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Julian
2010-03-25, 23:13
This is a practice test, not homework (I already know the answer, but I want to know how to find it)

An article in a journal reports that 34% of American fathers take no responsibility for child care. A researcher claims that the figure is higher for fathers in the town of Cheraw. A random sample of 225 fathers from Cheraw, yielded 97 who did not help with child care. Find the P-value for a test of the researcher's claim.

The answer is 0.0019.

I don't really know this section that well :( I kinda get the null/alternate hypothesis stuff, for example, H0 here is that mu (would it be mu in this problem?) = .34, and H1: mu(?) > .34

Yeah I'm kinda lost, and this book is a piece of shit that isn't really helping, any help would be appreciated, thanks!

edit: Here's 2 more problems I don't really get:

In a poll of 278 voters in a certain city, 67% said that they backed a bill. The margin of error in the poll was reported as 6 percentage points (w/ a 95% degree of confidence) Which statement is correct?
A) The sample size is too small to achieve the stated margin of error
B) For the given sample size, the margin of error should be larger than stated
C) The reported margin of error is consistent with the sample size (correct answer)
D) There is not enough info to determine whether the margin of error is consistent w/ the sample size
E) For the given sample size, the margin of error should be smaller than stated.

Also another problem, except it's a poll of 390, 77% backed it, ME is 5% w/ 95% degree of confidence, and the answer is "The stated margin of error could be achieved with a smaller sample size"

Don't really get what's going on with those 2, any help would be appreciated :x thanks again

Heavencloud
2010-03-26, 00:00
H0 = u=.34
Ha = u>.34

Is a significance level given? I'm trying to remember how to solve this but it doesn't feel like I have enough information =p

Moss
2010-03-26, 00:48
Damn dude, I was doing these exact problems like 2 months ago. I'll look at my old stuff in class tomorrow.

Weiing
2010-03-26, 01:12
Ok... Let me see if I can recall stats looong time ago.

I don't know what formulae you have learned, but I was using this for my calculations:

N= 4p(1-p)/m²

Where N = sample size; p = percentage; m = margin of error


In a poll of 278 voters in a certain city, 67% said that they backed a bill. The margin of error in the poll was reported as 6 percentage points (w/ a 95% degree of confidence) Which statement is correct?
A) The sample size is too small to achieve the stated margin of error
B) For the given sample size, the margin of error should be larger than stated
C) The reported margin of error is consistent with the sample size (correct answer)
D) There is not enough info to determine whether the margin of error is consistent w/ the sample size
E) For the given sample size, the margin of error should be smaller than stated.


key: p=.67 m=.06

sticking it into the formula above:
N= (4)(.67)(1-.67)/(.06)²
= (4)(.67)(.33)/(.0036)
= .8844/.0036
= 245.667

thus, 245 is the minimum sample size needed for that margin of error. Since 278 is more than 245, then it is consistent (?????)
edit: punched in to see the sample size needed for a 5% margin of error, and you would need 353.76. so 278 falls within the range of 245-353


Also another problem, except it's a poll of 390, 77% backed it, ME is 5% w/ 95% degree of confidence, and the answer is "The stated margin of error could be achieved with a smaller sample size

key: p=.77 m=.05

N=(4)(.77)(1-.77)/(.05)²
= (4)(.77)(.23)/(.0025)
= .7084/.0025
= 283.36

Thus, the minimum sample size needed for that margin of error with that confidence interval is 283 people. 390 exceeds 283, so could have been done with fewer.

HOWEVER, I haven't done stats in over 3 years now. So I may be totally wrong and pulling stuff out of my butt. I sat here for a while punching in random numbers to try to figure it out so... I tried!

CDF
2010-03-26, 06:42
This is not a thorough explanation but may help:

1) Assuming you already covered the relevant material, the p-value to obtain is based off a test statistic that was already presented in lecture. Your textbook should have a section about "test for one proportion" or "test for a binomial proportion" where the normal approximation to the binomial distribution is applied and the corresponding test statistic given. The following is a brief summary of what is involved in the problem, implicitly or otherwise.

Stating the null and alternative hypotheses for a test of a single proportion:

http://latex.codecogs.com/gif.latex?H_0: \pi \leq .34
http://latex.codecogs.com/gif.latex?H_1: \pi > .34

Finding the sample proportion, which estimates the "true" proportion:

http://latex.codecogs.com/gif.latex?\hat{\pi} = 97/225

Obtaining the variance of the sample proportion given the null hypothesis (under which the proportion is .34):

http://latex.codecogs.com/gif.latex?Var(\hat{\pi}) = \frac{\pi(1-\pi)}{n}

You wouldn't use the standard error (based on 97/225) because the problem asks for a p-value, which is based on the null hypothesis.

