Question

a car company claims that its new suv gets better gas mileage than its competitors suv. a random sample of 42 of its suvs has a mean gas mileage of 13.9 miles per gallon (mpg). the population standard deviation is known to be 0.6 mpg. a random sample of 36 competitors suvs has a mean gas mileage of 13.5 mpg. the population standard deviation for the competitor is known to be 1.1 mpg. test the companys claim at the 0.01 level of significance. let the car companys suvs be population 1 and let the competitors suvs be population 2. step 2 of 3: compute the value of the test statistic. round your answer to two decimal places.

Explanation:

Step1: Identify the formula for two - sample z - test statistic

For two independent samples with known population standard deviations, the z - test statistic is given by $z=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}}$. Since we are testing if $\mu_1>\mu_2$, under the null hypothesis $\mu_1 - \mu_2 = 0$. So the formula simplifies to $z=\frac{\bar{x}_1-\bar{x}_2}{\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}}$.

Step2: Substitute the given values

We are given that $\bar{x}_1 = 13.9$, $\bar{x}_2=13.5$, $\sigma_1 = 0.6$, $n_1 = 42$, $\sigma_2 = 1.1$, $n_2 = 36$.
First, calculate the denominator:
$\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}=\sqrt{\frac{0.6^{2}}{42}+\frac{1.1^{2}}{36}}$
$=\sqrt{\frac{0.36}{42}+\frac{1.21}{36}}$
$=\sqrt{0.00857 + 0.03361}$
$=\sqrt{0.04218}\approx0.2054$.
Then, calculate the numerator: $\bar{x}_1-\bar{x}_2=13.9 - 13.5=0.4$.
Now, find the z - value: $z=\frac{0.4}{0.2054}\approx1.95$.

Answer:

$1.95$