Question

consider the following hypotheses
$h_{0}:p = 0.23$
$h_{1}:p>0.23$
given that $hat{p}=0.3$, $n = 130$, and $\alpha=0.01$, answer the following questions
a. determine the critical value(s), the test statistic. what conclusion should be drawn?
b. determine the p - value for this test
(note: $x=hat{p}n$)
a. determine the critical value(s) of the test statistic.
$z_{\alpha}=\square$
(use a comma to separate answers as needed. round to two decimal places as needed.)
calculate the test statistic
$z_{0}=\square$ (round to two decimal places as needed.)
what conclusion should be drawn?
$\bigcirc$ a. do not reject $h_{0}$. there is insufficient evidence that $p>0.23$
$\bigcirc$ b. reject $h_{0}$. there is insufficient evidence that $p>0.23$
$\bigcirc$ c. do not reject $h_{0}$. there is sufficient evidence that $p>0.23$
$\bigcirc$ d. reject $h_{0}$. there is sufficient evidence that $p>0.23$

Explanation:

Step1: Find the critical value

For a right - tailed test with \(\alpha = 0.01\), we look up the \(z\) - value in the standard normal distribution table. The critical value \(z_{\alpha}\) is the value such that \(P(Z>z_{\alpha})=\alpha\). From the standard normal table, \(z_{0.01} = 2.33\).

Step2: Calculate the test statistic

The formula for the test statistic \(z_{0}\) in a one - sample proportion test is \(z_{0}=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}\)
Given \(\hat{p}=0.3\), \(p = 0.23\), and \(n = 130\)
First, calculate the denominator \(\sqrt{\frac{p(1 - p)}{n}}=\sqrt{\frac{0.23\times(1 - 0.23)}{130}}=\sqrt{\frac{0.23\times0.77}{130}}\)
\(=\sqrt{\frac{0.1771}{130}}\approx\sqrt{0.0013623}\approx0.0369\)
Then, \(z_{0}=\frac{0.3 - 0.23}{0.0369}=\frac{0.07}{0.0369}\approx1.90\)

Step3: Draw a conclusion

Since the test statistic \(z_{0}=1.90 0.23\)

Answer:

The critical value \(z_{\alpha}=2.33\), the test statistic \(z_{0}=1.90\), and the conclusion is A. Do not reject \(H_{0}\). There is insufficient evidence that \(p>0.23\)