In a study of 11,000 car crashes, it was found that 5720 of them occurred within 5 miles of home (based on data from Progressive Insurance). Use a 0.01 significance level to test the claim that more than 50% of car crashes occur within 5 miles of home. Are the results questionable because they are based on a survey sponsored by an insurance company?

Step-by-step explanation:

We would set up the hypothesis test.

a) For the null hypothesis,

p = 0.5

For the alternative hypothesis,

p > 0.5

Considering the population proportion, probability of success, p = 0.5

q = probability of failure = 1 - p

q = 1 - 0.5 = 0.5

Considering the sample,

Sample proportion, P = x/n

Where

x = number of success = 5720

n = number of samples = 11000

P = 5720/11000 = 0.52

We would determine the test statistic which is the z score

z = (P - p)/√pq/n

z = (0.52 - 0.5)/√(0.5 × 0.5)/11000 = 4.195

Using a z test score calculator, the probability value is 0.000014

Since alpha, 0.01 > than the p value, 0.000014, we would reject the null hypothesis.

it means that there is sufficient evidence for us to conclude that more than 50% of car crashes occur within 5 miles of home.

The results are not questionable

6l = 6000 ml

300ml/6000ml

= 300/6000

= 3/60

= 1/20

= 1:20