Question

The National Transportation Safety Board publishes statistics on the number of automobile crashes that people in various age groups have. A researcher claims the population proportion of young drivers, ages 18 – 24, having accidents is greater than 12%. Her study examined 1000 young drivers ages 18 – 24 and found that 134 had an accident this year. Test at α = .05. a) State H0 and Ha b) Find the critical value, show work c) Find the appropriate test statistic, show work d) Do a P - value test e) Make a decision to reject or fail to reject the null hypothesis and interpret the results.

Answer 1

Answer:

a) Null hypothesis:$p\leq 0.12$  

Alternative hypothesis:$p > 0.12$  

b) $z_{\alpha}=1.64$

c) $z=\frac{0.134 -0.12}{\sqrt{\frac{0.12(1-0.12)}{1000}}}=1.362$  

d) $p_v =P(z>1.362)=0.0866$  

e) For this case since the statistic is lower than the critical value and the p value higher than the significance level we have enough evidence to FAIL to reject the null hypothesis so then we don't have information to conclude that the true proportion is higher than 0.12

Step-by-step explanation:

Information given

n=1000 represent the random sample selected

X=134 represent the number of young drivers ages 18 – 24 that had an accident

$\hat p=\frac{134}{1000}=0.134$ estimated proportion of young drivers ages 18 – 24 that had an accident

$p_o=0.12$ is the value that we want to verify

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic

$p_v{/tex} represent the p valuePart aWe want to verify if the population proportion of young drivers, ages 18 – 24, having accidents is greater than 12%: Null hypothesis:[tex]p\leq 0.12$  

Alternative hypothesis:$p > 0.12$  

The statistic would be given by:

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

Part b

For this case since we are conducting a right tailed test we need to find a critical value in the normal standard distribution who accumulates 0.05 of the area in the right and we got:

$z_{\alpha}=1.64$

Part c

For this case the statistic would be given by:

$z=\frac{0.134 -0.12}{\sqrt{\frac{0.12(1-0.12)}{1000}}}=1.362$  

Part d

The p value can be calculated with the following probability:

$p_v =P(z>1.362)=0.0866$  

Part e

For this case since the statistic is lower than the critical value and the p value higher than the significance level we have enough evidence to FAIL to reject the null hypothesis so then we don't have information to conclude that the true proportion is higher than 0.12