Question

a researcher compares two compounds (1 and 2) used in the manufacture of car tires that are designed to reduce braking distances for suvs equipped with the tires. the mean braking distance for suvs equipped with tires made with compound 1 is 74 feet with a population standard deviation of 13.4. the mean braking distance for suvs equipped with tires made with compound 2 is 77 feet with a population standard deviation of 14.3. suppose that a sample of 41 braking tests are performed for each compound. using these results, test the claim that the braking distance for suvs equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used. let μ1 be the true mean braking distance corresponding to compound 1 and μ2 be the true mean braking distance corresponding to compound 2. use the 0.05 level of significance. step 2 of 5: compute the value of the test statistic. round your answer to two decimal places. answer how to enter your answer (opens in new window)

Explanation:

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

The formula for the two - sample z - test statistic when the population standard deviations $\sigma_1$ and $\sigma_2$ are known is $z=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}}$. Here, we want to test if $\mu_1<\mu_2$, so under the null hypothesis $H_0:\mu_1\geq\mu_2$ and the alternative hypothesis $H_1:\mu_1 < \mu_2$, we can assume $\mu_1-\mu_2 = 0$ for the test statistic calculation. We are given $\bar{x}_1 = 74$, $\bar{x}_2=77$, $\sigma_1 = 13.4$, $\sigma_2 = 14.3$, and $n_1=n_2 = 41$.

Step2: Substitute the values into the formula

$z=\frac{(74 - 77)-0}{\sqrt{\frac{13.4^{2}}{41}+\frac{14.3^{2}}{41}}}$. First, calculate the denominator: $\frac{13.4^{2}}{41}+\frac{14.3^{2}}{41}=\frac{13.4^{2}+14.3^{2}}{41}=\frac{179.56 + 204.49}{41}=\frac{384.05}{41}\approx9.367$. Then, $\sqrt{\frac{384.05}{41}}\approx3.06$. And the numerator is $74 - 77=-3$. So, $z=\frac{-3}{3.06}\approx - 0.98$.

Answer:

$-0.98$