Question

In a study of cell phone usage and brain hemispheric? dominance, an Internet survey was? e-mailed to 6983 subjects randomly selected from an online group involved with ears. There were 1322 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.

Answer 1

`p_v =P(z<-2.298)=0.0107`

If we compare the pvalue and the significance level given
`\alpha=0.01` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion of surveys returned is not significantly less than 0.2 or 20% .

Explanation:

1) Data given and notation

n=6983 represent the random sample taken

X=1322 represent the number of surveys returned

`\hat p=(1322)/(6983)=0.189` estimated proportion of surveys returned

`p_o=0.2` is the value that we want to test

`\alpha=0.01` represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

`p_v` represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the return rate is less than 20% or 0.2.:

Null hypothesis:
`p\geq 0.2`

Alternative hypothesis:
`p <0.2`

When we conduct a proportion test we need to use the z statistic, and the is given by:

`z=\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}}` (1)

The One-Sample Proportion Test is used to assess whether a population proportion
`\hat p` is significantly different from a hypothesized value
`p_o`.

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

`z=\frac{0.189 -0.2}{\sqrt{(0.2(1-0.2))/(6983)}}=-2.298`

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided
`\alpha=0.01`. The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

`p_v =P(z<-2.298)=0.0107`

If we compare the pvalue and the significance level given
`\alpha=0.01` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion of surveys returned is not significantly less than 0.2 or 20% .