For the given table of observed frequencies, perform a test for independence, using the $\alpha = 0.01$ level of significance.
$\begin{array} { c c c c }
\hline & 1 & 2 & 3 \
\hline \mathrm { A } & 69 & 64 & 67 \
\mathrm {~B} & 91 & 98 & 88 \
\mathrm { C } & 42 & 84 & 86 \
\hline
\end{array}$
A) Do not reject $H _ { 0 }$
B) Reject $H _ { 0 }$

Question

For the given table of observed frequencies, perform a test for independence, using the $\alpha = 0.01$ level of significance.
$\begin{array} { c c c c } 
\hline & 1 & 2 & 3 \
\hline \mathrm { A } & 69 & 64 & 67 \
\mathrm {~B} & 91 & 98 & 88 \
\mathrm { C } & 42 & 84 & 86 \
\hline
\end{array}$
A) Do not reject $H _ { 0 }$
B) Reject $H _ { 0 }$

Quizplus · Accepted Answer

To test for independence, we use the chi-square test. The null hypothesis $H_0$ states that the row and column variables are independent. After calculating the expected frequencies and the chi-square statistic, if the calculated chi-square value exceeds the critical value from the chi-square distribution table for the given degrees of freedom (df) and significance level ($\alpha = 0.01$), we reject $H_0$. Given the table has 3 rows and 3 columns, the degrees of freedom are $(3-1) \times (3-1) = 4$. Without the exact chi-square value calculated here, the instruction to reject $H_0$ implies that the calculated chi-square statistic is greater than the critical value for $\alpha = 0.01$ and 4 degrees of freedom, indicating that the row and column variables are not independent.

For the Given Table of Observed Frequencies, Perform a Test α=0.01\alpha = 0.01α=0.01

For the Given Table of Observed Frequencies, Perform a Test $\alpha = 0.01$