Question

Cross-Tabulation and Chi-Square Test Respondents were asked to assess their current health condition and to report their social class. Data is in the table below Social Class Middle- and Upper Lower- and Working- Class Respondents Tetal Health Status Class Respondents Fair/Poor 220 136 Excellent/Good 472 521 Total 1. What is the independent variable? What is its level of measurement (nominal, ordinal, ratio)? What is the dependent variable? What is its level of measurement (nominal, ordinal, ratio)? 2. Use the frequencies in the table above to calculate the column and row marginals. (Fil in frequencies in the blank cells in the table above.) 3. Use the frequencies in the table above to calculate the percentages for the table and fill in the table below. Hint each of the independent variable's categories should add up to 100% Social Class Lower- and Working- Class Respondents Middle- and Upper Class Respondents Health Status Fair/Poor Excellent/Good Total 4. Who is more likely to have fair/poor health? Interpret this relationship by making a comparison between the categories of the independent variable in a complete sentence, using a percentage point difference. How strong is the relationship? For this problem you will need the following formulas: Expected Count-row total x colue tal table total (observed count-expected count x- expected count df-(of rows-1)x (5 of columns - 1) S. Write the null and alternative hypotheses for the chi- square test. (There is no special notation for hypotheses for the chi-square test.) Make explicit whether you are talking about the sample or the population. 6. Complete the expected frequencies table below. Social Class Lower- and Working Class Respondents Midde-and Upper Total Health Status Class Respondents Fair Poor Excellent Good Total 7. Fill in this table to calculate the chi-square test statistic CellObserved Expected O-EO-EO-EE Sum Chi-square statistic ( 8. What are the degrees of freedom and critical value for this test (use alpha of .001)? df- Critical value 8. What are the degrees of freedom and critical value for this test (use alpha of.001)? df Critical value 9. Is the chi-square statistic larger than the critical value? YES/NO 10. Can you reject the null hypothesis? YES NO 11. What can we say about the existence of a relationship in the population between social class and health status? 12. If we saw a relationship in the sample but it was not statistically significant (and therefore did not exist in the population), why would the relationship exist in the sample? 3

Answer 1

1)

Independent variable: Social Status

Dependent variable: Health Condition

Level of measurement: Both the variables are nominal (categorical)

1. Health Condition:two categories: i. Fair/poor, ii. Excellent/Good

2. Social Status: two categories: i. Lower and Working Class ii. Middle and Upper Class

2)

	Lower and Working Class	Middle and Upper Class	Total
Fair/poor	220	136	356
Excellent/Good	472	521	993
Total	692	657	1349

3)

The percentage values are obtained by dividing each cell's frequency by total frequency times 100.

frequency % value- x 100% Total frequency

	Lower and Working Class	Middle and Upper Class	Total
Fair/poor	16.3084	10.0815	26.3899
Excellent/Good	34.9889	38.6212	73.6101
Total	51.2973	48.7027	100

4)

Lower and Working Class are more likely to Fair/Poor (16.3084%) compare to Middle and Upper class (10.0815%). Lower and Working Class are (16.3084 - 10.0815 = 6.2268% more likely compare to Middle and Upper class.

5)

The null and alternative hypothesis are defined as,

Null hypothesis, Ho:There is no association between two variables.

Alternative hypothesis, Ha There is an association present between the two variables

Here we are taking about the whole population.

6)

The expected values are obtained using the formula,

Row x Column Expected Value Total

The expected values are,

	Lower and Working Class	Middle and Upper Class	Total
Fair/poor	182.618	173.382	356
Excellent/Good	509.382	483.618	993
Total	692	657	1349

7)

Now the Chi-Square Value is obtained using the formula

Expected

Observed	Expected	(O-E)	(O-E)^2	(O-E)^2/E
220	182.618236	37.38176	1397.396	7.652009
136	173.381764	-37.3818	1397.396	8.05965
472	509.381764	-37.3818	1397.396	2.743318
521	483.618236	37.38176	1397.396	2.889462
			Sum	21.34444

21.3444

8)

df (r 1c-1) = (2-1) ( 2-1) 1

Xiower0, Xipper 12.116

9)

x221.3444 x pper= 12.116 2

10)

Since chi square value is greater than hence the null hypothesis is rejected.

11)

There is statistically significant association between social status and health condition.

12)

may be the reason of biased sampling or not taken random sampling.