Answer to Question #294629 in Statistics and Probability for Yaku

Question

b) two types of engines, A and B, were compared. Gas mileage, in miles per gallon, was measured. 50 experiments were conducted using engine type A and 75 experiments were done with engine type B. The gasoline used and other conditions were held constant. The average gas mileage was 36 miles per gallon for engine A and 42 miles per gallon for engine B. Find a 95% condence interval on $\mu$_B - $\mu$_A, where $\mu$_B and $\mu$_Aare true mean gas mileages for engines A and B, respectively. population standard deviations are 6 and 8 for engines A and B, respectively. What can you conclude?

Answer 1

Summary statistics:

Engine $A: \quad n_{1}=50 ; \quad \bar{x}_{1}=36 ; \quad \sigma_{1}=6$

Engine $B: \quad n_{2}=75 ; \quad \bar{x}_{2}=42 ; \quad \sigma_{2}=8$

Confidence level, $1-\alpha=95 \%=0.95$

Implies, level of significance, $\alpha=1-0.95=0.05$

Here, population standard deviations are known, so z-tabulated value $z_{\alpha / 2}$ to find the confidence interval for $\mu_{1}-\mu_{2}$ .

At $\alpha=0.05$ , the two-tailed critical value from the z-table, $z_{\alpha / 2}=1.96$

Then, the 95% confidence interval for $\mu_{1}-\mu_{2}$ is,

$\begin{aligned} &\left(\bar{x}_{1}-\bar{x}_{2}\right) \pm z_{\alpha / 2} \sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}} \\ &(36-42) \pm(1.96) \sqrt{\frac{6^{2}}{50}+\frac{8^{2}}{75}} \\ &-6 \pm 2.4585 \\ &(-8.4585,-3.5415) \end{aligned}$

So, the 95% confidence interval on the difference of population mean gas mileages for engines A and B is $(-8.4585,-3.5415).$