91Ó°ÊÓ

The data for theshifts with and without death and whether Gilbert was working is provided.

The level of significance is 0.01.

Compute theexpected frequencies using the formula stated below,

\(E = \frac{{\left( {row\;total} \right)\left( {column\;total} \right)}}{{\left( {grand\;total} \right)}}\)

The counts for total rows and columns are,

Theexpected frequency tableis represented as,

The hypotheses are formulated as,

\({H_0}:\)The deaths on shifts are independent of whether Gilbert was working.

\({H_1}:\)The deaths on shifts are dependent on whether Gilbert was working.

The value of the test statisticis computed as,

\(\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \\ = \frac{{{{\left( {40 - 11.5893} \right)}^2}}}{{11.5893}} + \frac{{{{\left( {217 - 245.4107} \right)}^2}}}{{245.4107}} + ... + \frac{{{{\left( {1350 - 1321.5893} \right)}^2}}}{{1321.5893}}\\ = 86.4809\\ \approx 86.481\end{aligned}\)

Therefore, the value of the test statistic is 86.481.

The degrees of freedomare computed as,

\(\begin{aligned}{c}\left( {r - 1} \right)\left( {c - 1} \right) = \left( {2 - 1} \right)\left( {2 - 1} \right)\\ = 1\end{aligned}\)

Therefore, the degrees of freedom are 1.

The critical value for 1 degrees of freedom and at 0.01 level of significance is 6.635.

Therefore, the critical value is 6.635.

The P-value is computed as 0.000.

The critical (6.635) is less than the value of the test statistic (86.481). In this case, the null hypothesis is rejected.

Therefore, the decision is to reject the null hypothesis.

There isinsufficient evidence to support the claimthat deaths on shifts are independent of whether Gilbert was working or not.

The results suggest that Gilbert cannot be considered as innocence for the deaths on the basis of results.