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 \(19.6\) and standard deviation \(0.4\). Under the same conditions, high-power brand B gasoline gave a sample of \(16\) values with mean \(20.2\) and standard deviation \(0.6\). Is the mileage of B significantly better than that of A? Assume normality. Test the hypothesis using both \(P\)-value and fixed significance level with \(\alpha=0.05\) approaches (if possible).

This question is a type of problem in finding a difference between two means. Before solving anything, we must state our hypotheses first. Since the claim is that the mileage of brand B gasoline is better than brand A gasoline, the following would be our null and alternative hypotheses:

\(H_0:\mu_A-\mu_B=0\)

\(H_a:\mu_A<\mu_B\)

After stating the hypotheses, let’s write down all the given to make it easier for us to plug in the given values in the following formulas to be solved.

Given:

\(n=16\)

\(\bar{x}_A=19.6,\bar{x}_B=20.2\)

\(\sigma_A=0.4,\sigma_B=0.6\)

\(\Delta=0\)

\(\alpha=0.05\)

After writing down all the given, we can now use the formula below to find \(Z_0\).

\[ Z_0=\frac{\bar{x}_A-\bar{x}_B-\Delta}{\sqrt{\frac{\sigma_A^2}{n_A}+\frac{\sigma_B^2}{n_B}}} \]

\[ =\frac{\bar{x}_A-\bar{x}_B-\Delta}{\sqrt\frac{\sigma_A^2+\sigma_B^2}{n}} \]

\[ =\frac{19.6-20.2-0}{\sqrt\frac{0.4^2+0.6^2}{16}} \]

\[ =-3.328201177\approx-3.33 \]

After finding the \(Z_0\), we will now find the \(P\)-value in the Standard Normal Cumulative Probability Table.

\[ P-value=\Phi(-3.33)=0.000434 \] We can also use code chunks in R to find its value.

P <- pnorm(q=-3.33);
P

## [1] 0.0004342299

We now compare it to the given significance level \(\alpha=0.05\). Since \(0.000434<0.05\), we reject the null hypothesis.

\(\therefore\) There is enough evidence to say that the mileage of brand B gasoline is significantly better than brand A gasoline.