Question

In a study of 420,095 Danish cell phone users, 135 subjects developed cancer of the brain or nervous system (based on data from the Journal of the National Cancer Institute as reported in USA Today). Test the claim of a somewhat common belief that such cancers are affected by cell phone use. That is, test the claim that cell phone users develop cancer of the brain or nervous system at a rate that is different from the rate of 0.0340% for people who do not use cell phones. Because this issue has such great importance, use a 0.005 significance level. Based on these results, should cell phone users be concerned about cancer of the brain or nervous system?

Answer 1

Answer:

$z=\frac{0.0003214 -0.00034}{\sqrt{\frac{0.00034(1-0.00034)}{420095}}}=-0.654$  

$p_v =2*P(Z$  

So the p value obtained was a very high value and using the significance level given $\alpha=0.005$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say that the true proportion not differs significantly from the specified value of 0.00034 or 0.034%.

Step-by-step explanation:

1) Data given and notation  

n=420095 represent the random sample taken

X=135 represent the subjects developed cancer of the brain or nervous system (based on data from the Journal of the National Cancer Institute as reported in USA Today)

$\hat p=\frac{135}{420095}=0.0003214$ estimated proportion of subjects developed cancer of the brain or nervous system (based on data from the Journal of the National Cancer Institute as reported in USA Today)

$p_o=0.00034$ is the value that we want to test

$\alpha=0.005$ represent the significance level

Confidence=99.5% or 0.995

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 brain or nervous system at a rate that is different from the rate of 0.0340% :  

Null hypothesis:$p=0.00034$  

Alternative hypothesis:$p \neq 0.00034$  

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

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (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:  

$z=\frac{0.0003214 -0.00034}{\sqrt{\frac{0.00034(1-0.00034)}{420095}}}=-0.654$  

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 $\alpha=0.005$. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

$p_v =2*P(Z$  

So the p value obtained was a very high value and using the significance level given $\alpha=0.005$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say that the true proportion not differs significantly from the specified value of 0.00034 or 0.034%.