Question

This Test: 60 pts possible In a study of 11,000 car crashe was found that 5700 of them occurred within mles of home based on a company data 2001 significance level to test the camat more than 50% of car crashes within 5 mies of home Use this information to answer the following questions. HD70.5 OEM #05 О, но Hip-05 Hyp:05

Answer 1

Step 1:

NULL hypothesis : P=0.5

он

Alternative hypothesis$H_{1}$: P>0.5

Step 2: $\alpha$ =0.01

Step 3:

Let n=sample size=11,000=number car crashes.

X=Number of car crashes occur with in 5 miles=5709

p =x/n=5709/11000=0.519

the test statistic is

P Z P P(1-P)/n

Where Z Is standrad normal variate.

$Z=\frac{0.5-0.519}{\sqrt{0.519\left ( 1-0.519 \right )/11000}}$

   = -0.0003

Step 4: the critical value of Z at level of significance 0.01 is 2.33

The calculated value of Z Is less than critical value of Z.

We accept NULL Hypothesis.

P=0.5

There no sufficient evidence to accept claim.