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 3 of 5: find the $p$-value associated with the test statistic. round your answer to four decimal places. answer how to enter your answer (opens in new window) $p =$

Explanation:

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

The formula for the two - sample z - test statistic 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, $\bar{x}_1 = 74$, $\bar{x}_2=77$, $\mu_1-\mu_2<0$ (claim), $\sigma_1 = 13.4$, $\sigma_2 = 14.3$, $n_1=n_2 = 41$. Since the null hypothesis is $H_0:\mu_1-\mu_2\geq0$ and the alternative hypothesis is $H_1:\mu_1-\mu_2<0$, we calculate the test statistic $z$.
\[

$$\begin{align*} z&=\frac{(74 - 77)-0}{\sqrt{\frac{13.4^{2}}{41}+\frac{14.3^{2}}{41}}}\\ &=\frac{- 3}{\sqrt{\frac{179.56}{41}+\frac{204.49}{41}}}\\ &=\frac{-3}{\sqrt{\frac{179.56 + 204.49}{41}}}\\ &=\frac{-3}{\sqrt{\frac{384.05}{41}}}\\ &=\frac{-3}{\sqrt{9.367073}}\\ &=\frac{-3}{3.060564}\\ &\approx - 0.98 \end{align*}$$

\]

Step2: Find the p - value

Since this is a left - tailed test, the p - value is the probability of getting a z - score less than the calculated test statistic. We use the standard normal distribution table (or z - table). The p - value corresponding to $z=-0.98$ is $P(Z < - 0.98)$. Looking up in the standard normal table, $P(Z < - 0.98)=0.1635$.

Answer:

$0.1635$