Question

. To determine the effectiveness of a new safety control system in an industrial plant the number of accidents were measured for two weeks before and two weeks after its installation. The following data were obtained: 3 and 1 5 and 2 2 and 0 3 and 2 3 and 2 3 and 0 0 and 2 4 and 3 1 and 3 6 and 4 4 and 1 1 and 0 Use the paired sample sign test to test, at level 0.05, whether or not the new system is effective.

Answer 1

$H_0 : μ=0 \\ H_1 : μ≠0$

Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).

$s= \sqrt{\frac{\sum (d - \bar{d})^2}{n-1}} \\ = \sqrt{\frac{32.25}{12-1}} \\ = 1.7122 \\ SE = \frac{s}{\sqrt{n}} \\ 0.4943 \\ df = n -1 = 11 \\ t = \frac{(x_1-x_2)-D}{SE} \\ = \frac{125-0}{0.4943} \\ = 2.53$

where d_i is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 11 degrees of freedom is more extreme than 2.53; that is, less than - 2.53 or greater than 2.53.

P-value = P(t < - 2.53) + P(t > 2.53)

Use the excel formula , "=T.DIST.2T(2.53,11)" for finding p-values.

P-value = 0.0280

Interpret results. Since the P-value (0.0280) is less than the significance level (0.05), we have to reject the null hypothesis.

Reject H₀. Hence the new system is effective.