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

false

step-by-step explanation:

a.) list the factors of 40.

1/2/4/5/8/10/20/40

b.) list the factors of 40 that represent the number of cans that could be on each shelf.

5

c.) how many different arrangements of cans are possible?

just one because if you put more on one shelf you will go way over 40 and the cabinet will probably break.

d.) what are the possible arrangements of the cans?

put 5 cans on each self to reach 40 cans in total.

e.) which arrangement uses all the shelves?

5 because you can put 5 cans on each shelf, and since there are 8 shelfs you are equally going to get 40 cans.