Only answers to red questions needed! Speed of a randomly chosen car in a specif

Question

Only answers to red questions needed! Speed of a randomly chosen car in a specific section is a random variable X. We know that = 8 . Speed for the cars is independent of each other. and Speed limit is 60 km/h. How big share of the drivers were speeding? What is the probability that a randomly chosen car had speed between 60 and 70 km/h ? = 60 and empirical In a speed control we check n = 30 cars. Results have the average Y standard deviation S 8.4. Can we use this data to assert that average speed in the road section have increased (and therefore is more than 56 km/h)? Do a null hypothesis and test it it with significance of 0.05. Which hypothesis will you choose? Ho: 256 to H 1: 56 56 What is your conclusion? A Accept Ho B Discard Ho C Neither-there isn't enough data for conclusion

Explanation / Answer

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 56
standard Deviation ( sd )= 8
a.
P(X > 60) = (60-56)/8
= 4/8 = 0.5
= P ( Z >0.5) From Standard Normal Table
= 0.3085
30.85% are speeding more than 60 Km/h
b.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 60) = (60-56)/8
= 4/8 = 0.5
= P ( Z <0.5) From Standard Normal Table
= 0.6915
P(X < 70) = (70-56)/8
= 14/8 = 1.75
= P ( Z <1.75) From Standard Normal Table
= 0.9599
P(60 < X < 70) = 0.9599-0.6915 = 0.2685

c.
Given that,
population mean(u)=56
sample mean, x =60
standard deviation, s =8.4
number (n)=30
null, Ho: <=56
alternate, H1: >56
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.6991
since our test is right-tailed
reject Ho, if to > 1.6991
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =60-56/(8.4/sqrt(30))
to =2.608
| to | =2.608
critical value
the value of |t | with n-1 = 29 d.f is 1.6991
we got |to| =2.608 & | t | =1.6991
make decision
hence value of | to | > | t | and here we reject Ho
p-value :right tail - Ha : ( p > 2.6082 ) = 0.00712
hence value of p0.05 > 0.00712,here we reject Ho
ANSWERS
---------------
null, Ho: <=56 , alternate, H1: >56
test statistic: 2.608
critical value: 1.6991
decision: reject Ho
p-value: 0.00712

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