Question

in a study of cell phone usage and brain hemispheric dominance, an internet survey was e - mailed to 6989 subjects randomly selected from an online group involved with ears. there were 1300 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.
the test statistic is ( z=-2.93 ) (round to two decimal places as needed )
the p - value is (square) (round to three decimal places as needed )

Explanation:

Step1: Determine the type of test

Since the claim is that the return rate \(p\) is less than \(0.2\), this is a left - tailed test.

Step2: Calculate the P - value

For a left - tailed test with test statistic \(z=-2.93\), the P - value is the probability that \(Z < - 2.93\).
Using the standard normal distribution table or a calculator with a normal distribution function (e.g., in Excel: NORM.S.DIST(-2.93,TRUE) or in R: pnorm(-2.93)), we find that \(P(Z < - 2.93)\)

$$P(Z < - 2.93)=0.002$$

Answer:

\(0.002\)