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Suppose 5 automobiles are randomly selected. One gallon of regular unleaded gaso

ID: 3320552 • Letter: S

Question

Suppose 5 automobiles are randomly selected. One gallon of regular unleaded gasoline in placed into the tank of each car, and each car is driven around a racetrack at a constant speed until all gas is used. The process is repeated with one gallon of unleaded premium gasoline.    The number of miles each car travelled, on each type of gasoline, is recorded. Let 1 = population mean of Gasoline 1, let 2 = population mean of Gasoline 2.       The data is as follows:

A researcher is interested in testing whether, in a large population of cars, the mean mileages are different when using the two types of gasoline. Assume independent samples.                                                         

At the 10% significance level, perform the appropriate hypothesis test to first make an inference that the population variances of miles run for each type of gasoline are to be assumed equal or unequal.  

Your conclusion is:

Do Not Reject the null; there is not sufficient evidence the population variances differ. Also, the p-value associated with this test is approx 0.80.

Do Not Reject the null; there is sufficient evidence the population variances differ. Also, the p-value associated with this test is approx 0.40.

Reject the null; there is sufficient evidence the population variances differ. Also, the p-value associated with this test is approx 0.40.

none of the above

Car Gasoline 1 Gasoline 2 1 10 21 2 18 24 3 12 20 4 11 23 5 13 27

Explanation / Answer

The statistical software output for this problem is:

Two sample variance hypothesis test:
12 : Variance of Gasoline 1
22 : Variance of Gasoline 2
12/22 : Ratio of two variances
H0 : 12/22 = 1
HA : 12/22 1

Hypothesis test results:

Hence,

Do Not Reject the null; there is not sufficient evidence the population variances differ. Also, the p-value associated with this test is approx 0.80.

Option A is correct.

Ratio Num. DF Den. DF Sample Ratio F-Stat P-value 12/22 4 4 1.2933333 1.2933333 0.8092
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