Question

In a study of

11 comma 00011,000

car​ crashes, it was found that

5544

of them occurred within 5 miles of home​ (based on insurance company​ data). Use a 0.01 significance level to test the claim that more than​ 50% of car crashes occur within 5 miles of home. Use this information to answer the following questions.

z =

​(Round to two decimal places as​ needed.)

c. What is the​ P-value?

​P-value=

​(Round to four decimal places as​ needed.)

d. What is the​ conclusion?

e. Are the results questionable because they are based on a survey sponsored by an insurance​ company?

Answer 1

Data:    

n = 11000   

p = 0.5   

p' = 5544/11000 = 0.504   

Hypotheses:   

Ho: p ≤ 0.5   

Ha: p > 0.5   

Decision Rule:   

α = 0.01   

Critical z- score = 2.326347874

Reject Ho if z > 2.326347874

Test Statistic:   

SE = √{(p (1 - p)/n} = √(0.5 * (1 - 0.5)/√11000) = 0.004767313

z = (p' - p)/SE = (0.504 - 0.5)/0.00476731294622796 = 0.84

p- value = 0.2007

Decision (in terms of the hypotheses):

Since 0.8390471 < 2.326347874 we fail to reject Ho

Conclusion (in terms of the problem):

There is no sufficient evidence that the proportion is greater than 50%

The results are not questionable because we have used a strict value for α (0.01)