Question

Suppose that a response can fall into one of k = 5 categories with probabilities
p1, p2,   , p5
and that n = 300 responses produced these category counts.

Category 1 2 3 4 5
Observed Count   46 61 72 51 70

(a) Are the five categories equally likely to occur? How would you test this hypothesis?
H0: p1 = p2 = p3 = p4 = p5 =

1
5

Ha: At least one pi is different from

1
5

.
H0: At least one pi is different from 0.
Ha: p1 = p2 = p3 = p4 = p5 = 0
    
H0: At least one pi is different from

1
5

.
Ha: p1 = p2 = p3 = p4 = p5 =

1
5

H0: p1 = p2 = p3 = p4 = p5 = 0
Ha: At least one pi is different from 0.
H0: p1 = p2 = p3 = p4 = p5 = 1
Ha: At least one pi is different from 1.
(b) If you were to test this hypothesis using the chi-square statistic, how many degrees of freedom would the test have?
degrees of freedom

(c) Find the critical value of ?2 that defines the rejection region with ? = 0.05. (Round your answer to three decimal places.)
?20.05 =

(d) Calculate the observed value of the test statistic.
?2 =  

(e) Conduct the test and state your conclusions.
There is sufficient evidence to indicate that at least one category is more likely to occur than the others.There is sufficient evidence to indicate that there is not at least one category more likely to occur than the others.    There is insufficient evidence to indicate that at least one category is more likely to occur than the others.There is insufficient evidence to indicate that there is not at least one category more likely to occur than the others.

Answer 1

(a) Are the five categories equally likely to occur? How would you test this hypothesis?

Null Hypothesis

Alternative Hypothesis H1:  At least one pi is different from

b) Degrees of freedom = n-1= 5-1= 4

c) The critical value of Chi square for 4 df at 5% significance level is 9.488

d) Under H0, the expected frequencies are

=(46 +61 +72 +51 +70)/5 =300/5 = 60

The Calculation table for Chi square test statistic is

	Obs. Freq. ( O )	Exp freq. (E )	(O-E)^2/E
	46	60	3.266667
	61	60	0.016667
	72	60	2.4
	51	60	1.35
	70	60	1.666667
Total	300	300	8.7

The test statistic is  

e) Since chi square calculated is less than Chi square tabulated, Fail to Reject H0.

Answer: There is insufficient evidence to indicate that at least one category is more likely to occur than the others.