Disease No Disease Totals Exposed 280 720 1000 Non-Exposed 260 740 1000 Totals 5
ID: 3296374 • Letter: D
Question
Disease
No Disease
Totals
Exposed
280
720
1000
Non-Exposed
260
740
1000
Totals
500
1500
2000 (=n)
Which statement describes the result of the chi-square test for this data?
a. The chi-square shows significance and suggests an association
b. The chi-square shows no significance and does not suggest an association
c. The chi-square proves that the exposure causes the outcome
d. The chi-square indicates the magnitude of the cause/effect
Disease
No Disease
Totals
Exposed
280
720
1000
Non-Exposed
260
740
1000
Totals
500
1500
2000 (=n)
Explanation / Answer
Solution:- (B)
The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
H0: There is an association between exposure and disease.
Ha: There is no association between exposure and disease.
Formulate an analysis plan. For this analysis, let the significance level be 0.05. Using sample data, we will conduct a chi-square test for independence.
Analyze sample data. Applying the chi-square test for independence to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.
DF = (r - 1) * (c - 1) = (2 - 1) * (2 - 1) = 1
Er,c = (nr * nc) / n
E1,1 = (1000 * 500) / 2000 = 250
E1,2 = (1000 * 1500) / 2000 = 750
E2,1 = (1000 * 500) / 2000 = 250
E2,2 = (1000 * 1500) / 2000 = 750
2 = [ (Or,c - Er,c)2 / Er,c ]
2 = 1.0147
where DF is the degrees of freedom, r is the number of levels of gender, c is the number of levels of the voting preference, nr is the number of observations from level r of gender, nc is the number of observations from level c of voting preference, n is the number of observations in the sample, Er,c is the expected frequency count when gender is level r and voting preference is level c, and Or,c is the observed frequency count when gender is level r voting preference is level c.
The chi-square statistic is 1.0147.
The p-value is 0.313776. This result is not significant at p < .05.
We use the Chi-Square Distribution Calculator to find P(2 > 1.0147) = 0.313776.
Interpret results. Since the P-value (0.313776) is more than the significance level (0.05), we cannot reject the null hypothesis. Thus, we conclude that there is no relationship between gender and voting preference.