Question

1) Suppose there is a random sample of n observations, divided into four groups. The table below summarizes the count of observations that were seen in each group.

Group 1	Group 2	Group 3	Group 4
16	46	51	37

We are interested in testing the null hypothesis H0:p1=p2=p3=p4=0.25, against the alternative hypothesis HA: At least one proportion is incorrect

a) What is the expected count for each of the groups?

Expected: (number)

b) What is the value of the test statistic? Round your response to at least 2 decimal places.  

c) What are the appropriate degrees of freedom?

d) What is the P-value? Round to at least 4 decimal places

e) What conclusion can be made at the 5% level of significance?

A		There is evidence that at least one proportion is not equal to 0.25.
B		There is no significant evidence that any of the proportions is not equal to 0.25.

2) Suppose there is a random sample of 1,159 observations, divided into four groups. The table below summarizes the observations that were seen in each group.

Group 1	Group 2	Group 3	Group 4
556	197	127	279

We are interested in testing the Null hypothesis Observed=Expected, under the assumption that the expected proportions are  .50, .20, .10, and .20 for the 4 groups, respectively.

a) What are the expected values?

Group 1	Group 2	Group 3	Group 4

b) What is the value of the test statistic? Round your response to at least 3 decimal places.

c) What is the P-value for the test? Round your response to at least 4 decimal places.

d) What conclusion can be made at the 5% level of significance?

A		There is evidence against the null hypothesis, and therefore at least one of the observed proportions is not the same as the expected proportions.
B		There is no significant evidence against the null hypothesis, and therefore we do not have evidence the observed and expected proportions are different.

Answer 1

Question 1

a) What is the expected count for each of the groups?

Group	Observed	Expected
1	16	150*0.25 = 37.5
2	46	150*0.25 = 37.5
3	51	150*0.25 = 37.5
4	37	150*0.25 = 37.5
Total	150	150

b) What is the value of the test statistic?

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

Group	O	E	(O - E)^2	(O - E)^2/E
1	16	37.5	462.25	12.32666667
2	46	37.5	72.25	1.926666667
3	51	37.5	182.25	4.86
4	37	37.5	0.25	0.006666667
Total	150	150		19.12

Chi square = ∑[(O – E)^2/E]

Test statistic = Chi square = 19.12

c) What are the appropriate degrees of freedom?

We are given N = 4

Degrees of freedom = df = N – 1 = 4 – 1 = 3

d) What is the P-value?

P-value = 0.000258215

(by using Chi square table)

e) What conclusion can be made at the 5% level of significance?

P-value < α = 0.05

So, we reject the null hypothesis

There is evidence that at least one proportion is not equal to 0.25.

Question 2

a) What are the expected values?

Group	O	Exp. Prop.	Expected
1	556	0.5	1159*0.5 = 579.5
2	197	0.2	1159*0.2 = 231.8
3	127	0.1	1159*0.1 = 115.9
4	279	0.2	1159*0.2 = 231.8
Total	1159	1	1159

b) What is the value of the test statistic?

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

Group	O	Exp. Prop.	E	(O - E)^2	(O - E)^2/E
1	556	0.5	579.5	552.25	0.952976704
2	197	0.2	231.8	1211.04	5.224503883
3	127	0.1	115.9	123.21	1.063071613
4	279	0.2	231.8	2227.84	9.611044003
Total	1159	1	1159		16.8515962

Chi square = ∑[(O – E)^2/E]

Test statistic = Chi square = 16.8515962

c) What is the P-value for the test?

We have N = 4, df = N – 1 = 3

P-value = 0.000758185

(By using Chi square table)

d) What conclusion can be made at the 5% level of significance?

P-value < α = 0.05

So, we reject the null hypothesis

There is evidence against the null hypothesis, and therefore at least one of the observed proportions is not the same as the expected proportions.