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

galactus

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=0.25, z=1.96 for a 95% CI,

Plug in your values and solve for n.

jaystang44

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 ....

jaystang44

round up i believe right?????

120.4735


yea I rounded up to 121 thats right
thanks again galactus

jaystang44

Two more study questions I dont 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 DIDNT know that didnt 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)

galactus

For #2:

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

Use the formula

You got n=400 and

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.

jaystang44

thanks I understand a little better but where is the z-score soming from I have 1.96 writen in my notes too but I cant see where I got it from?

jaystang44

So I believe the anwser for my number 1 is A the margin of error will decrease bc less ppl are in the sample.

Is that correct in thinking that way?

morgaine300

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.

morgaine300

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



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?