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  1. #16
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    Each expert rated both wines so this is paired situation.

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    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/handb...n3/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.

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    so...

    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 when I look up .005--I still get 3.106. So am I still doing it wrong?

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    Sorry- my mistake - you are correct.

  5. #20
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    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?

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    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?)

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    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

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

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    I am so frustrated with this stuff

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    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

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    Thanks

    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.

  12. #27
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    need help


    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

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