Question

This le 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) 0 34 Suppose a random sample of 11 trafic alles in catalogas rests in 51 that we spesie BAC Does the suggest that the region has a higher proportion of traffic fatalities involving a positive BAC than the country at the 01 level of significance? Because npo (-P)-010, the sample size is 5% of the population size and the sample the requirements for testing the hypothesis (Round to one decimal place as needed) What are the null and alterative hypotheses? Ho versus H, (Type integers or decimals. Do not found) Find the test statistico TO (Round to two decimal places as needed Find the p-value P value - (Round to three decimal places as needed) Determine the conclusion for this hypothesis test Choose the correct answer below O A. Since P valueco do not reject the null hypothesis and conclude that there is not sufficient evidence that the region has a higher proportion of wat tans imoving a BAC than the country OB. Since P-value do not reject the null hypothesis and conclude that there is not sufficient evidence that the region has a higher proportion of traffic fatis in a positive BAC than the country OC. Since P-value o reject the nuli hypothesis and conclude that there is sollicent evidence that the region has a higher proportion of actualtos vingapore BAC than the country O D. Since P-value > reject the null hypothesis and conclude that there is sufficient evidence that the region has a proportion or states ogspoove BAC than the county Click to select your answers)

Answer 1

Answer:

Where np(1-p)= 111*0.34*(1-0.34)=24.9084 $\geq$ 10 , the sample size is less than 5% of the population size.  Therefore sampling distribution of sample proportion is approximately Normal.

To test,

=0.34 V/s > 0.34

Given that,

n=111 , x=51 , p₀=0.34

$\because \hat{p}=$x/n = 51 / 111 = 0.459

Use $\alpha =$ 0.1

Test statistic,

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

z=((0.459-0.34)/sqrt(0.34*(1-0.34)/111)

z= 2.6466

# Test statistic=z=2.65

P value= p(z>2.6466)

=1-p(z<2.6466)

=1-0.9959 ( From standard Normal table0

=0.004

# Conclusion:

Since P- value < $\alpha$ , reject the null hypothesis and conclude that there is sufficient evidence that the region has a higher proportion of traffic fatalities involving a positive BAC than country.