Question

The following are the average weekly losses of worker-hours due to accidents in 10 industrial plants before and after a certain safety program was put into operation:
BEFORE
Plant 1: 45
Plant 2: 73
Plant 3: 46
Plant 4: 124
Plant 5: 33
Plant 6: 57
Plant 7: 83
Plant 8: 34
Plant 9: 26
Plant 10: 17
AFTER
Plant 1: 36
Plant 2: 60
Plant 3: 44
Plant 4: 119
Plant 5: 35
Plant 6: 51
Plant 7: 77
Plant 8: 29
Plant 9: 24
Plant 10: 11
Conduct a paired t test to analyze this problem. Use the 0.05 level of significance to test whether the safety program is effective.
The t statistic for this test is
The number of degrees of freedom for this test is
The null hypothesis must be rejected at level alpha = 0.05. We conclude that the industrial safety program is effective. The evidence is strong since a computer calculation gives the P value .9985. Enter 0 if this statement is FALSE or 1 otherwise

Answer 1

Answer:

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-5.2 -0}{\frac{4.077}{\sqrt{10}}}=-4.0332$  

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

$df=n-1=10-1=9$  

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

$p_v =2*P(t_{(9)}$  

The p value is lower than the significance level given $\alpha=0.05$, so then we can conclude that we can reject the null hypothesis that no difference between the two measures. So there is a significant difference between the measurements before and after

The null hypothesis must be rejected at level alpha = 0.05. We conclude that the industrial safety program is effective. The evidence is strong since a computer calculation gives the P value .9985. Enter 0 if this statement is FALSE or 1 otherwise

This STATEMENT is FALSE (1) since the p value is 0.0030 and not 0.9985

Step-by-step explanation:

Previous concepts

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.  

Solution to the problem

Let put some notation  

x=test value before , y = test value after  

x: 45,73,46,124,33,57,83,34,26,17  

y: 36,60,44,119,35,51,77,29,24,11

The system of hypothesis for this case are:  

Null hypothesis: $\mu_y- \mu_x = 0$  

Alternative hypothesis: $\mu_y -\mu_x \neq 0$  

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

d: -9,-13,-2, -5, 2, -6, -6, -5, -2, -6  

The second step is calculate the mean difference  

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{-52}{10}=-5.2$  

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

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =4.077$  

The 4 step is calculate the statistic given by :  

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-5.2 -0}{\frac{4.077}{\sqrt{10}}}=-4.0332$  

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

$df=n-1=10-1=9$  

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

$p_v =2*P(t_{(9)}$  

The p value is lower than the significance level given $\alpha=0.05$, so then we can conclude that we can reject the null hypothesis that no difference between the two measures. So there is a significant difference between the measurements before and after

The null hypothesis must be rejected at level alpha = 0.05. We conclude that the industrial safety program is effective. The evidence is strong since a computer calculation gives the P value .9985. Enter 0 if this statement is FALSE or 1 otherwise

This STATEMENT is FALSE (1) since the p value is 0.0030 and not 0.9985