Question

The U.S. Department of Transportation, National Highway Traffic Safety Administration, reported that 77% of all fatally injured automobile drivers were intoxicated. A random sample of 55 records of automobile driver fatalities in a certain county showed that 34 involved an intoxicated driver. Do these data indicate that the population proportion of driver fatalities related to alcohol is less than 77% in Kit Carson County? Use α = 0.01. Solve the problem using both the critical region method and the P-value method. Since the sampling distribution of is the normal distribution, you can use critical values from the standard normal distribution (Round any intermediate calculations to at least four decimal places. Round the test statistic and critical value to two decimal places. Round the P-value to four decimal places.)

Answer 1

Solution :

Population proportion, P = 77%

Sample size is given by n = 55

Sample proportion = $\frac{x}{n} = \frac{34}{55}= 0.61$

Null hypothesis

The population proportion of the fatalities of the drivers that are related to alcohol = 77%

$H_0 : P = 77$ %

$H_0 : P = 0.77$

Alternate hypothesis, $H_a : P < 0.77$

Therefore, the test statistics = $\frac{0.61 - 0.77}{\sqrt{\frac{0.7 \times 0.23}{55}}}$

                                                = -3.2

Critical value = -1.64

and the rejection zone : test zone, -1.645

Therefore, we fail rejecting $H_0$

p value 0.0582

rejection zone :: p value , 0.01

So we fail to reject $H_0$

Thus the data does not indicate the population proportion of the drivers fatalities which are related to the alcohols which is less than 77% in the county of Kit Carson.