TABLE 12-3
A computer used by a 24-hour banking service is supposed to randomly assign each transaction to one of 5 memory locations. A check at the end of a day's transactions gave the counts shown in the table to each of the 5 memory locations, along with the number of reported errors.
$$\begin{array}{l}
\begin{array} { l l l l l r }
\hline \text { Memory Location: } & 1 & 2 & 3 & 4 & 5 \
\hline \text { Number of Transactions: } & 82 & 100 & 74 & 92&102\
\text { Number of Reported Errors: } & 11 & 12 & 6 & 9 & 10
\end{array}\
\begin{array} { l l l l l l }

\end{array}
\end{array}$$ The bank manager wanted to test whether the proportion of errors in transactions assigned to each of the 5 memory locations differ.
-Referring to Table 12-3, the degrees of freedom of the test statistic is
A) 10.
B) 4.
C) 8.
D) 448.

Question

TABLE 12-3
A computer used by a 24-hour banking service is supposed to randomly assign each transaction to one of 5 memory locations. A check at the end of a day's transactions gave the counts shown in the table to each of the 5 memory locations, along with the number of reported errors.
 $$\begin{array}{l}
\begin{array} { l l l l l r } 
\hline \text { Memory Location: } & 1 & 2 & 3 & 4 & 5 \
\hline \text { Number of Transactions: } & 82 & 100 & 74 & 92&102\
\text { Number of Reported Errors: } & 11 & 12 & 6 & 9 & 10
\end{array}\
\begin{array} { l l l l l l }

\end{array}
\end{array}$$ The bank manager wanted to test whether the proportion of errors in transactions assigned to each of the 5 memory locations differ.
-Referring to Table 12-3, the degrees of freedom of the test statistic is
A) 10.
B) 4.
C) 8.
D) 448.

Quizplus · Accepted Answer

The degrees of freedom for a chi-square test for homogeneity or independence is calculated as (number of categories - 1). Here, there are 5 memory locations, so the degrees of freedom is 5 - 1 = 4.

TABLE 12-3 a Computer Used by a 24-Hour Banking Service

TABLE 12-3
a Computer Used by a 24-Hour Banking Service