Question

A random sample of 100 likely voters in a small city produced 59 voters in favor of Candidate A. The observed value of the standardized test statistic for performing a test of $\mathrm{H}0:\mathrm{p}=0.5H_0:p=0.5\text{versus Ha:}\mathrm{p}>0.5H_a:p>0.5$

is which of the following?

a)$z=\frac{0.59-0.5}{\sqrt{\frac{0.59(0.41)}{100}}}$

b)$z=\frac{0.59-0.5}{\sqrt{\frac{0.5(0.5)}{100}}}$

c)$z=\frac{0.5-0.59}{\sqrt{\frac{0.59(0.41)}{100}}}$

d)$z=\frac{0.5-0.59}{\sqrt{\frac{0.5(0.5)}{100}}}$

e)$z=\frac{0.59-0.5}{\sqrt{100}}$

Short answer

$\text{The test statistic is}\frac{0.59-0.50}{\sqrt{\frac{0.30(1-0.300}{100}}}\text{.}$

Step-by-step solution

Step 1:Given information

Number of trials $\left(n\right)=100$

Number of events$\left(x\right)=59$

Level of significance$\left(\alpha \right)=0.05$

The null and alternative hypotheses are:

$H_0:p=0.5$

$H_a:p>0.5$

Step 2:Explaination

The formula to compute the test statistic is:

$z=\frac{\hat{p}-p_0}{\sqrt{\frac{P_0\left(1-p_0\right)}{n}}}$

Where,

$p_0$ is the population proportion

$\hat{p}$ is the sample proportion

Now,

The sample proportion can be computed as:

$\text{Sample proportion}\left(\hat{p}\right)=\frac{x}{n}$

$=\frac{59}{100}=0.59$

Now, test statistic can be calculated as:

$z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}$

$=\frac{0.59-0.50}{\sqrt{\frac{0.50(1-0.50)}{100}}}$

$\text{Thus, the test statistic is}\frac{0.59-0.50}{\sqrt{\frac{0\text{S5(1-0.000)}}{100}}}\text{.}$

Hence, the correct option is (d).