the manufacturer of a new racecar engine claims that the proportion of engine failures due to overheating…

the manufacturer of a new racecar engine claims that the proportion of engine failures due to overheating for this new engine, ($p_1$), will be no higher than the proportion of engine failures due to overheating of the old engines, ($p_2$). to test this statement, nascar took a random sample of 115 of the new racecar engines and 165 of the old engines. they found that 12 of the new racecar engines and 6 of the old engines failed due to overheating during the test. does nascar have enough evidence to reject the manufacturers claim about the new racecar engine? use a significance level of $\\alpha = 0.01$ for the test.\n\nstep 4 of 6: compute the value of the test statistic. round your answer to two decimal places.

the manufacturer of a new racecar engine claims that the proportion of engine failures due to overheating for this new engine, ($p_1$), will be no higher than the proportion of engine failures due to overheating of the old engines, ($p_2$). to test this statement, nascar took a random sample of 115 of the new racecar engines and 165 of the old engines. they found that 12 of the new racecar engines and 6 of the old engines failed due to overheating during the test. does nascar have enough evidence to reject the manufacturers claim about the new racecar engine? use a significance level of $\\alpha = 0.01$ for the test.\n\nstep 4 of 6: compute the value of the test statistic. round your answer to two decimal places.

Answer

Explanation:

Step1: Identify sample sizes and successes

$n_1 = 115, x_1 = 12$ (new engines); $n_2 = 165, x_2 = 6$ (old engines).

Step2: Calculate sample proportions

$\hat{p}_1 = \frac{12}{115} \approx 0.1043$; $\hat{p}_2 = \frac{6}{165} \approx 0.0364$.

Step3: Calculate pooled proportion

$$\bar{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{12 + 6}{115 + 165} = \frac{18}{280} \approx 0.0643$$

Step4: Calculate standard error

$$SE = \sqrt{\bar{p}(1 - \bar{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)} = \sqrt{0.0643(0.9357) \left( \frac{1}{115} + \frac{1}{165} \right)} \approx 0.0298$$

Step5: Compute the test statistic

$$z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.1043 - 0.0364}{0.0298} \approx 2.28$$

Answer:

2.28