Question

Twelve sets of identical twins were enrolled in a study to measure the effect of home environment on certain social attitudes. One twin in each set was randomly assigned to a foster environment or a home environment. The twin assigned to the foster environment went to live with a low income family on state welfare for a period of 1 year. At the end of the year, an attitude survey was administered.

Their results follow.

ID	1	2	3	4	5	6	7	8	9	10	11	12
Foster	83	75	72	76	88	80	72	85	70	78	77	71
Home	65	67	75	77	69	65	73	78	70	72	73	79

Perform a dependent samples t-test (paired t-test) to examine if the hypothesis that living in the foster environment leads to higher scores on the attitudinal survey. Use a 5% significance level.

(a)	State the null hypothesis and the alternate hypothesis.
	AnswerH0:μd ≤ 0 ; H1: μd > 0H0:μd = 0 ; H1: μd ≠ 0H0:μd ≥ 0 ; H1: μd < 0H0:μd >0 ; H1: μd = 0

(b)	Compute the value of the test statistic.    (state your answer accurate to 2 decimal places)
	Test Statistic t = Answer

(c)	State the Degrees of Freedom?
	df = Answer

(d)	Look up the critical value tc   (state to 3dp as given in your t-tables)
	tc = Answer

Answer 1

a)

Answer:

Explanation:

The null hypothesis is defined as there is no difference in the attitude score between the identical twins such that the mean difference of score is zero and the alternative hypothesis tests the claim that the mean difference of score is greater than zero as shown below,

b)

Answer:

Test Statistic t = 2.15

Explanation:

The t-statistic is obtained using the formula,

where the difference, D = Foster - Home.

The sample mean and the sample standard deviation values are obtained in excel using the function =AVERAGE() and =STDEV(). The screenshot is shown below,

Now,

c)

Answer:

df = 11

Explanation:

Degrees of Freedom = n - 1 = 12 - 1 = 11

d)

Answer:

Explanation:

The t critical value is obtained from the t critical value table for the significance level = 0.05, the df = 11 and for one-tailed distribution,