You are conducting a multinomial hypothesis test $(\alpha=0.05)$ for the claim that all 5 categories are equally likely to be selected. Complete the table.
\begin{tabular}{|c|c|c|}
\hline Category & \begin{tabular}{c}
Observed \\
Frequency
\end{tabular} & \begin{tabular}{c}
Expected \\
Frequency
\end{tabular} \\
\hline A & 8 & $11.4 \checkmark \sigma^{8}$ \\
\hline B & 14 & $11.4 \checkmark \sigma^{8}$ \\
\hline C & 5 & $11.4 \checkmark 0^{8}$ \\
\hline D & 5 & $11.4 \checkmark 0^{8}$ \\
\hline E & 25 & $11.4 \checkmark \sigma^{\circ}$ \\
\hline
\end{tabular}
Report all answers accurate to three decimal places. But retain unrounded numbers for future calculations.
What is the chi-square test-statistic for this data? (Report answer accurate to three decimal places, and remember to use the unrounded Pearson residuals in your calculations.)
\[
\chi^{2}=25.018
\]
of
What are the degrees of freedom for this test?
d.f. $=4$
$\sigma^{\circ}$
What is the p-value for this sample? (Report answer accurate to four decimal places.)
\[
\text { p-value }=
\]
The $p$-value is...
less than (or equal to) $\alpha$
Answer
Final Answer: The p-value for this sample is \(\boxed{4.989 \times 10^{-5}}\).
Step 1 ：We are conducting a multinomial hypothesis test $(\alpha=0.05)$ for the claim that all 5 categories are equally likely to be selected. The observed and expected frequencies for each category are given in the table.
Step 2 ：The chi-square test-statistic for this data is \(\chi^{2}=25.018\).
Step 3 ：The degrees of freedom for this test is \(d.f. = 4\).
Step 4 ：The p-value is the probability that a chi-square statistic would be as extreme as, or more extreme than, the observed statistic, given the null hypothesis.
Step 5 ：Using the chi-square cumulative distribution function (CDF), we calculate the p-value.
Step 6 ：Final Answer: The p-value for this sample is \(\boxed{4.989 \times 10^{-5}}\).
