Question

To compare the popularity rates of two different political candidates in their respective states, a random sample of 212 citizens from State A yielded 145 who were in favor of their state's candidate whereas 128 citizens out of the 200 selected from State B were in favor of their state's candidate. We wish to use a significance level of 0.05 to test whether there exists a difference between the popularity rates of these two candidates.

Answer 1

Answer:

There exists no difference between the popularity rates of these two candidates.

Step-by-step explanation:

Given : a random sample of 212 citizens from State A yielded 145 who were in favor of their state's candidate whereas 128 citizens out of the 200 selected from State B were in favor of their state's candidate.

To Find :  We wish to use a significance level of 0.05 to test whether there exists a difference between the popularity rates of these two candidates.

Solution:

A random sample of 212 citizens from State A yielded 145 who were in favor of their state's candidate

So, $n_1=212, y_1=145$

128 citizens out of the 200 selected from State B were in favor of their state's candidate.

$n_2=200, y_2=128$

We will use Comparing Two Proportions

$\widehat{p_1}=\frac{y_1}{n_1}$

$\widehat{p_1}=\frac{145}{212}$

$\widehat{p_1}=0.68$

$\widehat{p_2}=\frac{y_2}{n_2}$

$\widehat{p_2}=\frac{128}{200}$

$\widehat{p_2}=0.64$

Let p_1 and p_2 be the probabilities of the popularity rates of these two candidates.

$H_0:p_1=p_2\\H_a:p_1\neqp_2$

$\widehat{p}=\frac{y_1+y_2}{n_1+n_2} =\frac{145+128}{212+200}=0.66$

Formula of test statistic :$\frac{\widehat{p_1}-\widehat{p_2}}{\sqrt{\widehat{p}(1-\widehat{p})(\frac{1}{n_1}+\frac{1}{n_2})}}$

Substitute the values  

 test statistic :$\frac{0.68-0.64}{\sqrt{0.66(1-0.66)(\frac{1}{212}+\frac{1}{200})}}$

 test statistic : 0.856

refer z table for p value

p value = 0.8023

α = 0.05

Since p value >α

So, we accept the null hypothesis

Hence there exists no difference between the popularity rates of these two candidates.