The average gasoline price of one of the major oil companies in Europe has been $1.25 per liter. Recently, the company has undertaken severa

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.

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1 week 2021-11-21T13:04:43+00:00 1 Answer 0 views 0

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    2021-11-21T13:06:38+00:00

    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<-2.5) =0.00621

    d) Null hypothesis:  \mu \geq 1.25

    Alternative hypothesis: \mu<1.25

    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<-2.5) = 0.00621

    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<1.25

    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<-2.5) =0.00621

    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<1.25

    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<-2.5) = 0.00621

    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.

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