Question

1A)

You are conducting a multinomial Goodness of Fit hypothesis test for the claim that the 4 categories occur with the following frequencies: H:PA = 0.1; P = 0.3; Pc = 0.1; Pp = 0.5 Complete the table. Report all answers accurate to three decimal places. Observed Expected Category Frequency Frequency А 17 B 43 с 12 D 48 What is the chi-square test-statistic for this data? x = What is the P-value? 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 Report all answers accurate to three decimal places.

1B)

1C)

1D)

Answer 1

We would be looking at the first question on hypothesis test here all parts as:

The expected value for each category here is computed as:
E_i = (Sum of all frequencies)*proportion of share of each category.

Category	O_i	E_i	(O_i - E_i)^2/E_i
A	17	12	2.0833
B	43	36	1.3611
C	12	12	0.0000
D	48	60	2.4000
	120	120	5.8444

Therefore the chi square test statistic value here is computed as:

$\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} = 5.8444$

Therefore 5.8444 is the required chi square test statistic value here.

b) for k - 1 = 3 degrees of freedom, the p-value here is obtained from chi square distribution tables here as:

$p =P(\chi^2_3 > 5.8444) = 0.1194$

Therefore 0.1194 is the required p-value here.

As the p-value here is 0.1194 > 0.05 which is the level of significance, therefore the test is not significant here and we cannot reject the null hypothesis here. therefore we dont have sufficient evidence here and therefore we cannot reject the null hypothesis here.