Question

A new sales training program has been instituted at a rent-to-own company. Prior to the training, 10 employees were tested on their knowledge of products offered by the company. Once the training was completed, the employees were tested again in an effort to determine whether the training program was effective. Scores are known to be normally distributed. The sample scores on the tests are listed next. Use pretest score as µ1 for population 1 and posttest score as µ2 for population 2, or µD as the mean of the difference calculated as pretest score minus posttest score.Col1 Pretest Score 66 94 87 84 76 88 Col2 Posttest Score 75 100 93 85 75 90Which of the following are the appropriate hypotheses to determine if the training increases scores?
A. H0: µ1 – µ2? 0, HA: µ1 – µ2< 0
B. H0: µ1 – µ2? 0, HA: µ1 – µ2> 0
C. H0: µD ? 0, HA: µD <0
D. H0: µD ? 0, HA: µD >0

Answer 1

Answer:

C. H0: µD = 0, HA: µD <0

Step-by-step explanation:

We assume that the paired sample data are simple random samples and that the differences have a distribution that is approximately normal. So for this case is better apply a paired t test.  

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.  

Let put some notation  

x=test value for pretest , y = test value for posttest  

x: 66, 94, 87, 84, 76, 88

y: 75, 100, 93, 85, 75, 90  

The system of hypothesis for this case are:  

Null hypothesis: $\mu_D \geq 0$  

Alternative hypothesis: $\mu_D < 0$  

Because the difference D is defined as Pretest-Postest. And we want to see if the postets score is higher than the pretest with th training.

The first step is calculate the difference $d_i=x_i-y_i$ and we obtain this:  

d: -9, -6, -6, -1, 1, -2

The second step is calculate the mean difference  

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{-23}{6}=-3.833$  

The third step would be calculate the standard deviation for the differences, and we got:  

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =3.763$  

The 4 step is calculate the statistic given by :  

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-3.833 -0}{\frac{3.763}{\sqrt{6}}}=-2.494$  

The next step is calculate the degrees of freedom given by:  

$df=n-1=6-1=5$  

Now we can calculate the p value, since we have a left tailed test the p value is given by:  

$p_v =P(t_{(5)}$  

The p value is lower than the significance level assumed $\alpha=0.05$, so then we can conclude that we can reject the null hypothesis. So we can say that the differences between Pretest and Posttest $\mu_{pretest}-\mu_{postest}$ are less than 0.