In recent year, some professional baseball players complained that umpires were calling more strikes than the average of 22 with a standard deviation of 4. A manager in Atlanta decided to determine if this was true. A random sample of 10 games was selected and the mean and standard deviation was 23.2 and 5.5 respectively. Test null hypothesis at the .05 level of significance.

Calculate the z-score corresponding to your specific mean (23.2) and standard deviation (5.5). Then consult a table of values for a standard normal distribution to see what the cutoff for a .05 level of significance is.

Set up your hypotheses and run the test.



(claim)

The test statistic is

What did you get? Do you reject or not reject the claim.

I see what I did. I set my null and alternative up backwards. THis is OBVIOUSLY my first class! Thanks

How do I find the p-value?

The p-value is the area in the tail. In your case it is about .2451

Because 1-0.7549=0.2451

So , is the p value the same as the rejection region?

Sort of.

A z-score or t-score is the point on the line you are testing against. Using a p-value is doing by the percent of the area instead. But it's the area for your sample, not for the critical value. The rejection region for yours is that .05.

The p-value is the area corresponding to your sample numbers. You have to know how to use those charts to get the area equivalent to the t-score of 2.57. Since all probabilities add up to 1, if you take 1 - that number, it gets you the area left over on the outsides in the tails. That's for your sample.

The p-value for your sample has to be less than the .05 in order to reject.

Are you required to do p-values? If not, I personally wouldn't mess with it. Just adds to the confusion, and you're supposed to get the same answer whether to reject or not.

Quote:

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Originally Posted by morgaine300

You have to know how to use those charts to get the area equivalent to the t-score of 2.57.

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So sorry. I've got your other problem in my head. I don't know what the z/t-score of this one was. (Can't find my book anywhere, so I can do the math but can't use my charts to get these scores, and I only have 2 of them memorized.)

The point was it's the area that corresponds to the test statistic for this problem.

Faced with rising fax costs, a firm issued a guideline that transmissions of 10 pages or more should be sent by 2-day mail instead. Exceptions are allowed, but they want the average to be 10 or below.The firm examined 35 randomly chosen fax transmissions during the next year, yielding a sample mean of 14.44 with a standard deviation of 4.45 pages. (a) At the .01 level of significance, is the true mean greater than 10? (b) Use Excel to find the right-tail p-value.

H:u<10
H:u>10
14.44-10/4.45/skrt35

Faced with rising fax costs, a firm issued a guideline that transmissions of 10 pages or more should be sent by 2-day mail instead. Exceptions are allowed, but they want the average to be 10 or below.The firm examined 35 randomly chosen fax transmissions during the next year, yielding a sample mean of 14.44 with a standard deviation of 4.45 pages. (a) At the .01 level of significance, is the true mean greater than 10? (b) Use Excel to find the right-tail p-value.

H:u<10
H:u>10
14.44-10/4.45/skrt35

is this on the right path

You should start you own thread. More are likely to see it.

Anyway:




The test stat is 5.9028 and the critical value is 2.3263.

The p-value is small. p<.01

Should you reject or not reject?