Question

the following are the average weekly losses of work-hours due to accidents in 10 industrial plants before and after a certain safety program was put into operation: 45 and 36, 73 and 60, 46 and 44, 124 and 119, 33 and 35 57 and 51, 83 and 77, 34 and 29, 26 and 24, 17 and 11 can we say with 95% confidence that the new safety program made a difference in the mean weekly loss of work-hours due to accidents? justify your answer.

Answer 1

To determine if the new safety program made a difference in the mean weekly loss of work-hours due to accidents, we can perform a paired t-test. With 95% confidence, we cannot say that the new safety program made a difference in the mean weekly loss of work-hours due to accidents.

To determine if the new safety program made a difference in the mean weekly loss of work-hours due to accidents, we can perform a paired t-test. The null hypothesis is that the mean difference is equal to zero, and the alternative hypothesis is that the mean difference is not equal to zero. We will use a 95% confidence level. Here are the steps:

1. Calculate the differences between the before and after values: -9, -13, -2, -5, 2, -6, -6, -5, -2, -6.
2. Calculate the mean difference: -5.6.
3. Calculate the standard deviation of the differences: 4.15.
4. Calculate the t-statistic: t = (mean difference - hypothesized mean difference) / (standard deviation of differences / sqrt(sample size)). In this case, the hypothesized mean difference is zero, so the t-statistic is -1.34.
5. Find the critical value for a two-tailed test with a 95% confidence level and 9 degrees of freedom. The critical value is approximately 2.26.
6. Compare the t-statistic to the critical value. Since -1.34 is not greater than 2.26, we fail to reject the null hypothesis.

Therefore, with 95% confidence, we cannot say that the new safety program made a difference in the mean weekly loss of work-hours due to accidents.