Question

In 1976, the Department of Labor conducted a randomized controlled experiment to study the effect of an income supplement on prisoners released from certain prisons in Texas and Georgia. This problem is modelled on that experiment, but with simplified data. The control group received no supplement, and the intervention group received an income equivalent to unemployment insurance. The rates of return to prison for subsequent crimes were similar in the two groups, and the investigators suspected that income supplement might have reduced the amount of time prisoners worked after release. In the first year after release, 62 participants assigned to the treatment group averaged 15 hours of paid work per week, with standard deviation 6.0. The 31 participants assigned to the control group worked on average 16.8 hours per week, with standard deviation 3.0. Did income support reduce the amount that ex-prisoners worked? Please conduct a hypothesis test procedure to answer the question

Answer 1

For Treatment Group:

$\small \bar X_1=15$, $\small s_1=6.0$ , $\small n_1=62$

For Control Group:

$\small \bar X_2=16.8$, $\small s_2=3.0$ , $\small n_2=31$

(1) Null and Alternative Hypotheses

$\small Ho: \mu_1= \mu_2$

$\small Ha: \mu_1<\mu_2$

2)Test Statistics

$\small t = \frac{\bar X_1 - \bar X_2}{\sqrt{ \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}(\frac{1}{n_1}+\frac{1}{n_2}) } }$

$\small = \frac{ 15 - 16.8}{\sqrt{ \frac{(62-1)6^2 + (31-1)3^2}{ 62+31-2}(\frac{1}{ 62}+\frac{1}{ 31}) } } = -1.572$

3) P-value

The p-value is p = 0.0597

4) Decision about the null hypothesis

As significance level is not know we assume it to be a, so $\small \alpha$ = a.

If p = 0.0597 > a , then the null hypothesis is not rejected.

But if p = 0.0597 < a , then the null hypothesis is rejected.