Question

1. An experimenter examining decision-making asked 10 children to solve as many problems as they could in 10 minutes. One group (5 subjects) was told that this was a test of their innate problem-solving ability; a second group (5 subjects) was told that this was just a time filling task. The data for each group is presented below. Innate ability: 4,5,8, 3,7 Time-filling task: 11, 6,9,7,9 Does the mean number of problems solved vary with the experimental condition? i.e., is there a difference in the means for the two groups? When addressing this question, use a significance level of.05. (a) State the null and alternative hypotheses that are implied by the question that is posed. (b) Assume that, in the population, it is reasonable to believe that number of problems solved is normally distributed. You can also assume that the variances are equal. Given these assumptions, compute the appropriate test statistic to test the null hypothesis from part A.

Answer 1

(a)
Null hypothesis H0: Mean number of problems solved by students of Innate ability group is equal to that of Time-filling task group.
Alternative hypothesis H0: Mean number of problems solved by students of Innate ability group is not equal to that of Time-filling task group.

(b)

Mean number of problems solved by students of Innate ability group, m1 = (4 + 5 + 8 + 3 + 7) / 5 = 5.4

Mean number of problems solved by students of Time-filling task group, m2 = (11 + 6 + 9 + 7 + 9) / 5 = 8.4

Variance of problems solved by students of Innate ability group, s1² = [(4 - 5.4)² + (5 - 5.4)² + (8 - 5.4)² + (3 - 5.4)² + (7 - 5.4)² ] / 4= 4.3

Variance of problems solved by students of Time-filling task group, s2² = [(11 - 8.4)² + (6 - 8.4)² + (9 - 8.4)² + (7 - 8.4)² + (9 - 8.4)² ] / 4 = 3.8

As, we are testing for the difference of Means from Independent Samples and it is assumed that the variances of the two groups are equal, we will calculate pooled variance to determine the standard error of the difference in means.

Pooled variance, sp² = [(n1-1)s1² + (n2 - 1)s2² ] / (n1 + n2 - 2)

= [(5 - 1) * 4.3 + (5-1) * 3.8] / (5 + 5 -2)

=  4.05

Standard error of the difference in means = $\sqrt{s_p^2 ((1/n_1)+(1/n_2))} = \sqrt{4.05 * ((1/5)+(1/5))} = 1.272792$

Test statistic, t = (Difference in means) / Standard error = (8.4 - 5.4) / 1.272792 = 2.357

(c)

Degree of freedom = n1 + n2 - 2 = 5 + 5 - 2 = 8

Critical value of t at significance level of 0.05 and df = 8 is, 2.306

Decision rule - We reject the null hypothesis H0, is the test statistic is greater than 2.306.

(d)

As, the observed test statistic (2.357) is greater than the critical value, we reject null hypothesis H0.

(e)

As, we reject the null hypothesis, there is significant evidence that mean number of problems solved by students of Innate ability group is not equal to that of Time-filling task group.