjects (nonconfined).

After that, the same group of subjects were placed in solitary confinement for seven days and their alpha-wave frequencies were measured again in Hertz.

The data from both confined and nonconfined groups of the experiment are recorded in the following table.

Subject #	Confined	Nonconfined
1	7	8
2	10	9
3	9	10
4	9	10
5	10	11
6	9	11
7	11	11
8	12	11
9	11	12
10	12	14

Use a chi-square test to investigate if the solitary confinement reduces the alpha-wave frequencies of the brain. In performing this test, state appropriate hypothesis (define any symbols used). Calculate the value of the test statistic, P-value of this test and write a meaningful conclusion.

(While a paired t-test is what would usually be used for this question, we are specifically asking you to perform a chi-square test).

Answer 1

ANSWER:

Confined	Confined (Expected)	squared distance	Non confined	Non confined (Expected)	squared distance	Total
7	100 * 15 / 207 = 7.24	(7.24 - 7)^2 / 7.24 = 0.007	8	107 * 15 / 207 = 7.76	(7.76 - 8)^2 / 7.76 = 0.005	15
10	100 * 19 / 207 = 9.17	(9.17 - 10)^2 / 9.17 = 0.075	9	107 * 19 / 207 = 9.83	(9.83 - 9)^2 / 9.83 = 0.07	19
9	100 * 19 / 207 = 9.17	(9.17 - 9)^2 / 9.17 = 0.003	10	107 * 19 / 207 = 9.83	(9.83 - 10)^2 / 9.83 = 0.002	19
9	100 * 19 / 207 = 9.17	(9.17 - 9)^2 / 9.17 = 0.003	10	107 * 19 / 207 = 9.83	(9.83 - 10)^2 / 9.83 = 0.002	19
10	100 * 21 / 207 = 10.14	(10.14 - 10)^2 / 10.14 = 0.001	11	107 * 21 / 207 = 10.86	(10.86 - 11)^2 / 10.86 = 0.002	21
9	100 * 20 / 207 = 9.66	(9.66 - 9)^2 / 9.66 = 0.045	11	107 * 20 / 207 = 10.34	(10.34 - 11)^2 / 10.34 = 0.015	20
11	100 * 22 / 207 = 10.62	(10.62 - 11)^2 / 10.62 = 0.013	11	107 * 22 / 207 = 11.38	(11.38 - 11)^2 / 11.38 = 0.015	22
12	100 * 23 / 207 = 11.11	(11.11 - 12)^2 / 11.11 = 0.071	11	107 * 23 / 207 = 11.89	(11.89 - 11)^2 / 11.89 = 0.07	23
11	100 * 23 / 207 = 11.11	(11.11 - 11)^2 / 11.11 = 0.001	12	107 * 23 / 207 = 11.89	(11.89 - 12)^2 / 11.89 = 0.001	23
12	100 * 26 / 207 = 12.56	(12.56 - 12)^2 / 12.56 = 0.024	14	107 * 26 / 207 = 13.44	(13.44 - 14)^2 / 13.44 = 0.02	26
Total =100			Total = 107			Total = 207

Chi square test statistic = sum of squared distances

= 0.5 approximately which is way less than critical chi square value needed to reject the null hypothesis

Hence, using chi square test, we do not reject the null hypothesis.

The difference between the two test results is because of different methodology used in testing. Whereas, the t test uses mean of the groups to reach a conclusion, the chi square test identifies relationship between two variables using deviation from the expected value.