Question

In a study of cell phone usage and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 6967 subjects randomly selected from an online group involved with ears. There were 1308 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.

Answer 1

Following are the response to the given question:

Explanation:

Testing hypothesis:

`H_0: p \geq 0.2 \\\\H_a: p < 0.2`

When it's a lower tailed test:

`\hat{p}=(x)/(n)=(1308)/(6967)\approx 0.1877\\\\n= 6967\\\\`

claimed proportion,
`P = 0.2`

Significance level,
`\alpha= 0.01`

calculating statistics :

`\hat{p}, \sigma_{\hat{p}}=\sqrt{(P * (1-p))/(n)}\\\\`

`=\sqrt{(0.2 * (1-0.2))/(6967)}\\\\ =\sqrt{(0.2 -0.04)/(6967)}\\\\ =\sqrt{(0.16)/(6967)} \\\\ =√(2.2965)\approx 1.51`