A researcher collects crash data at 12 intersections in NYC and found an average crash Due: Wednesday, 12/6 rate of 4.76 crashes/year with a standard deviation of 0.47 crashes/year a. Is the crash data at these 12 intersections different than the city average of 4.56 crashes/year at a 20% level of significance? Define your null and alternative hypothesis. Show your work. b. Is the crash data at these 12 intersections different than the city average of 4.56 crashes/year at a 5% level of significance? Define your null and alternative hypothesis. Show your work.

Given:

Sample mean, n = 12

Sample mean of crashes per year, X' = 4.76

Std deviation, s.d = 0.47

a) Given:

u = 4.56

Significance level = 20% = 0.2

For null and alternative hypothesis, we have:

H0 : u = 4.56

Ha : u ≠ 4.56

This is a two-tailed test.

Given a small sample size and standard deviation was not given, we test if the crash data at the 12 intersections is different from the city average of 4.56 using the t-test.

= 1.4741

Degree of freedom, df = n - 1

df = 12 - 1 = 11

Therefore from the distribution table at

T.DIST.2T( 1.4741, 11)

p-value = 0.1685

Since the p-value(0.1685) is less than level of significance (0.2) we reject the null hypothesis, H0.

We can say there is enough evidence to show that the crash data at the 12 intersections at 20% significance level is different from the city average of 4.56

b) Given:

u = 4.56

Significance level = 5% = 0.05

This is the same as part (a), apart from a different level of significance.

H0 : u = 4.56

Ha : u ≠ 4.56

= 1.4741

Degree of freedom, df = n - 1

df = 12 - 1 = 11

Therefore from the distribution table at

T.DIST.2T( 1.4741, 11)

p-value = 0.1685

Since p-value(0.1685) is greater than level of significance (0.05), we fail to reject the null hypothesis.

We can say there is not enough statistical evidence that ctash data at the 12 intersections at 5% significance level is different from the city average (4.56)

abde or 1,2,4, 5 is the answer

step-by-step explanation: