In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6975 subjects randomly selected from an online group involved with ears. There were 1326 surveys returned. Use a 0.01 significance level to test the claim that the return rate is less than $20 \%$. Use the P-value method and use the normal distribution as an approximation to the binomial distribution.
Identify the null hypothesis and alternative hypothesis.
A.
\[
\begin{array}{l}
H_{0}: p=0.2 \\
H_{1}: p \neq 0.2
\end{array}
\]
C.
\[
\begin{array}{l}
H_{0}: p \neq 0.2 \\
H_{1}: p=0.2
\end{array}
\]
E. $H_{0}: p=0.2$
\[
H_{1}: p< 0.2
\]
\[
\begin{array}{l}
H_{0}: p=0.2 \\
H_{1}: p< 0.2
\end{array}
\]
The test statistic is $z=-2.07$ '.
(Round to two decimal places as needed.)
The P-value is $0.019^{\top}$.
(Round to three decimal places as needed.)
B. $H_{0}: p< 0.2$
\[
H_{1}: p=0.2
\]
D.
\[
\begin{array}{l}
H_{0}: p=0.2 \\
H_{1}: p> 0.2
\end{array}
\]
F.
\[
\begin{array}{l}
H_{0}: p> 0.2 \\
H_{1}: p=0.2
\end{array}
\]
Because the $P$-value is the significance level, the null hypothesis. There is evidence to support the claim that the return rate is less than $20 \%$.
Answer
Final Answer: \(\boxed{H_{0}: p=0.2, H_{1}: p<0.2}\)
Step 1 ：Identify the null hypothesis and alternative hypothesis.
Step 2 ：The null hypothesis is usually a statement of no effect or no difference and is the assumption that any kind of difference or significance you see in a set of data is due to chance.
Step 3 ：The alternative hypothesis is what you might believe to be true or hope to prove true.
Step 4 ：In this case, the null hypothesis is that the return rate is 20% (p=0.2), and the alternative hypothesis is that the return rate is less than 20% (p<0.2). This is because we are testing the claim that the return rate is less than 20%.
Step 5 ：So, the correct answer is: \(H_{0}: p=0.2, H_{1}: p<0.2\)
Step 6 ：Final Answer: \(\boxed{H_{0}: p=0.2, H_{1}: p<0.2}\)
