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When consumers apply for credit, their credit is rated using FICO scores. The average credit rating is 700. Credit ratings are given below for a sample of applicants for car loans. Use this sample data to test the null hypothesis at the .05 level of significance.
661 595 769 734 725 690
Ok, why question is do I set the alternative hypothesis to be H: mu =700?
You would never set the alternative as an =. It's either >, <, or =/ (not equal). This one is not real clear but is this a two-tailed? It doesn't say what you're actually testing. i.e. Are you testing if it's greater than 700, less than 700, or that the applicants just don't show it's 700? If this is all the info you have, I'd go with the latter.
In which case, null is mu = 700, and alternative is mu does not equal 700. (Sorry, I've not fully learned the math functions here yet.) So you're using alpha=.05 as a two-tailed, meaning you've got to split it until alpha/2 = .025 for each tail. And, you only have six in your sample, meaning this should be a t-score, not a z-score.
EDIT: Actually, whether you'll use t score will depend on the book and what's available in the problem. I've seen this done differently in different books. Some use t for anything under 30. I originally learned that it's t if it's under 30 and you do not have a population standard deviation (sigma), and only have sample st.dev. Unless you've left something out, you don't have a population one.
I am so confused. This is all the information the instructor gave me. This is an ONLINE stats class. It is awful and I am struggling. I do not even know how to calculate this test statistic.
I use a stats macro for Excel. Most calculators have nice stats packages. The test statistic for the mean is given by the formula
Using the given data and a two tailed test, I get
a sample mean of 695.67
A sample SD of 61.77
Critical value: +/- 2.57
test statistic: -0.1718
p value: 0.8703
I got the sample mean and SD but beyond that, I am lost. I appreciate your help.
The "critical value" is the point where you hit the 95%, i.e. for a 2-tail, 2.5% from each end. That would be gotten off whatever charts you're using and they can be quite different in different books. From what galactus gave you as the critical value, I'll assume he (she?) used a t-score, cause z at 95% would be 1.96.
If your sample passes that point either direction, then you're in the rejection region.
But you also have to get a t-score for the sample that was actually done. Galactus gave the equation for that. A t-score is treated exactly the same as a z-score. So it's the same equation. And since you have a sample standard deviation, you replace the sigma with that. (The 61.77.) Those can be used interchangeably in that equation. And you're comparing your sample mean with the null mean.
i.e. 695.67 - 700 / (61.77/sqrt of 6) = - .17
That is the t-score for the sample itself.
Now if you draw a normal distribution, and put the -2.57 over on the left, and then put the -.17 on the left from your sample, how does the sample compare? Is that in the reject region?
Quote:
Originally Posted by christide
Have you emailed the instructor for further explanation?
It does fall in the rejection region therefore I would reject the null, correct?
To mowerman, my instructor is almost impossible to get in touch with. I have tried several several times
The rejection region is in the tail, on both sides, the area past the -2.57 and + 2.57 on either side.Quote:
Originally Posted by christide
Does -.17 fall into those tails?
The idea is that if you take all possible samples you could, 5% of them would have means falling outside this range. So there's a 95% probability that for any given sample you take, its mean will be inside this range. So we only reject if it's outside that range, and we're 95% sure of being correct about it.
Your test statistic from your sample is very near the center, not in those tails.
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