Need help 4.A survey of 35 museums asked to provide the performance measure used most often for internal evaluation. A summary of the results is provided in the table. The data were analyzed using a chi-square test for a multinomial experiment. The results are shown in the accompanying table. Complete parts a and b.Click the icon to view the results.a. Is there evidence to indicate that one performance measure is used more often than any of the others? Test using a 0.1.Determine the null and alternative hypotheses for testing whether one performance measure is used more often than any of the other types. Choose the correct answer below.O A. Ho: P₁P₂ = P3 P4 P5 = 0.20Ha: P₁P₂ P3 P4 P5O B. Ho: P₁P₂ # P3P4 P5H₁: At least one of the multinomial probabilities equals its hypothesized value.O C. Ho: P₁ P₂ = P3P4 P5 = 0.20Ha: At least one of the multinomial probabilities does not equal its hypothesized value.O D. Ho: At least one of the multinomial probabilities differs from its hypothesized value.Ha: P₁ P₂ = P3 P4 P5 = 0.20Identify the test statistic.x² =(Round to three decimal places as needed.)Identify the p-value.(Round to three decimal places as needed.)Choose the correct conclusion below.O A. Reject Ho. There is insufficient evidence that at least one performance measure is used more often than any of the other types.O C. Fail to reject Ho. There is insufficient evidence that at least one performance measure is used more often than any of the other types.b. Find a 90% confidence interval for the proportion of museums worldwide that use total visitors as their performance measure. Interpret the result.Select the correct choice below and fill in the answer boxes to complete your choice.(Round to three decimal places as needed.)andO A. There is 90% confidence that the true proportion of museums that use total visitors as their performance measure is betweenOB. The proportion of museums that respond with total visitors as their performance measure is betweenandOC. The proportion of samples that have 90% of museums using total visitors as their performance measure is betweenand1: Chi-square test for a multinomial experiment resultsChi-Square Goodness-of-Fit Test for Observed Counts in Variable: NUMBERUsing category names in PERFORMCategoryObservedTotal visitors9Paying visitors3Big shows7Funds raised5Members11DFChi-Sq45.714286N35Test Proportion0.20.20.20.20.2P-value0.222OB. Reject Ho. There is sufficient evidence that at least one performance measure is used more often than any of the other types.O D. Fail to reject Ho. There is sufficient evidence that at least one performance measure is used more often than any of the other types.Expected7Contribution to Chi-SQ0.5714292.2857140.0000000.5714292.2857147777PerformanceMeasureTotal visitorsPaying visitorsBig showsFunds raisedMembersNumber ofMuseumsX37511

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Accepted Answer

Solution: 
Given information: 
n= 35 Sample size 
 k= 5 Categories