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.

a)

b)

c)

d) Null hypothesis:

Alternative hypothesis:

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:

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)

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:

Alternative hypothesis:

Part a

The standard error for this case is given by:

Part b

The statistic for this hypothesis is given by:

And replacing the info provided we got:

Part c

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

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

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

Part d

Null hypothesis:

Alternative hypothesis:

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:

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

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.

b

step-by-step explanation:

u take $75 minus $60.the anwer/original($75)multiply by 100

there are 70 more legs than eyes at the dog show (196 – 126 = 70). dogs have two more legs than eyes, and humans have the same number of legs as eyes. thus 70 is twice the number of dogs at the show, so the number of dogs is half of 70, which is 35. to get the number of humans, subtract 35 from 63: 63 – 35 = 28.

step-by-step explanation: