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 \(\boxed{\text{Reject the Null Hypothesis}}\).