Question

(If some are cut off just answer what you can see)

2A)

You are conducting a multinomial hypothesis test (a = 0.05) for the claim that all 5 categories are equally likely to be selected. Complete the table. Observed Expected Squared Category Pearson Frequency Frequency Residual A 25 B 14 с 9 D 22 E 20 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.) x = What are the degrees of freedom for this test? d.f. What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value = It may be best to use the ECHIDIST() function in a Spreadsheet to do this calculation. The p-value is... less than (or equal to) a greater than a This test statistic leads to a decision to... reject the null accept the null fail to reject the null accept the alternative

2B)

2C)

2D)

Answer 1

2A)

25 Category A B с D E Total Observed freqexpected freq Squared pears 18 2.7222222222 14 18 0.8888888889 9 18 4.5 22 18 0.8888888889 20 18 0.2222222222 90 90 9.222

Here chi-square= 9.222

df=5-1= 4

P-value= 0.0558,

p-value is greater than alpha.

Therefore fail to reject null hypothsis.

2B)

Category A B с D E Total Observed freqexpected freq Squared pears 9.6 0.7041666667 7 9.6 0.7041666667 19 9.6 9.2041666667 8 9.6 0.2666666667 7 9.6 0.7041666667 48 11.583 48

Here chi-square= 11.583

df=5-1= 4

P-value= 0.0207,

p-value is less than or equal to alpha.

Therefore reject null hypothsis.

2C)

Category A B с D E Total Observed freq expected freq Pearson reside Squared pears 25 15.2 2.514 6.3184210526 17 15.2 0.462 0.2131578947 15 15.2 -0.051 0.0026315789 8 15.2 -1.847 3.4105263158 11 15.2 -1.077 1.1605263158 76 76 11.105

Here chi-square= 11.105

critical value= chisq(0.1,4) = 7.779

Here, Chi-sauare> 7.779 , reject null hypothsis.

2D)

13 Category A B с D E Total Observed freq expected freq Pearson residi Squared pears 19 18 0.236 0.055696 18 -1.179 1.3888888889 18 18 0 0 15 18 -0.707 0.5 25 18 1.65 2.7222222222 90 90 4.667

Here chi-square= 4.667

critical value= chisq(0.1,4) = 11.143

Here, Chi-sauare<11.143 , fail to reject null hypothsis.