Question

you are conducting a study to see if the proportion of voters who prefer candidate a is significantly less than 0.27. you use a significance level of $alpha = 0.005$.
$h_0: p = 0.27$
$h_1: p < 0.27$
you obtain a sample of size $n = 600$ in which there are 165 successes.
what is the test - statistic for this sample? (report answer accurate to two decimal places.)
test statistic=
what is the p - value for this sample? (report answer accurate to four decimal places.)
p - value=
the p - value is...
less than (or equal to) $alpha$
greater than $alpha$
this test statistic leads to a decision to...
reject the null
accept the null
fail to reject the null

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 165$ and $n=600$. So $\hat{p}=\frac{165}{600}=0.275$.

Step2: Calculate test - statistic

The formula for the test - statistic $z$ in a one - sample proportion test is $z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}$, with $p_0 = 0.27$, $\hat{p}=0.275$, and $n = 600$.
\[

$$\begin{align*} z&=\frac{0.275 - 0.27}{\sqrt{\frac{0.27\times(1 - 0.27)}{600}}}\\ &=\frac{0.005}{\sqrt{\frac{0.27\times0.73}{600}}}\\ &=\frac{0.005}{\sqrt{\frac{0.1971}{600}}}\\ &=\frac{0.005}{\sqrt{0.0003285}}\\ &=\frac{0.005}{0.0181246}\\ &\approx0.28 \end{align*}$$

\]

Step3: Calculate p - value

Since this is a left - tailed test ($H_1:p<0.27$), the p - value is $P(Z < 0.28)$. Looking up in the standard normal table, $P(Z < 0.28)=0.6103$.

Step4: Compare p - value and significance level

The significance level $\alpha = 0.005$. Since $p - value=0.6103>0.005=\alpha$, we fail to reject the null hypothesis.

Answer:

test statistic = 0.28
p - value = 0.6103
The p - value is greater than $\alpha$
This test statistic leads to a decision to fail to reject the null