Question

These two groups are two samples representing the population of workers in the economy. We want to know if the workers who take the training (treatment sample) have higher earnings than the group that do not take the training (control sample). If we find that the trained workers have higher earnings it would indicate that the training is effective.[1]In terms of statistics, we will do a hypothesis test on the difference between the mean earnings in the treatment population and the control population.

[1]There is a better way of testing the effectiveness of this training program with the available data. We will see how on the next homework.

Evaluating the Effectiveness of a Workforce Training Program Description: The following question is about a hypothesis test for the difference between two population means. It can be used for several purposes including testing experiments based on samples. We will skip the math and focus on the results and interpretation of the test. Suppose we want to test the effectiveness of a job training program on the earnings of workers. We want to design an experiment where we randomly assign a group of workers to take the training and compare it to another group that do not take the training. Call the group of individuals randomly selected to take the training the treatment sample and the group of individuals not selected to take the training the control sample. These two groups are two samples representing the population of workers in the economy. We want to know if the workers who take the training (treatment sample) have higher earnings than the group that do not take the training (control sample). If we find that the trained workers have higher earnings it would indicate that the training is effective.1 In terms of statistics, 1 There is a better way of testing the effectiveness of this training program with the available data. We will see how on the next homework. we will do a hypothesis test on the difference between the mean earnings in the treatment population and the control population. Is the training program effective? Test the following hypothesis at the 10% significance leve treatment_control treatmentcontrol > 0 GI where, utreatment is the mean earnings for the treatment population (workers with the training) and μcontrol is the mean earnings for the control population (workers without the training). We will be working with the mean and standard deviation of variable re78 Read the alternative hypothesis carefully. It implies that the treatment group (workers with the training) have higher earnings than the control group (workers without) Rejection of the null hypothesis thus provides evidence supporting the effectiveness of the training program.2 The formulas for the test statistic and degrees of freedom are different than hypothesis tests for one population mean. We will skip the calculations and focus on the interpretation of the test. The five steps of hypothesis testing are the same. 1. Show the five steps of hypothesis testing using the calculations from the Excel output shown in the table below; where you have the test statistic, critical values, degrees of freedom and p-value. Conclude, is the training program effective? (Note: The Excel output gives the one-sided and two-sided critical values and p-values Choose the appropriate one.) Hatmentycontrol the interpretation of the test. The five steps of hypothesis testing are the same. 1. Show the five steps of hypothesis testing using the calculations from the Excel output shown in the table below; where you have the test statistic, critical values, degrees of freedom and p-value. Conclude, is the training program effective? (Note: The Excel output gives the one-sided and two-sided critical values and p-values. Choose the appropriate one.) treatmentontrol treatmentcontrol a- 0.10 Ho: μ > 0 Mean and Variance in Control and Treatment: Stats Mean Variance Observations treatment 5976 4793895'7 297 control 5090 32696539 425 Stats calculated from Excel Stats Values Hypothesized Mean Difference 2 Again, read the previous footnote. 557 1.82 0.035 1.282 0.07 1.645 Use R to produce the output for the hypothesis test above. Write the R command and copy and paste the output. No need to redo the five steps. Note that the content from df t Stat P(T S t) one-tail t Critical one-tail P(T S t) two-tail t Critical two-tail 2. the R output is the same as above and your conclusions should be the same. z-Test: Hypothesis Test About Population Mean with Population

Answer 1

1) The five steps are as follows :

Step 1 : State the null hypotheses

H₀ : There is no significant difference in the two groups (treatment and control)

H_A : The mean of the treatment > mean of the control group

It is one sided as we are interested in seeing if the performance has improved after training

Step 2 : Decide on the level of significance , given as 10%

The critical value for t (one sided) = 1.282

(However I am not sure about the degree of freedom. It should be n1+ n2-2 = 297+425-2 = 720 It is shown as 557 ??)

Step 3 : Calculate the test statistic (unequal variances assumed)

t = mean (treatment) - mean (control) - 0 (under Ho)/ (S(tr)²/n1+ S(cntr)²/n2)²

where S (squared) is the respective sample variance

Step 4: Compare Calculated t to Critical Value of t

Step 5: As it is more reject the null hypthesis. ie performance has significantly improved

Also the p value is 3.5% for the one sided test. As it is less than 10% reject nulll hypothesis

2) As i dont have the data am pasting the R command

t.test (treatment-control, mu=0, alt="one sided". conf=0.9, Var.eq=T, paired =F)