Disease No Disease Totals Exposed 280 720 1000 Non-Exposed 260 740 1000 Totals 5

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.

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.