Help Please...

[Esadler38chick]

I have my book which is sort of like Greek to me, but I have read the chapter several times, and tried to work some of the problems, but the examples in the book are unclear. I don't want answers I just want to know if somebody could give me a few examples of how to do it. I have the Practical Business Statistics Book 5th ed and I am in Chapter 10. Maybe if somebody has this book they can help me?

[JohnM]

We don't have access to specific textbooks, so go ahead and post one of the questions that is confusing you, tell us what you've tried so far, and we'll try to give you helpful guidance!

[Esadler38chick]

Stress Levels were recorded during a true answer and a false answer given by each of six people in a study of lie-detecting equipment, based on the idea that the stress involved in telling a lie can be measured. THe results are shown:
PERSON TRUE ANSWER FALSE ANSWER
1 12.8 13.1
2 8.5 9.6
3 3.4 4.8
4 5.0 4.6
5 10.1 11.0
6 11.2 12.1

a) Was everyone's stress levels higher during false answers than during true answers?

My answer--No person 4 has a higher true stress level

b) Find the average stress levels for true and for false answers. Find the average change in stress level (false-true).

My answer--average true=8.5 average false=9.2
false-true=0.7

c) Find the appropriate standard error for the average difference. In particular, is this a paired or an unpaired situation?

My answer-- I believe it is a paired situation

but I don't understand how to get the standard error for the average difference. I used the formula for small-sample situiation (sorry I don't know how to type the equation out on here


d) Find the 95% two-sided confidence interval for the mean difference in stress level.

I can't do this becasue I don't have the answer from above.

I hope this helps

[JohnM]

a) and b) look fine to me

c) yes, it is paired

use the formula for the standard deviation (aka standard error) of the differences - just compute all 6 differences, then compute the standard deviation of them

d) the mean (average) difference is 0.7, and the standard deviation of the differences is computed in c above

-the two-sided 95% confidence interval around 0.7 would be:

= 0.7 +/- [ t(a/2,df) * s/sqrt(n) ]

where t(a/2,df) is the t-statistic for a/2 (which is .025) and 5 df (degrees of freedom, which is n-1)

s is the standard deviation
n is the sample size

the t-statistic for a/2 = .025 and df=5 is 2.571

= 0.7 +/- [ 2.571 * (s/sqrt(6)) ]

[Esadler38chick]

Thank you then my real problem is that I am unsure as to what the standard deviation (aka standard error) of the differences formula is. I found the standard deviation for both--true and false---true is 3.64 and false is 3.67.

And now what am I suppose to do? This is where I really keep getting confused.



Thank you for this help. I have more that I need help starting, I will post one when I get off of work.

[JohnM]

For each of the 6 persons, compute false-true (call it "delta") and compute the standard deviation of the "deltas," the same way you computed the std dev for the false and true numbers.

[Esadler38chick]

Thank you so much I got that one.

Now here is the one I just don't know what direction to go--I'm not sure where to start.

As part of a decision regarding a new product launch, you want to test whether or not a large enough percentage (10% or more) of the community would be interested in purchasing it. You will launch only if you find convincing evidence of such demand. A survey of 400 randomly selected people in the community finds that 13.0% are willing to try your proposed new product.

A) Why is a one-sided test appropriate here?



B) Identify the null and research hypotheses for a one-sided test using both words and mathematical symbols.

null hypothesis=.13
research hypothesis >=.10



C) Perform the test at the 5% signicance level and describe the result.



D) Perform the test at the 1% signicance level and describe the result.


E) State the p- value as either p>0.05, p<0.05, p<0.01, or p<0.001.



I guess my null and research hypotheses is right?

What I need help with is an explanation of how to do the significance levels and exactly how to do a one-sided test?

[JohnM]

A) think about the objective --> you really don't care if there is "too much" interest, you just want to know if there is "enough" interest

B) .13 is the sample percentage, the null and research hypotheses are always about the population

C) since you have a large sample size (n=400), you can use the normal distribution here to test whether the sample proportion 0.13 is significantly larger than .10, at a 5% significance level (alpha=.05)

compute z and look up its probability in the normal distribution tables

z = (p - P)/sp

small p = .13
Big P = .10
sp = sqrt(pq/n) where q = 1-p

if the probability of the z score is <= .05, then we reject the null hyp.

D) exactly the same as C, except set alpha=.01

if the probability of the z score is <= .01, then we reject the null hyp.

E) determine the p-value of the test --> this is whatever the probability is for the computed z score, and select which choice "fits the best"

[Esadler38chick]

I know I am being a pain, you totally lost me...Here is what I tried after your last post...

For part b--is the Ho:mu<=mu(o) Ho:mu<=.10
H1:mu>mu(o) H1:mu>.10


I computed z and got 1.79 I look that up on the table and it gives me .9633. That is more than .05 so does that mean I accept the null hypothesis?

