Question

A researcher decides to analyze the effects of nutrition on personality. He collects 7 pairs of identical twins and randomly assigns one twin from each pair to a controlled diet condition. The twins assigned to the other condition are allowed to eat whatever they please. The following are scores on a standardized personality inventory:

Personality inventory scores for twins in controlled diet condition:
10, 9, 4, 3, 8, 6, 7

Personality inventory scores for twins in eat-what-you-want condition:
16, 11, 9, 4, 5, 9, 12

NOTE: The order of the scores is important here. The first score in the controlled diet condition (10) should be paired with the first score from the other condition (16), etc.

Hint: This is a two-tailed test.

1. Null hypothesis:
2. Alternative hypothesis:
3. Statistical test (be specific!):
4. Significance level: alpha = .01
5. degrees of freedom:
6. Critical region (t-value):
7. Calculated t:
8. Decision:

Answer 1

This will an independent 2 samples t-test since both data sets have different individual. Test will be conducted for difference in population means.

Let twin 1 be the controlled diet and twin 2 be eat-what-you-want

be population mean personalities of twin 1 and twin 2 respectively

twin1	twin2
10	16
9	11
4	9
3	4
8	5
6	9
7	12

Statistical summary

  

Pooled variance =

=

Null hypothesis: (Personalities of both twins are same and the difference is '0')

Alternative hypothesis: (Personalities of both twins are different and the difference is not '0' )

Statistical test (be specific!): 2 sided t-test with equal variance for difference between population means

Significance level: alpha = .01

degrees of freedom:   = 12 (14-2)\

Critical region (t-value): Since this a 2 sided test Critical value = = =3.055 (found using t-tables)

Calculated t:      = -1.481

Decision: Since |T.S.| < C.V.

  We have insufficient evidence to reject the null hypothesis.

We can conclude that 0.01 level of significance the personalities of both twins is same under both dietary conditions.