Question

In auto racing a pit stop is is where a racing vehicle stops for new tires and repairs and other mechanical adjustments. The efficiency of a pit crew that makes these adjustments can effect the outcome of a race. A pit crew that its mean pit stop time ( for 4 new tireds and fuel) is 13 seconds. A random sample of 32 pit stops has a sample mean of 12.9 Seconds. Assume that the populations standard deviation is is 0.19 seconds. Using α= 0.01 level of significance, test the pit crew's claim. a. State the hypotheses. b. Find the critical value(s). c. Compute the test statistic. d. Make the decision. d. Summarize the results.

Answer 1

Answer:

a) $H_{0}$: t=13 seconds

   $H_{a}$: t<13 seconds

b) At α= 0.01, one-tailed critical value is -2.33

c) Test statistic is −2,98

d) since -2.98<-2.33, we can reject the null hypothesis. There is significant evidence that mean pit stop time for the pit crew is less than 13 seconds at α= 0.01.

Step-by-step explanation:

according to the web search, the question is missing some words, one part should be like this:

"A pit crew claims that its mean pit stop time ( for 4 new tires and fuel) is less than 13 seconds."

Let t be the mean pit stop time of the pit crew.

$H_{0}$: t=13 seconds

$H_{a}$: t<13 seconds

At α= 0.01, one-tailed critical value is -2.33

Test statistic can be calculated using the equation:

$z=\frac{X-M}{\frac{s}{\sqrt{N} } }$ where

• X is the sample mean pit stop time (12.9 sec)

• M is the mean pit stop time assumed under null hypothesis (13 sec)

• s is the population standard deviation (0.19 sec.)

• N is the sample size (32)

Then $z=\frac{12.9-13}{\frac{0.19}{\sqrt{32} } }$ ≈ −2,98

since -2.98<-2.33, we can reject the null hypothesis. There is significant evidence that mean pit stop time for the pit crew is less than 13 seconds at α= 0.01.