Question

In: Statistics and Probability

Iowa has decided to run a quasi-experiment in regard to a new program to reduce unemployment....

Iowa has decided to run a quasi-experiment in regard to a new program to reduce unemployment. Officials think that the new program will increase employment rates and earnings. Two hundred recipients are selected; 144 are randomly assigned to an unemployment program, and 81 are assigned to a control group. By follow-up interviews, the state finds out how much outside income per week is earned on average, by each individual with the following results:

Unemployment Program Control Group
Mean $140 $100
Std Dev 132 108
n 144 81


a. Present a hypothesis and a null hypothesis.  
b. Calculate your t-score (involving the overall standard error for the difference of means);
c. estimate your p-value (using n1+n2-2 for d.f.);
d. State a conclusion in plain English.

Solutions

Expert Solution

Solution:

a) Null hypothesis:

H0 : 1 = 2

That is, the new program will be same employment rates and earnings as control

Alternative hypothesis:

Ha : 1> 2

That is, the new program will increase employment rates and earnings.

b) EXCEL ( Mega stat) out put:

The value of standardized test statistic is t = 2.457

c) P- value = 0.0074

d) Since p - value(=0.0074) < (=0.05), then reject the null hypothesis.

Therefore , there is enough evidence to support the claim that the new program will increase employment rates and earnings.

please up vote the solution if it helped ...Thanks..


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