Question

QUESTION 10 A researcher is interested in knowing if caffeine consumption is related to concentration. She randomly assigns 200 students to two different groups: one group was given a highly caffeinated drink, and the other group was given placebo. Within an hour, she administers a test to both groups. The results of the test are categorized as Pass/Fail and are illustrated below. Pass Fail Caffeinated 83 17 Placebo 61 39 The calculated value of the appropriate test statistic for the above example is: 0 10 o 12 O 21 0 31

Answer 1

We can use Chi-square test for Association

Observed Frequencies:

Fail Total 17 100 Pass Caffeinated Placebo Total 83 61 100 39 56.0 144 200

Now We have to find out Expected Frequencies:

Ei = (Corresponding row total * Corresponding column total) / sample size

For First cell:

E1 = (100 * 144) / 200 = 72

E2 = (100 * 56) / 200 = 28

E3 = (100 * 144) / 200 = 72

E4 = (100 * 56) / 200 = 28

Therefore,

Expected Frequencies:

Pass Total Fail 72 72 28 Caffeinated Placebo Total 100 100 28 56.0 144 200

Test statistic:

$\chi ^{2} = \sum \frac{ (Oi-Ei)^{2}}{Ei}$

Where, Oi = Observed Frequencies

Ei = Expected Frequencies

Using the above Observed and Expected Frequency table, we can find the value of Chi-square.

The calculation of chi-square is given in the following table:

Computational table:

oi Ei 83 17 (Oi-E) 72 28 72 28 (Oi-Ei)^2 (Oi-Ei)^2/Ei 11 121 1.68 -11 121 4.32 -11 121 1.68 11 121 4.32 61 39 Total 200 200 0 484 12.00

Test statistic:

$\chi ^{2} = \sum \frac{(Oi-Ei)^{2}}{Ei} = 12$

ANSWER: B

B. 12