Question

in a study of cell phone usage and brain hemispheric dominance, an internet survey was e - mailed to 6970 subjec randomly selected from an online group involved with ears. there were 1331 surveys returned. use a 0.01 signifi level to test the claim that the return rate is less than 20%. use the p - value method and use the normal distributi an approximation to the binomial distribution. identify the null hypothesis and alternative hypothesis. a. $h_0:p
eq0.2$ $h_1:p = 0.2$ b. $h_0:pgt0.2$ $h_1:p = 0.2$ c. $h_0:p = 0.2$ $h_1:pgt0.2$ d. $h_0:p = 0.2$ $h_1:plt0.2$ e. $h_0:p = 0.2$ $h_1:p
eq0.2$ f. $h_0:plt0.2$ $h_1:p = 0.2$ the test statistic is $z=square$. (round to two decimal places as needed.)

Explanation:

Step1: Calculate the sample proportion \(\hat{p}\)

The sample proportion \(\hat{p}=\frac{x}{n}\), where \(x = 1331\) (number of successes, i.e., returned surveys) and \(n=6970\) (sample size). So, \(\hat{p}=\frac{1331}{6970}\approx0.191\).

Step2: Calculate the test - statistic \(z\)

The formula for the test - statistic in a one - sample proportion test is \(z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}\), where \(p = 0.2\) (the proportion in the null hypothesis).
Substitute the values:

$$
LATEXBLOCK0
$$

Answer:

The test statistic \(z\approx - 1.88\)