Also for part d--z is still 1.79 look it up on the z table and get .9633. That is still more than .01 so I accept the null hypothesis

Part e) p>0.05

I really hope I am doing some part of this right.

[JohnM]

You're not being a pain - that's what we're here for - if everyone understood this stuff, then we wouldn't have this great website:D

For the hypotheses, we're not testing the average (mu), we're testing the proportion (the population proportion is usually represented by "big" P and the sample proportion is usually represented by "small" p)

Ho: P = .10
Ha: P > .10

You computed z correctly, but what the z table gives you is the probability of getting that particular value of z or lower, so in order to find the probability of getting that z or higher, subtract .9633 from 1.

I talk about z tables in my post in the Examples section called "The Vaunted Normal Distribution"

[Esadler38chick]

You were right this is a great website--I am so GLAD I found it :)

I got another one...

Some frozen food dinners were randomly selected from, this weeks
production and destroyed in order to measure their actual calorie content. The claimed calorie content is 200. Here are the calorie counts for each dinner:

221 198 203 223 196 202 219 189 208 215 218 207


A) Estimate the mean calorie content you would have found had you been able to measure all packages produced this week.

Would that be 200

B) Approximately how different is the average calore content (for the sample) from the mean value for all dinners produced this week.

The sample average is 207.42, right? I take the mean of the above data.

C) Find the 99% confidence interval for the mean calorie content for all packages produced this week.



D) Is there a significant difference between claimed and measured calorie content? Justify your answer.


That is what I have so far. For the answer in part b I don't use for part c, but I do use it to compare in part d?

Also in part c this is a two-sided interval?

So I use Xbar-t(Stnd error) Xbar+t(Stnd error)

[JohnM]

A) Estimate the mean calorie content you would have found had you been able to measure all packages produced this week.

Would that be 200
Yes

B) Approximately how different is the average calore content (for the sample) from the mean value for all dinners produced this week.

The sample average is 207.42, right? I take the mean of the above data.
Yes, the difference would be 207.42-200=7.42

C) Find the 99% confidence interval for the mean calorie content for all packages produced this week.



D) Is there a significant difference between claimed and measured calorie content? Justify your answer.


That is what I have so far. For the answer in part b I don't use for part c, but I do use it to compare in part d?
yes
Also in part c this is a two-sided interval?
yes
So I use Xbar-t(Stnd error) Xbar+t(Stnd error)
yes

[Esadler38chick]

So xbar=207.42

t (look on table)=3.106 for sample size 12

Stnd error=10.82---


99% Confidence level

Xbar -t(Stnd error) Xbar-t(Stnd error)
207.42-(3.106)(10.82) 207.42-(3.106)(10.82)
173.81 241.03

And part d--I said yes, because you are expecting 200 and you could get anywhere from 173.81 to 241.03



I'm almost done, I have 2 more, but let me go and try them first.

[JohnM]

For the t statistic, you should get 3.497
(degrees of freedom = n-1 = 11)

standard error = std dev / sqrt(n) = 10.997/sqrt(12) = 3.175

99% conf interval = 207.42 +/- (3.497 * 3.175)
= (196.32, 218.52)

This is not significant, since the claimed population mean is within the confidence interval.

[Esadler38chick]

I keep forgetting to divide by the sqrt of n. I don't know where you got that t value from?


And yet another one I don't know where to start...


You are supervising an audit to decide whether any errors int eh recording of account transactions are "material errors" or not. Each account has a reported balance, whose accuracy can only be verified be careful and costly investigation; the account's error is defined as the difference between the reported balance and the actual balance. Note that the error is 0 for any accounts that is correctly reported. In practical terms, for this situation involving 12,000 accounts, the total error is material only if at least $5,000. The average error amount for 250 randomly selected accounts was found to be $0.25, and the standard deviation of the error amount was $193.05. YOu may assume that your reputation as an auditor is on the line, so you want to be fairly certain before declaring that the total error is not material.

A) Find the estimated total error based on your sample, and compare it to the material amount.


B) Identify the null and research hypotheses for a one-sided test of the population mean error per account and explain why a one-sided test is appropriate here.


C) Find the appropriate one sided 95% Confidence interval statement for the population mean error per account.


D) Find the t statistic.


E) Which hypothesis is accepted as a result of a one-sided test at the 5% level?


If you could point me in the right direction. Also a question I have is what exactly is the "t statistic?"


This is my last one...I think I did it right, I am unsure about part d.

A group of experts has rated your winery's two best varietals. Ratings are on a scale from 1 to 20, with higher numbers being better. The results are shown:

Expert/ Chardonnay / Cabernet Sauvignon
1 17.8 16.6
2 18.6 19.9
3 19.5 17.2
4 18.3 19.0
5 19.8 19.7
6 19.9 18.8
7 17.1 18.9
8 17.3 19.5
9 18.0 16.2
10 19.8 18.6

A) Is this a paired or unpaired situation?

