You are conducting a multinomial Goodness of Fit hypothesis test for the claim that the 4 categories occur with the following frequencies:
$H_{o}: p_{A}=0.3 ; p_{B}=0.4 ; p_{C}=0.15 ; p_{D}=0.15$

Complete the table. Report all answers accurate to three decimal places.
\begin{tabular}{|c||c|c|}
\hline Category & \begin{tabular}{c}
Observed \\
Frequency
\end{tabular} & \begin{tabular}{c}
Expected \\
Frequency
\end{tabular} \\
\hline A & 38 & 41.1 \\
\hline B & 44 & 54.8 \\
\hline C & 23 & 20.55 \\
\hline D & 32 & 20.55 \\
\hline
\end{tabular}

What is the chi-square test-statistic for this data? (2 decimal places)
\[
x^{2}=9.095
\]

What is the P-Value? (3 decimal places)
\[
\text { P.Value }=
\]

For significance level alpha 0.05 ,
What would be the conclusion of this hypothesis test?
Reject the Null Hypothesis
Fail to reject the Null Hypothesis

Step 1 ：The problem provides the observed and expected frequencies for four categories (A, B, C, D) and the null hypothesis $H_{o}: p_{A}=0.3 ; p_{B}=0.4 ; p_{C}=0.15 ; p_{D}=0.15$.

Step 2 ：The chi-square test-statistic for this data is calculated to be $x^{2}=9.095$.

Step 3 ：The P-Value is the probability that a chi-square statistic is more extreme than the observed value, given the degrees of freedom. In this case, the degrees of freedom is the number of categories minus 1, which is 4 - 1 = 3.

Step 4 ：Using the chi-square cumulative distribution function (CDF), the P-Value is calculated to be approximately 0.028.

Step 5 ：For a significance level alpha of 0.05, we compare the P-Value with the significance level. If the P-Value is less than the significance level, we reject the null hypothesis. If the P-Value is greater than the significance level, we fail to reject the null hypothesis.

Step 6 ：Since the P-Value is approximately 0.028, which is less than the significance level of 0.05, we reject the null hypothesis.

Step 7 ：Final Answer: The P-Value is approximately \(\boxed{0.028}\) and we \(\boxed{\text{Reject the Null Hypothesis}}\).