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.
a) Compute the standard error

b) Compute the test statistic

c) What is the p-value?

d) Develop appropriate hypothesis such as the reject of the null will support the contention that management efficiency measures had reduce gas prices in Europe.

e) At α = 0.05, what is your conclusion? Use critical value approach

f) Repeat the preceding hypothesis test using the p-value approach.

Answer 1

Answer:

a) $SE= \frac{\sigma}{\sqrt{n}}= \frac{0.14}{\sqrt{49}}= 0.02$

b) $t =\frac{1.20-1.25}{0.02}= -2.5$

c) $p_v =P(z$

d) Null hypothesis: $\mu \geq 1.25$

Alternative hypothesis: $\mu$

e) For this case we need to find a critical value who accumulate 0.05 of the area in the left at the normal standard distribution and we got:

$z_{crit}= -1.64$

Since our calculated value is lower than the critical value we have enough evidence to reject the null hypothesis at the significance level provided of 5%

f) $p_v = P(z$

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis at 5% of significance. Same result with the critical value approach.

Step-by-step explanation:

For this case we want to test is the true mean for the gasoline prices are significantly reduced from $1.25 per liter, so then the system of hypothesis are:

Null hypothesis: $\mu \geq 1.25$

Alternative hypothesis: $\mu$

Part a

The standard error for this case is given by:

$SE= \frac{\sigma}{\sqrt{n}}= \frac{0.14}{\sqrt{49}}= 0.02$

Part b

The statistic for this hypothesis is given by:

$t = \frac{\bar X -\mu}{SE}$

And replacing the info provided we got:

$t =\frac{1.20-1.25}{0.02}= -2.5$

Part c

Since we are conducting a left tailed test the p value would be given by:

$p_v =P(z$

And we can use the following excel code to find it:

=NORM.DIST(-2.5,0,1,TRUE)

Part d

Null hypothesis: $\mu \geq 1.25$

Alternative hypothesis: $\mu$

Part e

For this case we need to find a critical value who accumulate 0.05 of the area in the left at the normal standard distribution and we got:

$z_{crit}= -1.64$

Since our calculated value is lower than the critical value we have enough evidence to reject the null hypothesis at the significance level provided of 5%

Part f

$p_v = P(z$

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis at 5% of significance. Same result with the critical value approach.