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

For #1:

This one is pretty much set. Just subtract your confidence interval values from your mean.

For #2:

The formula for margin of error is E=z\frac{\sigma}{\sqrt{n}}

E=0.25, z=1.96 for a 95% CI, {\sigma}=1.4

Plug in your values and solve for n.

thanks I finally got the first one... its easy now that I did it lol... but thanks for number two that helps I for some reason had the wrong equation :confused:...

Round up I believe right??

120.4735


Yea I rounded up to 121 that's right
Thanks again galactus

Two more study questions I don't understand...


The heights of a simple random sample of 400 male high school sophomores in a Midwestern state are measured. The sample mean 66.2 inches. Suppose that the heights of male high school sophomores follow a normal distribution with standard deviation 4.1 inches.
Suppose the heights of a simple random sample of 100 male sophomores were measured rather than 400. Which of the following statements is true?

A. The margin of error for the 95% confidence interval would decrease.
B. The standard deviation o would decrease.
C. The margin of error for the 95% confidence interval would increase.
D. The margin of error for the 95% confidence interval would stay the same, because
the level of confidence has not changed.

SORRY Didn't know that didn't make it...


The heights of a simple random sample of 400 male high school sophomores in a Midwestern state are measured. The sample mean y=66.2 inches. Suppose that the heights of male high school sophomores follow a normal distribution with standard deviation o=4.1 inches. What is a 95% confidence interval for U ?
A. ((65.86, 66.54)
B. (59.46, 72.94)
C. (58.16, 74.24)
D. (65.80, 66.6)

For #2:

You have a 95% CI. That corresponds to a z-score of 1.96.

Use the formula E=z\frac{\sigma}{\sqrt{n}}

You got n=400 and \sigma=4.1

Use your given data and the formula to find your margin of error, then add and subtract it from you mean. Then you have your confidence interval.

Thanks I understand a little better but where is the z-score soming from I have 1.96 written in my notes too but I can't see where I got it from?

So I believe the answer for my number 1 is A the margin of error will decrease because less people are in the sample.

Is that correct in thinking that way?

The z score should be coming off a chart of some sort. How exactly you read it depends on the chart. You may have one (either within the chapter or in an appendix) that gives the z-score for some common intervals like 90%, 95%, 99% or whatever.

You may also have a chart that lists the 5%, i.e. 100-95%. Because it's a symmetric interval, then that 5% is split between the 2 tails. So you may have to use 2.5%. I've also seen some that list a one-tail column and a two-tail column, and then you can just use the 5% and look in the two-tail column.

It's hard to say without seeing it how to read your particular chart or charts. But that's where it comes from. (Actually I hear it comes from calculus, but what little I learned in high school is long forgotten, so you couldn't prove that by me. LOL.) 95% is so commonly used that I have that one memorized.

Sorry, didn't answer your other question. No. You can prove this to yourself mathematically. Since the s.d. for the sample mean is

\frac{4.1}{sqrt{400}}

Then putting in a smaller sample size will decrease your denominator, and therefore do what to the answer?

Then the error is that answer times the 1.96. So it will change likewise.

(See if you can figure that out.)

Now, if you want to start analyzing from a different point of view... think what a confidence interval is. Based on a sample, you're 95% sure about the interval. Literally, 95% have to be within that range. If the range decreases, you're not spread out as far. Can a smaller sample make you come up with a smaller range and still be 95% sure about it? A smaller range is like "tightening the reigns." Can you do that when you've got a smaller sample representing the population?

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

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 haven't figured out the whole math symbols thing I am sorry but (/=) is doesn't equal there combined I just don't know how to make it again I am SORRY!!

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?? :confused:

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

K sorry thanks

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????:confused:

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.

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.

Which question?

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????:confused:
This one.

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.

This is how I would have done this. Since it's a sample, you need to find what I call \sigma_{p-bar}. (p-bar is the same as p-hat, and I don't know how to do the symbol for that either ;) and that doesn't look how I mean -- i.e. sigma sub-p-bar) Your book could call this something else. This is the standard deviation for a sample proportion.

The equation I know for this is:

sqrt{\frac{p(1-p)}{n}}

I seem to recall there's another equation for this, but don't think I ever learned it. Again, that needs to be looked up in the text to see which one is being used.

Using this equation, I get:

sqrt{\frac{.6(.4)}{70}} = \,.0586

From there, you can use the same z equation. If you think of the z equation, rather than symbols, as:


z (or\, t) =\, \frac{point - {its \,mean}}{its \,standard \,deviation}

By says "its points" and "its standard deviation" I mean the ones that belong to the point you're using. i.e. is your point an x, an x-bar, or a p-bar? This can be applied whether you're doing a population, a sample mean, or a proportion. They all work exactly the same way. In this case, the point is the .5 you're finding the probability of, "its mean" is .6, and "its standard deviation" is the .0586 I figured out.

As for whether it could be a t, different books have different tests for this. The one I learned is np > .05 and np(1-p) > .05. Again, got to find that in the particular text being used.

Unfortunately, much of this is book-dependent. But this is the basic gist of it.