Obtaining the Z-statistic:

http://latex.codecogs.com/gif.latex?Z = \frac{\hat{\pi} - \pi }{\sqrt{Var(\hat{\pi})}}

A statement of the p-value (you may use a standard normal table in the back of the book to obtain the actual value):

http://latex.codecogs.com/gif.latex?P\left ( Z > \frac{\hat{\pi} - \pi}{\sqrt{Var(\hat{\pi} )}} \right )

2) You should have been given the sample size formula for polling ("yes/no" response) somewhere. The general formula is:

http://latex.codecogs.com/gif.latex?n = \frac{Z^2\pi(1-\pi)}{(ME)^2}

where Z is the Z-score corresponding to your chosen significance level. Z is 1.96 given a 5% significance level, so Z^2 is often approximated as 4. Since you usually don't have a good idea what the true proportion really is, the margin of error is usually based on pi = .5 because that gives the maximum n required to achieve a target margin of error (you want to be sure you achieve the desired M.E.), so n is often approximated as

http://latex.codecogs.com/gif.latex?n \approx \frac{1}{(ME)^2}

Use this formula to answer the question.

3) Refer to the general sample size formula:

http://latex.codecogs.com/gif.latex?n = \frac{Z^2\pi(1-\pi)}{(ME)^2}

Note that Z^2 is actually 3.841, not 4. Also, the true proportion could be less or greater than .5. Either or both could be invoked to reach the given conclusion.

Julian
2010-03-26, 09:26
Thanks, much appreciated!

Here's another one:

Assume that a simple random sample has been selected from a normally distributed population. State the final conclusion.

Test the claim that the mean age of the prison population in one city is less than 26 years. Sample data are summarized as n=25, xbar=24.4 years, and s=9.2 years. Use a significance level of alpha = 0.05.
H0: mu=26 Ha: mu<26
State your decision about H0.

(don't reject)

This problem is different than all the other problems of this type, so I'm not really sure how to go about it >_>. All the other problems have been just look at the p-value and alpha, and make a conclusion about it, which is easy, but I guess I gotta calculate the p-value for this one, and I have no idea how lol.

edit: If I did this right, I calculated t to be .8695, DF is 24, not sure how to go about calculating the p value.. :x Well, if I look at this crappy table I have, t for .1 is 1.318, so since it's less than that, it means the p-value is greater than .1 (so it's also greater than .05, the given alpha), so you don't reject?

Did I do that right? D:

Weiing
2010-03-26, 09:49
Yeah. I just calculated and got .8695 and DF of 24. Looking at my percentage points of t distribution, df of 24 with alpha of .05 shows 2.064. But since .8695 is less than that, can't reject the H0. So looks like you're right :D

Julian
2010-03-26, 10:24
Here's another:

In a survey of 5500 TV viewers, 20% said they watch network news programs. Find the standard error for the sample proportion.

(.0171)

I tried root(p(1-p)/n), which is (.2*.8/5500)^.5, but that gives .00539 <_<

EDIT: ok this one I have no clue how to do >_>

Scores on a certain test are normally distributed with a variance of 14. A researcher wishes to estimate the mean score achieved by all adults on the test. Find the sample size needed to assure with 98% confidence that the sample mean will not differ from the population mean by more than 2 units.

(20)

There's also a similar question to the above one:

The weekly earnings of students in one age group are normally distributed with a standard deviation of 81 dollars. A researcher wishes to estimate the mean weekly earnings of students in this age group. Find the sample size needed to assure with 98% confidence that the sample mean will not differ from the population mean by more than 5 dollars.

(1425)

These are the last questions in my practice exam, got everything else down, so this is it :D

Edit: trying to solve the above 2..

I basically have 5 +/- z * SE (or 2 for the first one) (z being 2.326 for 98% confidence)
Using SE = S/(n^.5) for standard deviation, and S=(variance/n)^.5 for the variance, I get 19 and 1420, which are... close enough? Especially compared to the other choices given. So I guess it's kinda right? idk <_<;

CDF
2010-03-26, 15:10
1) The reasoning on the lower bound of the p-value is fine. The t-statistic is actually -0.8695, though. Be careful if you actually have to report the p-value. To avoid making a "sign error," it could help to summarize what the p-value actually means (if this actually means anything to you):

http://latex.codecogs.com/gif.latex?P\left ( \bar{x} \leq 24.4 \right | H_0) = P(t\leq -0.8695 | H_0)

2) For n = 550, the standard error is 0.0171.

3) 20 and 1425 are based on the Z-score (2.32635) being rounded to the nearest hundredths place (2.33)