Question

One True Love by Educational Level A study is conducted in which people were asked whether they agreed or disagreed with the statement that there is only one true love for each person. The table below gives a two-way table showing the answers to this question as well as the education level of the respondents. A person's education is categorized as HS (high school degree or less), Some (some college), or College (college graduate or higher). Is the level of a person's education related to how the person feels about one true love? If there is a significant association between these two variables, describe how they are related. HS Some College Total Agree 360 165 200 725 Disagree 557 467 787 1811 Don't know 15 28 28 71 Total 932 660 1015 2607 Table 1 Educational level and belief in One True Love Round your answer for the chi-square statistic to one decimal place, and your answer for the p-value to three decimal places.

Answer 1

Solution:

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

Null hypothesis: H₀: Two categorical variables are independent.

Alternative hypothesis: H_a: Two categorical variables are dependent.

We assume level of significance = α = 0.05

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 = 9.487729

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Observed Frequencies
	Column variable
Row variable	HS	Some	College	Total
Agree	360	165	200	725
Disagree	557	467	787	1811
Don't know	15	28	28	71
Total	932	660	1015	2607

Expected Frequencies
	Column variable
Row variable	HS	Some	College	Total
Agree	259.1868	183.5443	282.2689	725
Disagree	647.4308	458.481	705.0882	1811
Don't know	25.38243	17.97468	27.64288	71
Total	932	660	1015	2607

Calculations
(O - E)
100.8132	-18.5443	-82.2689
-90.4308	8.518987	81.91178
-10.3824	10.02532	0.357115

(O - E)^2/E
39.21226	1.873614	23.97774
12.63104	0.15829	9.515886
4.246831	5.591585	0.004614

Test Statistic = Chi square = ∑[(O – E)^2/E] = 97.21186

Chi-square statistic = 97.2

P-value = 0.000

(By using Chi square table or excel)

P-value < α = 0.05

So, we reject the null hypothesis

Conclusion: Reject H₀

Does the test indicate an association between these two variables?

Answer: Yes

The test indicates that there is an association between these two variables because we reject the null hypothesis.