The average gasoline price of one of the major oil companies has been $2.20 per
ID: 3435444 • Letter: T
Question
The average gasoline price of one of the major oil companies has been $2.20 per gallon. Because of cost reduction measures, it is believed that there has been a significant reduction in the average price. In order to test this belief, we randomly selected a sample of 36 of the company's gas stations and determined that the average price for the stations in the sample was $2.14. Assume that the standard deviation of the population (s) is $0.12.
a.
State the null and the alternative hypotheses.
b.
Compute the test statistic.
c.
What is the p-value associated with the above sample results?
d.
At 95% confidence, test the company's claim.
I was given the answers below, but I am not completely sure on how to come up with them. Please help me by showing the work so I can completly understand!
a.
H0: m (greater than or equal to) 2.20
Ha: m < 2.20
b.
Z = -3
c.
p-value = almost zero (0.0013)
d.
p-value < .05; reject H0; the average price has been reduced.
a.
State the null and the alternative hypotheses.
b.
Compute the test statistic.
c.
What is the p-value associated with the above sample results?
d.
At 95% confidence, test the company's claim.
Explanation / Answer
Test Used: Z-Test For Single Mean
Set Up Hypothesis
Null Hypothesis H0: U=2.2
Alternate Hypothesis H1: U<2.2
Test Statistic
Population Mean(U)=2.2
Given That X(Mean)=2.14
Standard Deviation(S.D)=0.12
Number (n)=36
we use Test Statistic (Z) = x-U/(s.d/Sqrt(n))
Zo=2.14-2.2/(0.12/Sqrt(36)
Zo =-3
| Zo | =3
Critical Value
The Value of |Z a| at LOS 0.05% is 1.64
We got |Zo| =3 & | Z a | =1.64
Make Decision
Hence Value of | Zo | > | Z a| and Here we Reject Ho
P-Value : Left Tail - Ha : ( P < -3 ) = 0.0013
Hence Value of P0.05 > 0.0013, Here we Reject Ho
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