Question

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 () is $0.14.
1. At the 0.05 level of confidence, what is the correct statistical decision?

a. Reject the null hypothesis
b. Fail to reject the null hypotheses
c. Accept the null hypothesis
d. Cannot make a statistical decision with this information

The p-value for this problem is:

a. 0.4938
b. 0.0062
c. 0.0124
d. 0.05

Answer 1

Answer:

1. Option A) Reject the null hypothesis      

2. Option B)  0.0062

Step-by-step explanation:

We are given the following in the question:

1. Population mean, μ = $1.25 per liter

Sample mean, $\bar{x}$ = $1.20 per lite

Sample size, n = 49

Alpha, α = 0.05

Population standard deviation, σ = $0.14

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 1.25\text{ dollars per liter}\\H_A: \mu < 1.25\text{ dollars per liter}$

We use one-tailed z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{1.20 - 1.25}{\frac{0.14}{\sqrt{49}}}= -2.5$

Now, $z_{critical} \text{ at 0.05 level of significance } = -1.64$

Since,  

$z_{stat} < z_{critical}$

We reject the null hypothesis and accept the alternate hypothesis.

Option A) Reject the null hypothesis

Thus, there is enough evidence to support the claim that efficiency measures have reduced the prices.

2. We can calculate the p value from the standard table.

P-value = 0.0062

Option B)  0.0062