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.
identify the null hypothesis and alternative hypothesis.
a. ( h_0:p > 0.2 )
( h_1:p = 0.2 )
b. ( h_0:p
eq0.2 )
( 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
eq0.2 )
e. ( h_0:p = 0.2 )
( h_1:p > 0.2 )
f. ( h_0:p < 0.2 )
( h_1:p = 0.2 )
the test statistic is ( z=)

Explanation:

Step1: Calculate the sample proportion

The sample proportion \(\hat{p}=\frac{x}{n}\), where \(x = 1300\) (number of successes) and \(n=6989\) (sample size). So, \(\hat{p}=\frac{1300}{6989}\approx0.186\)

Step2: Calculate the test - statistic

The formula for the test - statistic \(z\) in a proportion test is \(z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}\), where \(p = 0.2\) (the proportion in the null hypothesis), \(\hat{p}\approx0.186\), and \(n = 6989\)

First, calculate the denominator \(\sqrt{\frac{p(1 - p)}{n}}=\sqrt{\frac{0.2\times(1 - 0.2)}{6989}}=\sqrt{\frac{0.2\times0.8}{6989}}=\sqrt{\frac{0.16}{6989}}\approx\sqrt{0.0000229}\approx0.00478\)

Then, calculate the numerator \(\hat{p}-p=0.186 - 0.2=- 0.014\)

So, \(z=\frac{-0.014}{0.00478}\approx - 2.93\)

Answer:

\(z\approx - 2.93\)