A company with a large fleet of cars hopes to keep gasoline

A company with a large fleet of cars hopes to keep gasoline costs down and sets a goal of attaining a fleet average of at least 28 mile per gallon (mpg). To see if the goal if being met, they check the gasoline usage for 50 company trips chosen at random, finding a mean of 27.02 mpg and a standard deviation of 3.9 mpg. Is this strong evidence that they have failed to attain their fuel economy goal? Write appropriate hypotheses. Are the necessary assumption to perform inference satisfied? Test the hypothesis and find the P-value. Explain what the P-value means in this context. Construct a 95% confidence interval for the mean mpg. State an appropriate conclusion.

Solution

Set Up Hypothesis
Null, H0: U>=28
Alternate, H1: U<28
Test Statistic
Population Mean(U)=28
Sample X(Mean)=27.02
Standard Deviation(S.D)=3.9
Number (n)=50
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =27.02-28/(3.9/Sqrt(50))
to =-1.777
| to | =1.777
Critical Value
The Value of |t | with n-1 = 49 d.f is 1.677
We got |to| =1.777 & | t | =1.677
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value :Left Tail -Ha : ( P < -1.7768 ) = 0.0409
Hence Value of P0.05 > 0.0409,Here we Reject Ho

They are failed to attain their fuel economy goal

[ANSWERS]
1. H0: U>=28
3. to =-1.777, ( P < -1.7768 ) = 0.0409
4.
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=27.02
Standard deviation( sd )=3.9
Sample Size(n)=50
Confidence Interval = [ 27.02 ± t a/2 ( 3.9/ Sqrt ( 50) ) ]
= [ 27.02 - 2.01 * (0.552) , 27.02 + 2.01 * (0.552) ]
= [ 25.911,28.129 ]

5. They are failed to attain their fuel economy goal