Question

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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 Category Frequency Frequency A 21 B 15 с 20 D 7 E 6 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.) 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

Answer 1

Answer:1

This is given that all the category having equal probability since be take 0.20 for each . and apply the test.

The following table is obtained:

Categories	Observed	Expected	(fo-fe)2/fe
A	21	69*0.20=13.8	(21-13.8)2/13.8 = 3.757
B	15	69*0.20=13.8	(15-13.8)2/13.8 = 0.104
C	20	69*0.20=13.8	(20-13.8)2/13.8 = 2.786
D	7	69*0.20=13.8	(7-13.8)2/13.8 = 3.351
E	6	69*0.20=13.8	(6-13.8)2/13.8 = 4.409
Sum =	69	69	14.406

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho : pı = 0.20, p2 = 0.20, P3 = 0.20, P4 = 0.20, p5 = 0.20 H.: Some of the population proportions differ from the values stated in the null hypothesis This corresponds to a Chi-Square test for Goodness of Fit. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, the number of degrees of freedom is df = 5 - 1 = 4, so then the rejection region for this test is R={x?:x>9.488}. (3) Test Statistics The Chi-Squared statistic is computed as follows: x = [ (0; -E;) = 3.757 +0.104+ 2.786 + 3.351 +4.409 = 14.406 E i=1 (4) Decision about the null hypothesis Since it is observed that x = 14.406 > x = 9.488, it is then concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that some of the population proportions differ from those stated in the null hypothesis, at the a = 0.05 significance level.

therefore results are

Chi square value = 14.406

d.f = 4

p value = 0.0061

Answer:2 a)

Categories	Observed	Expected	(fo-fe)2/fe
Category 1	19	146*0.13=18.98	(19-18.98)2/18.98 = 0
Category 2	33	146*0.14=20.44	(33-20.44)2/20.44 = 7.718
Category 3	27	146*0.13=18.98	(27-18.98)2/18.98 = 3.389
Category 4	21	146*0.24=35.04	(21-35.04)2/35.04 = 5.626
Category 5	15	146*0.20=29.2	(15-29.2)2/29.2 = 6.905
Category 6	31	146*0.16=23.36	(31-23.36)2/23.36 = 2.499
Sum =	146	146	26.137

b)

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho : p = 0.13, p2 = 0.14, p3 = 0.13, P4 = 0.24, p5 = 0.20, p6 = 0.16 H.: Some of the population proportions differ from the values stated in the null hypothesis This corresponds to a Chi-Square test for Goodness of Fit. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, the number of degrees of freedom is df = 6 - 1=5, so then the rejection region for this test is R = {x?: x’ > 11.07}. (3) Test Statistics The Chi-Squared statistic is computed as follows: 72 (0; - E;) E = 0 +7.718 + 3.389 + 5.626 + 6.905 +2.499 = 26.137 i=1 (4) Decision about the null hypothesis Since it is observed that x? = 26.137 > xé = 11.07, it is then concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that some of the population proportions differ from those stated in the null hypothesis, at the a = 0.05 significance level.

c) Final answer:

Chi square value = 26.137

p value = 0.0001