Question

Researchers studying children’s moral behavior have developed an altruism scale for 10-year-olds that reflects a child’s general level of helpfulness.  Scores on the altruism scale are normally distributed with mu = 95 and sigma= 14 (higher scores on the scale indicate a greater level of helpfulness).

Dr. Thornton wants to investigate whether exposing children to a videotape that shows a lost puppy in need of help will affect how helpful they are.  He selects a random sample of n = 36 10-year-olds, shows them the video, and then administers the altruism scale.  Dr. Thornton finds that the mean score for the children is M = 99.1.  

Test whether Dr. Thornton’s manipulation significantly increases altruism scores.

a. State the null and alternative hypotheses in sentence format (do not use symbol notation).

b. State the null and alternative hypotheses using statistical notation.

c. Identify the critical region(s) for an alpha = .05 level of significance for this sampling distribution.

d. Did Dr. Thornton’s manipulation significantly increase altruism scores?  Explain your answer.

Suppose that Dr. Thornton instead wants to test whether his manipulation has any effect on altruism scores.  Assume that he uses the same 36 children as noted above.  Restate the null and alternative hypotheses using symbol notation for this new test.

e. Identify the critical region(s) for an alpha = .05 level of significance for this sampling distribution.

f. Did Dr. Thornton find a significant effect of his manipulation?  Explain why the answer here is different than that for (d), above.

Answer 1

x̅ = 99.1, σ = 14, n = 36

a) Null and alternative hypothesis:

Ho: The mean altruism score is 95.

H1: The mean altruism score more than 95.

b) Null and Alternative hypothesis:

Ho : µ = 95

H1 : µ > 95

c) Critical value :

Right tailed critical value, z crit = ABS(NORM.S.INV(0.05) = 1.645

Reject Ho if z > 1.645

d) Test statistic:

z = (x̅- µ)/(σ/√n) = (99.1 - 95)/(14/√36) = 1.7571

Decision:

Reject the null hypothesis

Conclusion:

There is enough evidence to conclude that Dr. Thornton’s manipulation significantly increase altruism scores.

------------

Null and Alternative hypothesis:

Ho : µ = 95

H1 : µ ≠ 95

Critical value :

Two tailed critical value, z crit = ABS(NORM.S.INV(0.05/2)) = 1.960

Reject Ho if z < -1.96 or if z > 1.96

Test statistic:

z = (x̅- µ)/(σ/√n) = (99.1 - 95)/(14/√36) = 1.7571

Decision:

Do not reject the null hypothesis

Conclusion:

There is not enough evidence to conclude that Dr. Thornton has a significant effect of his manipulation.