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 72 feet, with a population standard deviation of 13.4. The mean braking distance for SUVs equipped with tires made with compound 2 is 75 feet, with a population standard deviation of 5.2. Suppose that a sample of 70 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 2 of 5: Compute the value of the test statistic. Round your answer to two decimal places.

Step 1 ：A researcher is comparing two compounds (1 and 2) used in the manufacture of car tires. The aim is to determine which compound reduces braking distances for SUVs equipped with the tires. The mean braking distance for SUVs equipped with tires made with compound 1 is 72 feet, with a population standard deviation of 13.4. The mean braking distance for SUVs equipped with tires made with compound 2 is 75 feet, with a population standard deviation of 5.2. A sample of 70 braking tests are performed for each compound. The researcher wants to 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. The level of significance used is 0.05.

Step 2 ：To compute the value of the test statistic, the formula for the z-score in hypothesis testing for two population means is used. The formula is: \[ z = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \] where: \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means, \(\mu_1\) and \(\mu_2\) are the population means, \(\sigma_1\) and \(\sigma_2\) are the population standard deviations, and \(n_1\) and \(n_2\) are the sample sizes.

Step 3 ：In this case, the claim being tested is that \(\mu_1 < \mu_2\), so the hypothesized difference \((\mu_1 - \mu_2)\) is 0. The sample sizes are both 70, and the other values are given in the problem.

Step 4 ：Substituting the given values into the formula, the test statistic is calculated as follows: \[ z = \frac{(72 - 75) - 0}{\sqrt{\frac{13.4^2}{70} + \frac{5.2^2}{70}}} \]

Step 5 ：The value of the test statistic, rounded to two decimal places, is \(\boxed{-1.75}\).