Question

A random sample of drug addicts in Seattle participated in a program to reduce drug dependency. Time 1 is a measure of the number of illegal drugs they took per day before participating in the program. Time 2 is a measure of the number of illegal drugs they took after participating in the program. You have been hired to evaluate the success of the program. You hypothesize that the average number of illegal drugs consumed by the addicts after participating in the program will decrease compared to the average number of illegal drugs consumed prior to participating in the program. Below are the data.

Interpret your answer using an alpha of .05.

Time 1 (drugs taken before program)	Time 2 (drugs taken after participating in the program)	Difference	D2
2.00	1.00	-1	1
3.00	4.00	1	1
3.00	3.00	0	0
4.00	5.00	1	1
4.00	3.00	-1	1
5.00	3.00	-2	4
Sum = 21	Sum = 19	Sum = -2	Sum = 8
Mean = 3.5	Mean = 3.17	Mean = -.33	Mean = 1.33

	a.	The obtained value does exceed the critical value of 2.015 at the .05 level of significance. The null hypothesis must not be accepted and it can be concluded that there is a significant difference between the two sets of scores and hence the program is successful.
	b.	The obtained value does not exceed the critical value of 2.015 at the .05 level of significance. The null hypothesis must be accepted and it can be concluded that there is no significant difference between the two sets of scores and hence the program is not successful.
	c.	No interpretation is needed since .67 is greater than .05.

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: μ_d = 0

Alternative hypothesis: μ_d ≠ 0

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).

s = sqrt [ (Σ(d_i - d)² / (n - 1) ]

s = 1.211

SE = s / sqrt(n)

S.E = 0.4944

DF = n - 1 = 6 -1

D.F = 5

t = [ (x₁ - x₂) - D ] / SE

t = 0.6742

where d_i is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 5 degrees of freedom is more extreme than 0.6742; that is, less than - 0.6742 or greater than 0.6742.

Thus, the P-value = 0.53

Interpret results. Since the P-value (0.53) is greater than the significance level (0.05), we have to accept the null hypothesis.

Reject H₀. The mean difference appears to differ from zero.

b) The obtained value does not exceed the critical value of 2.015 at the .05 level of significance. The null hypothesis must be accepted and it can be concluded that there is no significant difference between the two sets of scores and hence the program is not successful.