Question

3. Suppose a researcher believes that college faculty vote at higher rate than college students. She collects data from 200 college faculty and 200 college students using simple random sampling. If 167 of the faculty and 138 of the students voted in the 2008 Presidential election, (A) Find a 90% confidence interval for the difference between the proportion of faculty and the proportion of students who vote. (b) Is there enough evidence at the 3% level of significance to support the researcher's claim?

Answer 1

Group 1 (Faculty)

Proportion (p₁) = 167/200 = 0.835

Count (n₁) = 200

Group 2 (College Students)

Proportion (p₂) = 138/200 = 0.69

Count (n₂) = 200

p_c = (n₁*p₁ + n₂*p₂)/(n₁+n₂) = 0.763

Standard Error (SE) = (p_c*(1-p_c)/n₁ + p_c*(1-p_c)/n₂ )^1/2 = 0.043

Alpha = 0.1

Z_Critical = 1.64

a)

Hence,

90% CI for diff of Proportions = p₁- p₂ +/- SE*Z_Critical = 0.835 – 0.69 +/- 1.64 * 0.043 = {0.08,0.21}

b)

alpha = 0.03

Null and Alternate Hypothesis

H₀: µ₁ = µ₂ (Proportion of Faculty vote is same as College votes)

H_a: µ₁ > µ₂ (Proportion of Faculty vote is greater than College votes)

Test Statistic

t = (p₁ - p_{2 ­­}- (µ₁ - µ₂))/SE = (0.835-0.69)/0.043 = 3.41

p-value = TDIST(3.41,200+200-2,1) = 0.000361

Result

Since the p-value is less than 0.03, we reject the null hypothesis in favour of alternate hypothesis

Conclusion

Proportion of Faculty vote is greater than College votes