Question

Suppose there is a random sample of 1,083 observations, divided into four groups. The table below summarizes the observations that were seen in each group. Group 1 Group 2 Group 3 Group 4 531 184 101 267 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. What are the expected values? Group 1 Group 2 Group 3 Group 4 Number Number Number Number What is the value of the test statistic? Round your response to at least 3 decimal places. Number What is the P-value for the test? Round your response to at least 4 decimal places. Number

Answer 1

Solution:

Here, we have to use chi square test for goodness of fit.

Null hypothesis: H₀: Given observations of the data follows the expected proportions.

Alternative hypothesis: H_a: Given observations of the data do not follow the expected proportions.

We assume level of significance = α = 0.05

What are the expected values?

Expected values are given as below:

Group	Observed	Prop.	Expected
1	531	0.5	1083*0.5 = 541.5
2	184	0.2	1083*0.2 = 216.6
3	101	0.1	1083*0.1 = 108.3
4	267	0.2	1083*0.2 = 216.6
Total	1083	1	1083

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.

We are given

N = 4

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

α = 0.05

Critical value = 7.814727764

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Group	O	Prop.	E	(O - E)^2/E
1	531	0.5	541.5	0.203601108
2	184	0.2	216.6	4.906555863
3	101	0.1	108.3	0.492059095
4	267	0.2	216.6	11.72742382
Total	1083	1	1083	17.32963989

Chi square = ∑[(O – E)^2/E] = 17.32963989

Test statistic = 17.330

What is the P-value for the test?

P-value = 0.0006

(By using Chi square table or excel)

P-value < α = 0.05

So, we reject the null hypothesis

There is insufficient evidence to conclude that given observations of the data follows the expected proportions.