Question

in a study of cell phone usage and brain hemispheric dominance, an internet survey was e - mailed to 6970 subjects randomly selected from an online group involved with ears. there were 1331 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. ( h_0:p
eq0.2 )
( h_1:p = 0.2 )
b. ( h_0:p>0.2 )
( h_1:p = 0.2 )
c. ( h_0:p = 0.2 )
( h_1:p>0.2 )
d. ( h_0:p = 0.2 )
( h_1:p<0.2 )
e. ( h_0:p = 0.2 )
( h_1:p
eq0.2 )
f. ( h_0:p<0.2 )
( h_1:p = 0.2 )

the test statistic is ( z=-1.88 ).
(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 \(20\% = 0.2\), this is a left - tailed test.

Step2: Use the standard normal distribution to find the P - value

For a left - tailed test with test statistic \(z=-1.88\), the P - value is the probability that \(Z < - 1.88\).
Using the standard normal table or a calculator with a normal distribution function (e.g., in Excel: =NORM.S.DIST(-1.88,TRUE)), we find that \(P(Z < - 1.88)\)

$$P(Z < - 1.88)=0.030153$$

Answer:

\(0.030\)