Question

in a study of cell phone usage and brain hemispheric dominance, an internet survey was e - randomly selected from an online group involved with ears. there were 1335 surveys retur level to test the claim that the return rate is less than 20%. use the p - value method and use an approximation to the binomial distribution.

the test statistic is ( z=-1.85 )
(round to two decimal places as needed.)

the p - value is
(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 \(20\%=0.2\), this is a left - tailed test.

Step2: Calculate the P - value

For a left - tailed test with test statistic \(z=-1.85\), the P - value is the probability that \(Z < - 1.85\).
Using the standard normal distribution table or a calculator with a normal distribution function (\(P(Z < z)\) for \(z=-1.85\)), we know that \(P(Z < - 1.85)=0.032\) (from the standard normal table: \(P(Z < - 1.85)\) corresponds to the area to the left of \(z =-1.85\). Looking up in the standard normal table, \(P(Z < - 1.85)=0.0322\approx0.032\))

Answer:

The P - value is \(0.032\)