Question

a major oil company has developed a new gasoline additive that is supposed to increase mileage. to test this hypothesis, ten cars are randomly selected. the cars are driven both with and without the additive. the results are displayed in the following table. can it be concluded, from the data, that the gasoline additive does significantly increase mileage? let (d = (\text{gas mileage with additive})-(\text{gas mileage without additive})). use a significance level of (alpha = 0.05) for the test. assume that the gas mileages are normally distributed for the population of all cars both with and without the additive. step 2 of 5: find the value of the standard deviation of the paired differences. round your answer to two decimal places.

Explanation:

Step1: Calculate the paired - differences \(d\)

For each car, \(d_i=( \text{With additive})_i-( \text{Without additive})_i\).
For car 1: \(d_1 = 17.3−16.6 = 0.7\)
For car 2: \(d_2=13.2 - 11=2.2\)
For car 3: \(d_3 = 30−29.3 = 0.7\)
For car 4: \(d_4=29.5 - 28.5 = 1\)
For car 5: \(d_5=28.7 - 26.5 = 2.2\)
For car 6: \(d_6=23.2 - 20.1 = 3.1\)
For car 7: \(d_7=19.1 - 16 = 3.1\)
For car 8: \(d_8=28.4 - 25.4 = 3\)
For car 9: \(d_9=13.8 - 10.7 = 3.1\)
For car 10: \(d_{10}=20.4 - 19.3 = 1.1\)

Step2: Calculate the mean of the paired - differences \(\bar{d}\)

\(\bar{d}=\frac{\sum_{i = 1}^{n}d_i}{n}\), where \(n = 10\) and \(\sum_{i=1}^{10}d_i=0.7 + 2.2+0.7+1+2.2+3.1+3.1+3+3.1+1.1 = 19.3\)
\(\bar{d}=\frac{19.3}{10}=1.93\)

Step3: Calculate the squared differences \((d_i-\bar{d})^2\)

For \(d_1 = 0.7\): \((0.7 - 1.93)^2=(-1.23)^2 = 1.5129\)
For \(d_2 = 2.2\): \((2.2-1.93)^2=(0.27)^2 = 0.0729\)
For \(d_3 = 0.7\): \((0.7 - 1.93)^2=(-1.23)^2 = 1.5129\)
For \(d_4 = 1\): \((1 - 1.93)^2=(-0.93)^2 = 0.8649\)
For \(d_5 = 2.2\): \((2.2-1.93)^2=(0.27)^2 = 0.0729\)
For \(d_6 = 3.1\): \((3.1 - 1.93)^2=(1.17)^2 = 1.3689\)
For \(d_7 = 3.1\): \((3.1 - 1.93)^2=(1.17)^2 = 1.3689\)
For \(d_8 = 3\): \((3 - 1.93)^2=(1.07)^2 = 1.1449\)
For \(d_9 = 3.1\): \((3.1 - 1.93)^2=(1.17)^2 = 1.3689\)
For \(d_{10}=1.1\): \((1.1 - 1.93)^2=(-0.83)^2 = 0.6889\)
\(\sum_{i = 1}^{10}(d_i-\bar{d})^2=1.5129+0.0729+1.5129+0.8649+0.0729+1.3689+1.3689+1.1449+1.3689+0.6889 = 9.931\)

Step4: Calculate the standard deviation of the paired - differences \(s_d\)

The formula for the standard deviation of paired - differences is \(s_d=\sqrt{\frac{\sum_{i = 1}^{n}(d_i-\bar{d})^2}{n - 1}}\)
\(s_d=\sqrt{\frac{9.931}{9}}\approx1.05\)

Answer:

\(1.05\)