In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 69856985 subjects randomly

Question

kondor19780726 [428]

1 year ago

5

In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 69856985 subjects randomly

selected from an online group involved with ears. There were 13221322 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.

Mathematics

1 answer:

SVETLANKA909090 [29] · Answer 1 · 2020-08-08T07:24:27+03:00

Answer:

We can't conclude that the return rate is less than 20%.

Step-by-step explanation:

We have a sample of size n=6985 and a success rate of p=1322/6985=0.1892.

We have to test the claim that the return rate is less than p=0.2, so the null and alternative hypothesis are:

$H_0: \pi\geq0.2\\\\H_1: \pi$

The significance level is 0.01.

The standard deviation for this sample is

$\sigma=\sqrt{\frac{\pi(1-\pi)}{N} }= \sqrt{\frac{0.2(1-0.2)}{6985}}=0.0048$

Then, the test statistic is

$z=\frac{p-\pi}{\sigma} =\frac{0.1892-0.2}{0.0048}= -2.237$

We look up the P-value in a z-table for z=-2.237.

The P-value is 0.01264.

This P-value is greater than the significance level, so the effect is not significant and the null hypothesis can't be rejected.

We can't conclude that the return rate is less than 20%.

In a study of cell phone usage and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 69856985 subjects randomly

In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 69856985 subjects randomly