The average gasoline price of one of the major oil companies

The average gasoline price of one of the major oil companies in Europe has been $1.25 per liter. Recently, the company has undertaken several efficiency measures in order to reduce prices. Management is interested in determining whether their efficiency measures have actually reduced prices. A random sample of 49 of their gas stations is selected and the average price is determined to be $1.20 per liter. Furthermore, assume that the standard deviation of the population (sigma) is $0.14. Compute the standard error. Compute the test statistic. What is the p-value? Develop appropriate hypotheses such as the rejection of the null will support the contention that management efficiency measures had reduce gas prices in Europe. At alpha = 0.05 what is your conclusion? Use the critical value approach. Repeat the preceding hypothesis test using the p-value approach.

Solution

a.
Standard Error= sd/ Sqrt(n)
Where,
sd = Standard Deviation
n = Sample Size

Standard deviation( sd )=0.14
Sample Size(n)=49
Standard Error = ( 0.14/ Sqrt ( 49) )
= 0.02

Set Up Hypothesis
Null Hypothesis H0: U=1.25
Alternate, reduced price H1: U<1.25
b.
Test Statistic
Population Mean(U)=1.25
Given That X(Mean)=1.2
Standard Deviation(S.D)=0.14
Number (n)=49
we use Test Statistic (Z) = x-U/(s.d/Sqrt(n))
Zo=1.2-1.25/(0.14/Sqrt(49)
Zo =-2.5
| Zo | =2.5
d.
Critical Value
The Value of |Z | at LOS 0.05% is 1.64
We got |Zo| =2.5 & | Z | =1.64
e.
Make Decision
Hence Value of | Zo | > | Z | and Here we Reject Ho
f.
P-Value : Left Tail - Ha : ( P < -2.5 ) = 0.0062
Hence Value of P0.05 > 0.0062, Here we Reject Ho