Does It Pay to Plead Guilty? The accompanying table summarizes randomly selected sample data for San Francisco defendants in burglary cases (based on data from “Does It Pay to Plead Guilty? Differential Sentencing and the Functioning of the Criminal Courts,” by Brereton and Casper, Law and Society Review, Vol. 16, No. 1). All of the subjects had prior prison sentences. Use a 0.05 significance level to test the claim that the sentence (sent to prison or not sent to prison) is independent of the plea. If you were an attorney defending a guilty defendant, would these results suggest that you should encourage a guilty plea?

	Guilty Plea	Not Guilty Plea
Sent to Prison	392	58
Not Sent to Prison	564	14

Data are given on the number of subjects who got sentenced and who did not, depending on whether they pleaded guilty or not.

The chi-square test for independence of attributes is conducted to test the independence between the row variable (sentence) and the column variable (plea).

The null hypothesis is as follows:

\[{H_o}:\]The prison sentence is independent of the plea.

The alternative hypothesis is as follows:

\[{H_1}:\]The prison sentenceis not independent of the plea.

It is a right-tailed test

Observed Frequency:

The frequencies provided in the table are the observed frequencies. They are denoted by\(O\).

The following contingency table shows the observed frequency for each cell:

Expected Frequency:

The formula for computing the expected frequency for each cell is shown below:

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

The row total for the first row is computed below:

\(\begin{array}{c}Row\;Tota{l_1} = 392 + 58\\ = 450\end{array}\)

The row total for the second row is computed below:

\(\begin{array}{c}Row\;Tota{l_2} = 564 + 14\\ = 578\end{array}\)

The column total for the first column is computed below:

\(\begin{array}{c}Column\;Tota{l_1} = 392 + 564\\ = 956\end{array}\)

The column total for the secondcolumn is computed below:

\(\begin{array}{c}Column\;Tota{l_2} = 58 + 14\\ = 72\end{array}\)

The grand total is computed as follows:

\(\begin{array}{c}Grand\;Total = \left( {450 + 578} \right)\\ = \left( {956 + 72} \right)\\ = 1028\end{array}\)

Thus, the following table shows the expected frequencies for each of the corresponding observed frequencies:

The test statistic is computed below:

\[\begin{array}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}\;\;\; \sim } {\chi ^2}_{\left( {r - 1} \right)\left( {c - 1} \right)}\\ = \frac{{{{\left( {392 - 418.482} \right)}^2}}}{{418.482}} + \frac{{{{\left( {58 - 31.518} \right)}^2}}}{{31.518}} + \frac{{{{\left( {564 - 537.518} \right)}^2}}}{{537.518}} + \frac{{{{\left( {14 - 40.482} \right)}^2}}}{{40.482}}\\ = 42.557\end{array}\]

Thus,\({\chi ^2} = 42.557\).

Let r denote the number of rows in the contingency table.

Let cdenote the number of columns in the contingency table.

The degrees of freedom are computed below:

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

The critical value of\({\chi ^2}\)for 1 degree of freedom at 0.05 level of significance for a right-tailed test is equal to 3.8415.

The corresponding p-value is approximately equal to 0.000.

Since the value of the test statistic is greater than the critical value and the p-value is less than 0.05,the null hypothesis is rejected.

There is enough evidence to conclude that the prison sentence and the plea are not independent of each other.

Since the number of guilty defendants who are not sent to prison if they plead guilty than those who do not is substantially larger, the results encourage the guilty defendants to plead guilty.