In: Statistics and Probability
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.
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.
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