Question

The mother of a teenager has heard a claim that 29​% of teenagers who drive and use a cell phone reported texting while driving. She thinks that this rate is too high and wants to test the hypothesis that fewer than 29​% of these drivers have texted while driving. Her alternative hypothesis is that the percentage of teenagers who have texted when driving is less than 29​%. Upper H 0 : p equals 0.29 Upper H Subscript a Baseline : p less than 0.29 She polls 40 randomly selected​ teenagers, and 5 of them report having texted while​ driving, a proportion of 0.125. The​ p-value is 0.011. Explain the meaning of the​ p-value in the context of this question.

Answer 1

Answer:

$p_v =P(z$  

The p value is a value in order to reject or not the null hypothesis. If the p value is higher than the significance level we FAIL to reject the null hypothesis otherwise we have enough evidence to reject the null hypothesis.

Step-by-step explanation:

Data given and notation

n=40 represent the random sample taken

X=5 represent the people report having texted while​ driving

$\hat p=\frac{5}{40}=0.125$ estimated proportion of people report having texted while​ driving

$p_o=0.29$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

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

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.29.:  

Null hypothesis:$p\geq 0.29$  

Alternative hypothesis:$p < 0.29$  

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 $p_o$.

Calculate the statistic  

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

$z=\frac{0.125 -0.29}{\sqrt{\frac{0.29(1-0.29)}{40}}}=-2.30$  

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

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

$p_v =P(z$  

The p value is a value in order to reject or not the null hypothesis. If the p value is higher than the significance level we FAIL to reject the null hypothesis otherwise we have enough evidence to reject the null hypothesis.