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 = (gas mileage with additive)-(gas mileage without additive). use a significance level of α = 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 8 of 5: compute the value of the test statistic. round your answer to three decimal places.

Explanation:

Step1: Calculate the differences \(d\)

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 differences \(\bar{d}\)

\(\bar{d}=\frac{\sum_{i = 1}^{10}d_i}{n}\), where \(\sum_{i=1}^{10}d_i=0.7 + 2.2+0.7+1+2.2+3.1+3.1+3+3.1+1.1 = 20.3\) and \(n = 10\). So \(\bar{d}=\frac{20.3}{10}=2.03\).

Step3: Calculate the standard - deviation of the differences \(s_d\)

First, calculate \(\sum_{i = 1}^{10}(d_i-\bar{d})^2\).
\((d_1 - \bar{d})^2=(0.7 - 2.03)^2=(-1.33)^2 = 1.7689\); \((d_2 - \bar{d})^2=(2.2 - 2.03)^2=(0.17)^2 = 0.0289\); \((d_3 - \bar{d})^2=(0.7 - 2.03)^2=(-1.33)^2 = 1.7689\); \((d_4 - \bar{d})^2=(1 - 2.03)^2=(-1.03)^2 = 1.0609\); \((d_5 - \bar{d})^2=(2.2 - 2.03)^2=(0.17)^2 = 0.0289\); \((d_6 - \bar{d})^2=(3.1 - 2.03)^2=(1.07)^2 = 1.1449\); \((d_7 - \bar{d})^2=(3.1 - 2.03)^2=(1.07)^2 = 1.1449\); \((d_8 - \bar{d})^2=(3 - 2.03)^2=(0.97)^2 = 0.9409\); \((d_9 - \bar{d})^2=(3.1 - 2.03)^2=(1.07)^2 = 1.1449\); \((d_{10} - \bar{d})^2=(1.1 - 2.03)^2=(-0.93)^2 = 0.8649\).
\(\sum_{i = 1}^{10}(d_i-\bar{d})^2=1.7689+0.0289+1.7689+1.0609+0.0289+1.1449+1.1449+0.9409+1.1449+0.8649 = 9.899\).
\(s_d=\sqrt{\frac{\sum_{i = 1}^{10}(d_i-\bar{d})^2}{n - 1}}=\sqrt{\frac{9.899}{9}}\approx1.049\).

Step4: Calculate the test - statistic \(t\)

The formula for the test - statistic in a paired - samples \(t\) - test is \(t=\frac{\bar{d}-\mu_d}{s_d/\sqrt{n}}\), where \(\mu_d = 0\) (under the null hypothesis).
\(t=\frac{2.03-0}{1.049/\sqrt{10}}\approx\frac{2.03}{0.332}\approx6.114\).

Answer:

\(6.114\)