Question

Answer 1

The data of the problem is provided as follows:

$$ \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text { Production-line workers }\left(X_{1}\right) & 4 & 0 & 6 & 8 & 3 & 11 & 13 & 5 \\ \hline \text { Office workers }\left(X_{2}\right) & 9 & 2 & 7 & 1 & 4 & 7 & 9 & 8 \\ \hline \end{array} $$

Here, equal variance of t-test is used. Firstly, set the hypothesis as shown below:

State the null hypothesis:

\(H_{0}\) : There is no difference in days absent between the two groups of workers.

Against the alternative hypothesis:

\(H_{1}:\) There is significant difference in days absent between the two groups of workers.

The test statistic used is:

\(t=\frac{\left(\bar{x}_{1}-\bar{x}_{2}\right)-\left(\mu_{1}-\mu_{2}\right)}{\sqrt{s_{p}^{2}\left(\frac{1}{n_{1}}+\frac{1}{n_{2}}\right)}}\)

where \(s_{p}^{2}=\frac{\left(n_{1}-1\right) s_{1}^{2}+\left(n_{2}-1\right) s_{2}^{2}}{n_{1}+n_{2}-2}\)

and \(n_{1}\left(\right.\) or \(\left.n_{2}\right)=8\)

The mean of samples is:

\(\begin{aligned} \bar{X}_{1} &=\frac{\sum_{i=1}^{8} X_{i}}{n_{1}} \\ &=\frac{4+0+6 \ldots .+5}{8} \\ &=6.25 \end{aligned}\)

and

\(\bar{X}_{2}=\frac{\sum_{i=1}^{8} X_{i}}{n_{2}}\)

\(=\frac{9+2+7+\ldots+8}{8}\)

\(=5.88\)

The sample variances are:

\(\begin{aligned} s_{1}^{2} &=\frac{\sum_{i=1}^{8}\left(X_{i}-\bar{X}_{1}\right)}{n_{1}-1} \\ &=\frac{(4-6.25)^{2}+(0-6.25)^{2}+\ldots+(5-6.25)^{2}}{8-1} \\ &=18.21 \end{aligned}\)

and

\(s_{2}^{2}=\frac{\sum_{i=1}^{8}\left(X_{i}-\bar{X}_{2}\right)}{n_{2}-1}\)

\(=\frac{(4-6.25)^{2}+(0-6.25)^{2}+\ldots+(5-6.25)^{2}}{8-1}\)

\(=9.84\)

The pooled variance estimator is:

\(s_{p}^{2}=\frac{\left(n_{1}-1\right) s_{1}^{2}+\left(n_{2}-1\right) s_{2}^{2}}{n_{1}+n_{2}-2}\)

\(=\frac{(8-1)(18.21)+(8-1)(9.84)}{8+8-2}\)

\(=14.03\)

The number of degrees of test statistic is:

\(\begin{aligned} v &=n_{1}+n_{2}-2 \\ &=8+8-2 \\ &=14 \end{aligned}\)

Now, substitute the values in the test statistic as shown below :

\(t=\frac{\left(\bar{x}_{1}-\bar{x}_{2}\right)-\left(\mu_{1}-\mu_{2}\right)}{\sqrt{s_{p}^{2}\left(\frac{1}{n_{1}}+\frac{1}{n_{2}}\right)}}\)

\(=\frac{(6.25-5.88)-(0)}{\sqrt{(14.03)\left(\frac{1}{8}+\frac{1}{8}\right)}}\)

\(=0.20025\)

The rejection region is:

\(t_{a, v}=t_{0.10,14}\)

\(=1.761\)

From the above, it concludes that the calculated value of \(t\) is less than the tabulated value of \(t\), so null hypothesis is accepted. Therefore, it infers that there is no difference in days absent between the two groups of workers.