in a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6964 subjects randomly selected from an online group involved with ears. There were 1296 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 th normal distribution as an approximation to the binomial distribution. Identify the null hypothesis and alternative hypothesis. A. $\mathrm{H}_{0}: \mathrm{p}<0.2$ \[ H_{1}: p=0.2 \] c. $\mathrm{H}_{0}: \mathrm{p}=0.2$ $H_{1}: p>0.2$ E. \[ \begin{array}{l} H_{0}: p>0.2 \\ H_{1}: p=0.2 \end{array} \] The test statistic is $z=-2.90$. (Round to two decimal places as needed.) The $P$-value is $\square$. (Round to three decimal places as needed.) B. $\mathrm{H}_{0}: \mathrm{p} \neq 0.2$ \[ H_{1}: p=0.2 \] D. \[ \begin{array}{l} H_{0}: p=0.2 \\ H_{1}: p \neq 0.2 \end{array} \] F. \[ \begin{array}{l} H_{0}: p=0.2 \\ H_{1}: p<0.2 \end{array} \]
Solution
Step 1 ：Identify the null hypothesis and alternative hypothesis. 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. The alternative hypothesis is what you might believe to be true or hope to prove true. Given the problem, we are testing the claim that the return rate is less than 20%. Therefore, the null hypothesis should be that the return rate is equal to 20% and the alternative hypothesis should be that the return rate is less than 20%. The null hypothesis and alternative hypothesis are: \[H_{0}: p=0.2\] \[H_{1}: p<0.2\]
Step 2 ：The second part of the question is asking for the P-value. The P-value is the probability that, if the null hypothesis were true, we would observe a statistic at least as extreme as the one observed. To find the P-value, we can use the test statistic and the standard normal distribution. The test statistic is \(z = -2.9\).
Step 3 ：Calculate the P-value using the test statistic and the standard normal distribution. The P-value is approximately 0.0018658133003840375.
Step 4 ：Round the P-value to three decimal places. The P-value is \(\boxed{0.002}\).
