Question

According to a certain government agency for a large​ country, the proportion of fatal traffic accidents in the country in which the driver had a positive blood alcohol concentration​ (BAC) is 0.39. Suppose a random sample of 105 traffic fatalities in a certain region results in 54 that involved a positive BAC. Does the sample evidence suggest that the region has a higher proportion of traffic fatalities involving a positive BAC than the country at the alpha equals 0.1 level of​ significance?

Answer 1

Answer:

The sample evidence suggest that the region has a higher proportion of traffic fatalities involving a positive BAC than the country.

Step-by-step explanation:

In this case we have to perform a hypothesis test on a proportion.

The null and alternative hypothesis are:

$H_0:\pi\leq0.39\\\\H_1:\pi>0.39$

The level of significance is 0.1.

The standard deviation for the proportion can be calculated as:

$\sigma=\sqrt{\frac{\pi(1-\pi)}{N} }= \sqrt{\frac{0.39(1-0.39)}{105} }=\sqrt{\frac{0.2379}{105} }= 0.0476$

The proportion of the sample is

$p=\frac{54}{105}= 0.51$

The test statistic is

$z=\frac{p-\pi-0.5/N}{\sigma} =\frac{0.51-0.39-0.5/105}{0.0476} =\frac{ 0.1152 }{0.0476}=2.421$

The one-tail P-value for z=2.421 is $P(z>2.421)=0.00774$.

The P-value is smaller than the significance level, so the effect is significant.

The null hypothesis is rejected: the sample evidence suggest that the region has a higher proportion of traffic fatalities involving a positive BAC than the country.