Question

in a study of cell phone usage and brain hemispheric dominance, an internet survey was e - mailed to 6085 subjects randomly selected from an online group involved with ears. there were 1335 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:plt0.2 )
( h_1:p = 0.2 )
b. ( h_0:p = 0.2 )
( h_1:p
eq0.2 )
c. ( h_0:p
eq0.2 )
( h_1:p = 0.2 )
d. ( h_0:p = 0.2 )
( h_1:plt0.2 )
e. ( h_0:p = 0.2 )
( h_1:pgt0.2 )
f. ( h_0:pgt0.2 )
( h_1:p = 0.2 )
the test statistic is ( z=square )
(round to two decimal places as needed)

Explanation:

Step1: Identify the null and alternative hypotheses

The claim is that the return rate is less than \(20\%\) (or \(p < 0.2\)). The null hypothesis \(H_0\) is a statement of equality. So \(H_0:p = 0.2\). The alternative hypothesis \(H_1\) is the claim we are testing, so \(H_1:p<0.2\).

Step2: Calculate the sample proportion \(\hat{p}\)

The sample proportion \(\hat{p}=\frac{x}{n}\), where \(x = 1335\) (number of successes, i.e., returned surveys) and \(n=6085\) (sample size). So \(\hat{p}=\frac{1335}{6085}\approx0.22\)

Step3: Calculate the test - statistic \(z\)

The formula for the test - statistic \(z\) in a one - sample proportion test is \(z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}\)
Substitute \(p = 0.2\), \(\hat{p}\approx0.22\), and \(n = 6085\) into the formula:

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Answer:

The null hypothesis \(H_0:p = 0.2\) and the alternative hypothesis \(H_1:p<0.2\) (corresponding to option D). The test statistic \(z\approx3.90\)