Hypothesis Testing

[rlgordon624]

The manufacturer of Winston Tire Company (WTC) claims its new tires last for an average 40k miles. An independent testing agency road-tested 100 tires to substantiate the claim made by WTC. The sample mean was 39k miles with a sample SD of 5k miles. Using 5% significance, determine if there is a reason to reject the claim made by WTC and conclude that the tires last for less than 40k miles.


Ok, I am taking statistics and this class is really pushing to my limits. I don't understand a thing. I have tried to understand it but it does not click with me. I have managed to make it this far but this question is really getting me down. It is due tonight and I just don't understand it. Can anyone help me solve it. Please.
If it was not needed for my business degree I would not be taking it. If anyone can tell me what I need to do here that would be great.

 

[galactus]

You think they last less than 40,000, so set up the hypothesis like so:

\(\displaystyle H_{0}: \;\ {\mu}\geq 40,000\)

\(\displaystyle H_{a}: \;\ {\mu}<40,000 \;\ \text{claim}\)

You could even use:

\(\displaystyle H_{0}: \;\ {\mu}=40,000\)

\(\displaystyle H_{a}: \;\ {\mu}<40,000 \;\ \text{claim}\)

Now, use the formula to find the z-score:

\(\displaystyle z=\frac{(x-{\mu})\sqrt{n}}{\sigma}\)

\(\displaystyle x=39, \;\ {\mu}=40, \;\ {\sigma}=5, \;\ n=100\)

Now, what you get here is the test statistic. Compare this to the critical value you get from the \(\displaystyle {\alpha}=.05\)

The critical value is -1.645. The critical value can be found by looking up .05 in the body of the z table and noting the corresponding z score.

If the z score you get is less than -1.645, then it is in the rejection region and you reject the null hypothesis and their claim can not be substantiated. If it is more than -1.645, then it is not in the rejection region and you do not reject the null hypothesis and their claim holds water.