I think it is unpaired

B) Find the average rating for each varietal and the average difference in ratings (Chardonnay- Cabernet Sauvignon).

average Chardonnay= 18.61
average Cabernet=18.44
average Chardonnay-average Cabernet
18.61-18.44=.17


C) Find the appropriate standard error for the average difference.

This equals 1.58/sqrt 10=.4996

D) Find the 95% two-sided confidence interval for the mean difference in rating.

Xbar-t(stnd error) Xbar+t(stnd error)
.17-(2.262)(.4996) .17+(2.262)(.4996)
-.96 1.30

E) Test to see if the average ratings are significatly different. If they are significantly different. which varietal is superior?

[quark]

Each expert rated both wines so this is paired situation.

[JohnM]

The account auditing problem is very similar to the frozen dinner calories example. You have some examples - they follow basically the same logic.

The t-statistic is the t value you look up in the t-table. Remember, it's not based on sample size, but on degrees of freedom, which is n-1.

Here's a link to a t-table, so I'll try to show you how to find the correct value:
http://www.itl.nist.gov/div898/handbook/eda/section3/eda3672.htm

Let's say the sample size is 12 - that would give us degrees of freedom of 12-1=11.

Now, let's say we need to compute the 95% confidence limits. In the t-table, you'll find the value in the row=11, but in the column = .025 - why? Because in a two-sided confidence interval, you need to split the 5% between both sides of the distribution. So the t-value in this case would be 2.201.

[Esadler38chick]

If I take the 1% left from the 99% and divide by two then I get .005 and on the t table http://www.itl.nist.gov/div898/handb...n3/eda3672.htm (http://www.itl.nist.gov/div898/handb...n3/eda3672.) when I look up .005--I still get 3.106. So am I still doing it wrong?

[JohnM]

Sorry- my mistake - you are correct.

[Esadler38chick]

Here's what I got for account auditing problem---

info given:
n=250
acct error=reported-actual
total error is material if at least $5,000
average error for 250 (n)= $0.25
stnd deviation= $193.05


A) Find the estimated total error based on your sample, and compare it to the material amount.

is this the average error for the sample size times the number of accounts
$0.25*12,000=3,000

B) Identify the null and research hypotheses for a one-sided test of the population mean error per account and explain why a one-sided test is appropriate here.


Ho: total error is not material
Ha: total error is material
One sided becasue it is material if at least $5,000

C) Find the appropriate one sided 95% Confidence interval statement for the population mean error per account.

For this part, I did
Xbar-t(stnd error) which came out to .25-(1.645)(12.21)=-19.84---which I know is a wrong answer becasue it just doesn't make sense.


D) Find the t statistic.

The t statistic formula is t-stat=Xbar-mu(o)/stnd error

E) Which hypothesis is accepted as a result of a one-sided test at the 5% level?

[JohnM]

B) the hypotheses are "verbally" correct, but need to be put into numerical form

C) why don't you think this makes sense? (hint: could it be possible that the confidence interval includes an actual error in the negative direction; do all accounts have to have errors in the positive direction?)

[Esadler38chick]

B) Ho:mu<$5,000
Ha:mu>=$5,000

C) why don't you think this makes sense? (hint: could it be possible that the confidence interval includes an actual error in the negative direction; do all accounts have to have errors in the positive direction?)

I don't think it makes senses becasue the number is so low compared to the other information pertaining to the problem

[JohnM]

Well, not really --> the given standard deviation is pretty big....

[Esadler38chick]

I am so frustrated with this stuff:confused: :confused: :confused:

[JohnM]

Look at it this way:

average error for the accounts = 0.25
std dev for the errors = 193.05
n = 250

Now, picture a normal distribution curve, centered at 0.25. If you draw a bunch of random samples of 250 accounts and compute the mean each time, the values at each of the standard deviations of those means (or the standard errors of the means) would be:

-3 std devs = -36.60
-2 std devs = -24.39
-1 std devs = -12.18
mean = 0.25
+1 std devs = 12.23
+2 std devs = 24.44
+3 std devs = 36.65

[Esadler38chick]

Well thanks, for all your help, I really needed it. Those were my last problems, but I may be back--tomorrow we start regressions. I also will be back around exam time, I'm sure I'll need guidance then.

[afgafg]

can anyone please answer this question for me....i have this one question left n my assignment and if i m not able to do this i woul get failed ....so need help


Q: from past records of personal loans the sanguisunga bank has determined that 10% of borrowers default on their loans. it also found that, of those who default, 32% are unemployed while only 2% of those who do not default are unemployed.

a: what percentage of unemployed borrowers defualt?
b: what proportion of borrowers is unemployed?
c; what proportion of employed borrowers does not default?


thanks for the help