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Show your work clearly and completely to get full credit (Five-step process). NO WORK, NO CREDIT! The Mississippi Department of Transportation collected data on the number of cracks (called crack intensity) in an undivided two-lane highway using van-mounted state of the art video technology. The mean number of cracks found in a sample of eight 50-meter sections of the highway was x = 0.210, with a standard deviation of s = 0.105. Suppose the American Association of State Highway and Transportation Officials (AASHTO) recommends a maximum mean crack intensity of 0.100 for safety purposes. Test the hypothesis at 1% significance level that the true mean crack intensity of the Mississippi highway exceeds the AASHTO recommended maximum.

Explanation / Answer

Given that,
population mean(u)=0.1
sample mean, x =0.21
standard deviation, s =0.105
number (n)=50
null, Ho: <0.1
alternate, H1: >0.1
level of significance, = 0.01
from standard normal table,right tailed t /2 =2.405
since our test is right-tailed
reject Ho, if to > 2.405
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =0.21-0.1/(0.105/sqrt(50))
to =7.408
| to | =7.408
critical value
the value of |t | with n-1 = 49 d.f is 2.405
we got |to| =7.408 & | t | =2.405
make decision
hence value of | to | > | t | and here we reject Ho
p-value :right tail - Ha : ( p > 7.4078 ) = 0
hence value of p0.01 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: =0.1
alternate,it excceed the mississippi exceeds tha aasto H1: >0.1
test statistic: 7.408
critical value: 2.405
decision: reject Ho
p-value: 0
claim that excceed the mississippi exceeds tha aasto

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