[College - Stats Homework]

[Rubin0]

I have to present this stats problem to the class tomorrow. I am kind of lost as what to do.


The salesman at the tire store told Joe that the tires he was buying had a mean tread life of 30,000 miles with a standard deviation of 2,500 miles. Assume the tread life for tires is normally distributed. If the tires Joe bought woreout after 20,000 miles, what would you say to Joe if he claimed that he was not given the brand of tires he was told he was given but was given an inferior brand? Would your conclusion differ whether you use a one-tailed or two-tailed test?


Can someone please explain the steps to solving this problem? I have most of the forumulas in front of me, I'm just having a problem figuring out how and when to use them. As of right now I have a z-score of -4. I don't even know if this is correct. I am so lost.

[tsugomaru]

Re: [College - Stats Homework]

Well, what you need to do is a 9-step hypo test I believe. Here are the steps and the explanations:

1. u = 30,000 miles (This is the average distance that the tires are said to travel before wearing out)
2. Ho : u = 30,000 miles (This is where the tire's mean is 30,000 miles)
3. Ha : u < 30,000 miles (This states that the tire's true mean is actually below 30,000 miles, it's less then because you are testing if the tire is inferior, so it's a one-tailed test. It's only a two tailed test if you are testing whether or not the true average deviates from the mean)
4. Significance-Level = 0.05 (0.05 is the normal significance-level for a hypo-test. Because the results you receive won't actually threaten anyone's life, a 0.05 sig-level will suffice)
5. z = (x-bar - u) / (sd / sqrt(n)) (This is just the formula to calculate the z-score)
6. Assumptions (with every hypo test, you have to state your assumptions)
a. Not SRS - It's not a simple random sample because the only sample are Joe's tires.
b. The sample is not large enough, assuming that his car uses 4-tires. The Central limit theorem does not support a normal curve for the sample.
c. His 4 tires represent less than 10% of all tires.
7. z = (20,000 - 30,000) / (2500/ sqrt(4)) = -8
8. p-value is approximately 0 because the area left of -8 standard deviations is basically 0.
9. Reject Ho, evidence supports that these tires can't travel an average of 30,000 miles before wearing out.

However, you could argue that this was a bad test because the size of the sample wasn't larger than 30 and because it wasn't a simple random sample.

A one-tailed test only covers one tail of a normal curve, in my case, I chose to test the lower-tail. A two-tailed test covers both tails of a normal curve, so essentially, you're covering two areas, the area left of -8 sd and the area right of 8 sd. The p-value would be doubled, but in this case, it will still be 0 because 0 * 2 = 0.

I would tell Joe that he's probably right. It'd be kind of hard for a company to totally mess up a tire test like that.

~Tsugomaru

[Rubin0]

Re: [College - Stats Homework]

I'm trying to go through each step and figure this out so it sounds like I know what I'm talking about. Thank you very much. It helps a lot.


Edit: Thank you again. I completely undertand it and will have no trouble presenting this tomorrow morning.