Question

1. The manufacturer of a gasoline additive claims that the use of this additive increases gasoline mileage. A random sample of six cars was selected, and these cars were driven for 1 week without the gasoline additive and then for 1 week with the gasoline additive. The following table gives the miles per gallon for these cars without and with the gasoline additive.

Without	24.6	28.3	18.9	23.7	15.4	29.5
With	26.3	31.7	18.2	25.3	18.3	30.9

a. Assume that the population of paired differences is approximately normally distributed. You would like to test whether the use of gasoline additive increases the gasoline mileage. Let μ_d denote the mean difference in miles per gallon for cars with and without the gasoline additive. Calculate the test statistic. Round your intermediate calculations and final answer to the nearest hundredth.

b. Let μ_d denote the mean difference in miles per gallon for cars with and without the gasoline additive. The test statistic for this case is 2.97. Calculate the p-value. Round your final answer to the nearest ten thousandth (e.g, 0.1234).

Answer 1

a) We first compute the mean difference and the standard deviation of the differences as:

Without	24.6	28.3	18.9	23.7	15.4	29.5
With	26.3	31.7	18.2	25.3	18.3	30.9
d	-1.7	-3.4	0.7	-1.6	-2.9	-1.4	-10.3
(d - Mean(d))^2	0.00027778	2.83361111	5.84027778	0.01361111	1.40027778	0.10027778	10.1883333

The last column shows the sums for that particular row,

Using this, we obtain the standard deviation here as:

Therefore the test statistic now is computed here as:

Therefore -2.97 is the required test statistic value here. Note that it does not matter whether its negative or positive as it is a two tailed test.

b) As this is a two tailed test, the p-value here is computed from the  t distribution tables for n - 1 = 5 degrees of freedom as:

p = 2P(t₅ < -2.97) = 2*0.0156

= 0.0312

Therefore 0.0312 is the required p-value here.