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    Help Please...




    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?

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    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!

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    Please Help...

    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

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    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)) ]

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    Please help...

    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.

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    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.

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    Yeah

    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?

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    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"

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    Sorry

    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.

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    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

    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"

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    Ready for another one...

    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)

  12. #12
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    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

  13. #13
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    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.

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    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.

  15. #15
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    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?
    Last edited by Esadler38chick; 11-19-2005 at 11:28 PM. Reason: correction made

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