prob. and statistics

jaystang44 · Feb 27, 2008, 06:03 AM

first question......

The manager of a downtown restaurant is interested in how much customers spend there at lunch time. He estimated the mean expenditure by taking a sample of 75 diners. He found the mean amount spent on lunch for these 75 people was $9.74. He reported a 95% confidence interval of ($9.08, $10.40). He could have also reported his results by saying, "I am 95% confident that my estimate of $9.74 differs from the true mean amount spent for lunch at this restaurant by no more than $____." What margin of error goes in the blank? Be sure your answer is in the form of 0.27.

second question..........

The weights of adult chipmunks vary according to a normal distribution with standard deviation 1.4 ounces. The weights of adult chipmunks vary according to a normal distribution with standard deviation 1.4 ounces. Researchers suspect that the mean weights of populations of chipmunks depend on where they live. A team of biologists want to catch, weigh, and release a sample of chipmunks in Rocky Mountain National Park. Find the sample size required if they want a margin of error to be less than 0.25 ounces with 95% confidence.
A. 120 chipmunks
B. 44 chipmunks
C. 11 chipmunks
D. 121 chipmunks
E. 120.5 chipmunks

I can not get the right answer for either of these please help

Thanks

jaystang44 · Feb 29, 2008, 04:38 AM

So basically the smaller sample or range actually increases the margin of error. since there are less people to reach 95% making it have a bigger marrgin of error because each person kind of counts more toward the final result right? like with 100, 95 of those people have to be in that range leaving 5 people. to where with 400, you need 380 leaving 20.....

Well at least using the equation it makes since.......with using a smaller denominator it increases the answer and there fore making the error larger.

I kind of get the explination
Thanks Morgaine300

jaystang44 · Mar 3, 2008, 06:09 AM

A major car manufacturer wants to test a new engine to determine if it meets new air pollution standards. The mean emission ( U ) of all engines of this type must be approximately 20 parts per million of carbon. It it is higher than that, they will have to redesign parts of the engine. Ten engines are manufactured for testing purposes and the emission level of each is determined. Based on data collected over the years from a variety of engines, it seems reasonable to assume that emission levels are roughly normally distributed with O = 3 What is the appropriate null and alternative hypotheses?
A. Ho : U = 20 vs. Ha:U<20
B. Ho : U = 20 vs. Ha:U(/=)20
C. Ho : U = 20 vs. Ha:U>20

Sorry new to the site and havnt figured out the whole math symbols thing I am sorry but (/=) is doesnt equal there combined I just dont know how to make it again I am SORRY!!!!!!

jaystang44 · Mar 3, 2008, 06:17 AM

Suppose 60% of all people who get migraine headaches get some relief from ibuprofen. If a random sample of 70 migraine headache sufferers is given ibuprofen, use the normal distribution to approximate the probability that less than half of them will get some relief

^P <0.5) Give your answer to THREE decimal places. So, if your answer is 0.5656, you would give 0.566 as your answer.

That is p hat or proportion of success I do not know how to place it over the p. This is an extra credit challenge for me and I attempted it once using x is apoximatly normal but got it wrong. Any suggestions????

Curlyben · Mar 3, 2008, 06:33 AM

>Bunch of threads merged<
Please keep all of your homework to one thread<

jaystang44 · Mar 3, 2008, 06:42 AM

k sorry thanks

jamesbondurant · Mar 11, 2008, 11:41 PM

Quote:

Originally Posted by galactus

You can use a binomial. If you have a calculator, run it through that. Less than half means 34 or less get some relief.

$\sum_{k=0}^{34}C(70,k)(0.60)^{k}(0.40)^{70-k}$

You can also use ${\mu}=np; \;\ \sqrt{npq}={\sigma}$ . Where n=70, p=.60, and q=1-p.

After you figure those, use them in the formula $z=\frac{x-{\mu}}{\sigma}$

Don't forget your continuity correction. You should get an answer close to the above binomial summation.

I don't know how to put this into my calculator. Also, what do you mean by continuity correction? I am not familiar with that term. Just for reference, what was the answer to this question? I have a similar problem and I don't know how to do it.

jaystang44 · Mar 12, 2008, 04:43 AM

Which question?

jamesbondurant · Mar 12, 2008, 04:46 AM

Quote:

Originally Posted by jaystang44

Suppose 60% of all people who get migraine headaches get some relief from ibuprofen. If a random sample of 70 migraine headache sufferers is given ibuprofen, use the normal distribution to approximate the probability that less than half of them will get some relief

^P <0.5) Give your answer to THREE decimal places. So, if your answer is 0.5656, you would give 0.566 as your answer.

That is p hat or proportion of success I do not know how to place it over the p. This is an extra credit challenge for me and I attempted it once using x is apoximatly normal but got it wrong. Any suggestions????

This one.

morgaine300 · Mar 16, 2008, 01:40 PM

Just as some notes... James, just post your question into a new thread. (And I don't know what he means by continuity correction either. Not familiar with that term.)

And curlyben, these were actually separate questions (despite being related subject), and I for one would prefer having them in different threads cause all these different problems in one thread is confusing the heck out of me.



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