EXAMPLE 27 test for using the -value method
We report to two decimal places to allow the use of the table to calculate the -value.
The National Transportation Safety Board publishes statistics on the number of automobile crashes that people in various age groups have. Young people ages 18–24 have an accident rate of 12%, meaning that on average 12 out of every 100 young drivers per year had an accident. A researcher claims that the population proportion of young drivers having accidents is greater than 12%. Her study examined 1000 young drivers ages 18–24 and found that 134 had an accident this year. Perform the appropriate hypothesis test using the -value method with level of significance .
Solution
First, we check that both of our normality conditions are met. We are interested in whether the proportion has increased from 12%, so we have .
The normality conditions are met and we may proceed with the hypothesis test.
Step 1 State the hypotheses and the rejection rule.
Our hypotheses are
where represents the population proportion of young people ages 18–24 who had an accident. We reject the null hypothesis if the .
Step 2 Calculate .
Our sample proportion is . Because , the standard error of is
Thus, our test statistic is
548
Figure 9.46: FIGURE 36 -Value for a right-tailed test equals area to right of .That is, the sample proportion lies approximately 1.36 standard errors above the hypothesized proportion .
Step 3 Find the -value.
We have a right-tailed test, so our -value from Table 12 is . This is a Case 2 problem from Table 8 in Chapter 6 (page 355), where we find the tail area by subtracting the table area from 1 (Figure 36):
Step 4 State the conclusion and the interpretation.
The -value 0.0869 is not ≤ , so we do not reject . There is insufficient evidence that the population proportion of young people ages 18–24 who had an accident has increased.
NOW YOU CAN DO
Exercises 19–22.