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 $mu_1$ be the true mean braking distance corresponding to compound 1 and $mu_2$ be the true mean braking distance corresponding to compound 2. use the 0.05 level of significance. step 4 of 5: make the decision for the hypothesis test. answer reject null hypothesis fail to reject null hypothesis

Explanation:

Step1: Calculate the test - statistic

The formula for the two - sample z - test statistic for independent samples with known population standard deviations 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, $\bar{x}_1 = 74$, $\bar{x}_2=77$, $\sigma_1 = 13.4$, $\sigma_2 = 14.3$, $n_1=n_2 = 41$, and the null hypothesis $H_0:\mu_1\geq\mu_2$ (or $\mu_1-\mu_2\geq0$) and the alternative hypothesis $H_1:\mu_1<\mu_2$ (or $\mu_1 - \mu_2<0$). Substituting the values, we get $z=\frac{(74 - 77)-0}{\sqrt{\frac{13.4^{2}}{41}+\frac{14.3^{2}}{41}}}=\frac{- 3}{\sqrt{\frac{179.56}{41}+\frac{204.49}{41}}}=\frac{-3}{\sqrt{\frac{179.56 + 204.49}{41}}}=\frac{-3}{\sqrt{\frac{384.05}{41}}}=\frac{-3}{\sqrt{9.367073}}\approx\frac{-3}{3.06057}\approx - 0.98$.

Step2: Find the critical value

For a one - tailed test with a significance level of $\alpha = 0.05$ and the alternative hypothesis $H_1:\mu_1<\mu_2$, the critical value $z_{\alpha}$ is the z - score such that the area to the left of it under the standard normal curve is $\alpha$. Looking up in the standard normal table, $z_{0.05}=-1.645$.

Step3: Make the decision

Since the calculated test statistic $z\approx - 0.98$ and the critical value $z_{0.05}=-1.645$, and $-0.98>-1.645$ (the test statistic does not fall in the rejection region), we fail to reject the null hypothesis.

Answer:

Fail to Reject Null Hypothesis