Question 2

Brand A gasoline was used in \(16\) similar automobiles under identical conditions. The corresponding sample of \(16\) values (miles per gallon) had mean \(\bar{x_1} = 19.6\) and standard deviation \(s_1 = 0.4\). Under the same conditions, high-power brand B gasoline gave a sample of \(16\) values with mean \(\bar{x_2} = 20.2\) and standard deviation \(s_2 = 0.6\). Is the mileage of B significantly better than that of A? Assume normality. Test the hypothesis using both \(P\)-value and fixed significance level \(\alpha = 0.05\) approaches (if possible).

The question does not mention the variances of both populations, and the variances are not assumed equal.

Applying the 7-step procedure for Hypothesis Testing,

1. Parameter of Interest: The parameter of interest is the difference in mean miles per gallon, and \(\Delta_0 = 0\).


2. Null hypothesis: \(H_0 : \mu_A - \mu_B = 0\) or \(\mu_A = \mu_B\)


3. Alternative hypothesis: \(H_1 : \mu_A < \mu_B\). We want to reject \(H_0\) if brand B gasoline has better mileage than brand A under identical conditions.


4. Test Statistic:

The formula for the test statistic is given as:

\[\begin{aligned} t_0^* = \displaystyle\frac{\bar{x_1} - \bar{x_2} - \Delta_0}{\sqrt{\displaystyle\frac{s_1^2}{n_1} + \displaystyle\frac{s_2^2}{n_2}}} \end{aligned}\]


5. Reject \(H_0\) if:

The degrees of freedom on \(t_0^*\) are found to be

\[\begin{aligned} v = \displaystyle\frac{(\displaystyle\frac{s_1^2}{n_1} + \displaystyle\frac{s_2^2}{n_2})^2}{\displaystyle\frac{(s_1^2/n_1)^2}{n_1-1} + \displaystyle\frac{(s_2^2/n_2)^2}{n_2-1}} \end{aligned}\]

with values:

\(n_1 = n_2 = 16\)

\(s_1 = 0.4, s_2 = 0.6\)

Substituting these values:

\[\begin{aligned} v = \displaystyle\frac{(\displaystyle\frac{0.4^2}{16} + \displaystyle\frac{0.6^2}{16})^2}{\displaystyle\frac{(0.4^2/16)^2}{16-1} + \displaystyle\frac{(0.6^2/16)^2}{16-1}} = 26.13402062 \end{aligned}\]


Fixed Significance Level Test

qt(0.05, v)
## [1] -1.705295

Reject \(H_0\) if \(t_0^* < -t_{\alpha, v} = -1.705295\)


\(P\)-value Test

\(\alpha = 0.05\)

Reject \(H_0\) if \(P_0 < 0.05\)


6. Computations

\(n_1 = n_2 = 16\)

\(s_1 = 0.4, s_2 = 0.6\)

\(\bar{x_1} = 19.6, \bar{x_2} = 20.2\)

\(\Delta_0 = 0\)

Substituting these values to get \(t_0^*\):

\[\begin{aligned} t_0^* = \displaystyle\frac{19.6 - 20.2 - 0}{\sqrt{\displaystyle\frac{0.4^2}{16} + \displaystyle\frac{0.6^2}{16}}} = -3.328201177 \end{aligned}\]


7. Conclusion:

Fixed Significance Level Test

Recalling the condition for rejecting \(H_0\), \(t_0^* < -t_{\alpha, v}\), the \(T\)-value for \(\alpha = 0.05\) with \(v\) degrees of freedom is computed to be \(-1.705295\).

Since \(t_0^* = -3.328201177\) is less than \(-t_{0.05, v} = -1.705295\), there is sufficient evidence to reject the null hypothesis \(H_0 : \mu_A - \mu_B = 0\) or \(\mu_A = \mu_B\).


\(P\)-value Test

For the \(P\)-value test, the null hypothesis is rejected when the \(P\)-value is less than \(0.05\). \(P_0\) is computed to be:

pt(t, v)
## [1] 0.001302688

Since \(P_0 = 0.001302688 < \alpha = 0.05\), there is also sufficient evidence to reject the null hypothesis using the \(P\)-value test.

References:

D. C. Montgomery and G. C. Runger, in Applied Statistics and Probability for Engineers, 7th ed., Hoboken, NJ: Wiley, 2018, pp. 193–258.