Question

In: Statistics and Probability

on the excel Annual Income No H.S Degree H.S Degree Some College Total Low 65 45...

on the excel

Annual Income No H.S Degree H.S Degree Some College Total
Low 65 45 35 145
Medium 43 66 52 161
High 34 35 52 121
Total 142 146 139 427

1) state the null hypothesis for this study.

2) check whether all chi-square assumptions are met

3) calculate chi-square test statistics for the hypothetical data shown above.

(1) calculate total column percentages.

(2) calculate expected frequencies for all cells.

(3) calculate chi-square for each cell using formula (Oi-Ei)2 Ei

(4) calculate chi-square statistic

4) Identify degree of freedom for the data

5) Identify critical value for the data

6) Calculate the P-value for the data

7) Draw a conclusion referring to test statistic and p-value

Solutions

Expert Solution

Solution:

Here, we have to use chi square test for independence of two categorical variables.

1) State the null hypothesis for this study.

Null hypothesis: H0: The annual income and educational degree are independent from each other.

Alternative hypothesis: Ha: The annual income and educational degree are dependent.

We assume level of significance = α = 0.05

2) Check whether all chi-square assumptions are met

Required assumptions are given as below:

Given data is a categorical data.

Two variables should consist of two or more categorical and independent groups.

3) Calculate chi-square test statistics for the hypothetical data shown above.

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

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

E = row total * column total / Grand total

We are given

Number of rows = r = 3

Number of columns = c = 3

Degrees of freedom = df = (r – 1)*(c – 1) = 2*2 = 4

α = 0.05

Critical value = 9.487729037

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Observed Frequencies

Column variable

Row variable

No Degree

Degree

Some college

Total

Low

65

45

35

145

Medium

43

66

52

161

High

34

35

52

121

Total

142

146

139

427

Expected Frequencies

Column variable

Row variable

No Degree

Degree

Some college

Total

Low

48.22014052

49.578454

47.20140515

145

Medium

53.54098361

55.04918

52.40983607

161

High

40.23887588

41.372365

39.38875878

121

Total

142

146

139

427

Calculations

(O - E)

16.77986

-4.57845

-12.2014

-10.541

10.95082

-0.40984

-6.23888

-6.37237

12.61124

(O - E)^2/E

5.83913

0.42281

3.154022

2.075276

2.178424

0.003205

0.967313

0.981502

4.037787

Test statistic = Chi square = ∑[(O – E)^2/E] = 19.65946815

P-value = 0.000582933

(By using Chi square table or excel)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that the annual income and educational degree are dependent.


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