Question

In a study of cell phone usage and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 6997 subjects randomly selected from an online group involved with ears. There were 12961296 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

Answer:

$z=\frac{0.185 -0.2}{\sqrt{\frac{0.2(1-0.2)}{6997}}}=-3.137$  

$p_v =P(z$  

Since the p value is very low we have enough evidence to reject the null hypothesis at the significance lvel given of 0.01 so then we can conclude that the return rate ofr this case is significantly lower than 20%

Step-by-step explanation:

Data given

n=6997 represent the random sample taken

X=1296 represent the returned surveys

$\hat p=\frac{1296}{6997}=0.185$ estimated proportion of  returned surveys

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

z would represent the statistic (variable of interest)

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

System of hypothesis

We need to conduct a hypothesis in order to test the claim that the true proportion of returned surveys is less than 0.2.:  

Null hypothesis:$p\geq 0.2$  

Alternative hypothesis:$p < 0.2$  

The statistic is given by:

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

Calculate the statistic  

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

$z=\frac{0.185 -0.2}{\sqrt{\frac{0.2(1-0.2)}{6997}}}=-3.137$  

Statistical decision  

The next step would be calculate the p value for this test.  

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

$p_v =P(z$  

Since the p value is very low we have enough evidence to reject the null hypothesis at the significance lvel given of 0.01 so then we can conclude that the return rate ofr this case is significantly lower than 20%