## Related questions with answers

Test the null hypothesis of independence of the two classifications, $\mathrm{A}$ and $\mathrm{B}$, in the $3 \times 3$ contingency table displayed below. Test using $\alpha=.05$.

| | | | $\mathrm{B}$ | | |--------------|-------------------|------------------| | | | $\mathrm{B}_{1}$ | $\mathrm{~B}_{2}$ | $\mathrm{~B}_{3}$ | | | $\mathrm{A}_{1}$ | 40 | 72 | 42 | | $\mathrm{A}$ | $\mathrm{~A}_{2}$ | 63 | 53 | 70 | | | $\mathrm{~A}_{3}$ | 31 | 38 | 30 |

**b**. Estimate the percentage of the total number of responses that constitute each of the A classification totals.

Solution

VerifiedGiven:

$\begin{array}{ c | c c c | c } & B_1 & B_2 & B_3 & Total \\ \hline A_1 & 40 & 72 & 42 & 154 \\ A_2 & 63 & 53 & 70 & 186 \\ A_3 & 31 & 38 & 30 & 99 \\ \hline Total & 134 & 163 & 142 & 439 \end{array}$

We divide each count in the table by the corresponding row total:

$\begin{array}{ c | c c c | c } & B_1 & B_2 & B_3 & Total \\ \hline & & & \\ A_1 & \frac{40}{134}\approx 0.2985 & \frac{72}{163}\approx 0.4417 & \frac{42}{142}\approx 0.2958 & \frac{154}{439}\approx 0.3508 \\ & & & \\ A_2 & \frac{63}{134}\approx 0.4701 & \frac{53}{163}\approx 0.3252 & \frac{70}{142}\approx 0.4930 & \frac{186}{439}\approx 0.4237 \\ & & & \\ A_3 & \frac{31}{134}\approx 0.2313 & \frac{38}{163}\approx 0.2331 & \frac{30}{142}\approx 0.2113 & \frac{99}{439}\approx 0.2255 \\ & & & \\ \hline Total & 1 & 1 & 1 & 1 \end{array}$

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