Answer to Question #215150 in Statistics and Probability for Dk khan

Question

A candidate for mayor in a large city believes that he appeals to at least 10 per cent more of the educated voters than the uneducated voters. He hires the services of a poll-taking organization, and they find that 62 of 100 educated voters interviewed support the candidate, and 69 of 150 uneducated voters support him at the 0.05 significance level. 

Answer 1

p₁ = proportion of educated voters

$p_1= \frac{62}{100}=0.62$

p₂ = proportion of educated voters

$p_2= \frac{69}{150}=0.46 \\ H_0: p_1-p_2 ≥0.10 \\ H_1: p_1-p_2 <0.10 \\ p = \frac{62+69}{100+150} = \frac{131}{250}=0.524$

At 0.05 level of significance the critical value for a left-tailed test is $Z_c= -1.645$

The rejection region for this test is $Z<Z_c$

Test statistic

$Z = \frac{p_1-p_2}{\sqrt{ p(1-p) ( \frac{1}{n_1} + \frac{1}{n_2} ) }} \\ Z= \frac{0.62-0.46}{\sqrt{ 0.524(1-0.524) ( \frac{1}{100} + \frac{1}{150} ) }} =2.482 \\ Z = 2.482>Z_c = -1.645$

H₀ is not rejected.

There is enough evidence to claim that the candidate’s appeal to at least 10 per cent more of the educated voters than the uneducated voters at the 0.05 significance level.