Question

according to a certain government traffic safety 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 0.34. suppose a random sample of 102 traffic fatalities in a certain region with a large population results in 42 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 α = 0.05 level of significance?
start by determining whether the conditions for this hypothesis test are satisfied.
because ( np_0(1 - p_0) = 22.9>10 ), the sample size is less than 5% of the population size, and the sample is given to be random, the requirements for testing the hypothesis about the population proportion are satisfied.
(round to one decimal place as needed.)
what are the null and alternative hypotheses?
( h_0: p = 0.34 ) versus ( h_1: p>0.34 )
(type integers or decimals. do not round.)
find the test statistic, ( z_0 ).
( z_0=square ) (round to two decimal places as needed.)

Explanation:

Step1: Calculate the sample proportion $\hat{p}$

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 42$ (number of successes) and $n=102$ (sample size). So, $\hat{p}=\frac{42}{102}\approx0.412$

Step2: Calculate the test - statistic $z_{0}$

The formula for the test - statistic in a one - sample proportion test is $z_{0}=\frac{\hat{p}-p_{0}}{\sqrt{\frac{p_{0}(1 - p_{0})}{n}}}$

Given $p_{0}=0.34$, $n = 102$, and $\hat{p}\approx0.412$

First, calculate the denominator: $\sqrt{\frac{0.34\times(1 - 0.34)}{102}}=\sqrt{\frac{0.34\times0.66}{102}}=\sqrt{\frac{0.2244}{102}}\approx\sqrt{0.0022}\approx0.047$

Then, calculate the numerator: $\hat{p}-p_{0}=0.412 - 0.34=0.072$

So, $z_{0}=\frac{0.072}{0.047}\approx1.53$

Answer:

$z_{0}\approx1.53$