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Probability/Hypothesis testing help

  1. Jul 25, 2005 #1
    I need help on some questions: The first two seem really easy, but i cant seem to figure out the right approach:

    Marks in ECMB11 are normally distributed with a mean of 70% and standard deviation of 10%. Suppose an A is a mark of 80% or more.

    1. If you take a sample of 120 students, what is the probability that at least 20 will receive an A.
    2. What is the probability that average mark for 120 students will be at least 80%.

    -------------

    You are conducting a lower tailed test. Suppose alpha = 0.10, beta = 0.01, n = 16, s = 1460, and u subscript a (i guess population of a) = 2400. Determine u subscript o. (I guess population of o)

    I have no idea what they are even asking for in this question


    HELP is greatly appreciated.
     
  2. jcsd
  3. Jul 25, 2005 #2
    I'm not an expert at stats but I will take a stab at this...

    Notice that your mean is .7 and your SD is .1 so .8 is one SD from the mean. So what I am thinking is that you take

    [tex] 1 - ( 1/2 * 0.68 + .5)[/tex]

    to get the probability that any student will get .8 or more. Now that we know [tex]P(x \geq .8) [/tex] we can make it a binomial problem, either they are or they are not.

    [tex] _{120}C_{20}* (.1587)^{20}*(.8413)^{100} = .0946[/tex]

    I am :confused: about the rest of your questions....

    I just hope thats all correct.

    Regards
     
    Last edited: Jul 25, 2005
  4. Jul 26, 2005 #3
    Does anyone else know how to work these problems? I would like to see how they are worked out.... :smile:
     
  5. Jul 26, 2005 #4
    your answer to the first question seems right, but i think the formula you gave is for the probability of having exactly 20, where as i Need at least 20. But I think we can use a chart to get it. Im glad you were able to help on this one. Just wondering if the binomial logic part makes sense. Does seem right.

    After taht im.......................... :frown:
     
  6. Jul 26, 2005 #5

    EnumaElish

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    Are you certain that you have ua and uo, not [itex]\mu_a \text{ and } \mu_o[/itex]? [itex]\mu[/itex] (mu) is the symbol for distribution mean; subscript a denotes the alternative hypothesis and subscript o denotes the null hypothesis.
     
  7. Jul 26, 2005 #6
    thats what i meant (mu) im not sure how you guys get it to come out properly...

    isnt mu population mean as well? Oh wait, they are equal to each other....Uhh so yeah.....i guess i shuld have tried to type it like that. Hopefully can help me to solve the problems....
     
  8. Jul 26, 2005 #7
    i assumed that u would be interpreted as mu although i guess I should have made it clear.
     
  9. Jul 26, 2005 #8

    EnumaElish

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    Right. So the picture can be described as follows. You have two partially overlapping "bell curves" (two normal density plots or graphs). The point of their intersection is the critical x (xc), marked on the horizontal axis. The area under the rightmost bell curve to the LEFT of xc is alpha. The area under the leftmost bell curve and to the RIGHT of xc is beta. The mean of the leftmost curve is mua and the mean of the rightmost curve is muo.

    I hope this helps.
     
  10. Jul 27, 2005 #9

    EnumaElish

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    If you look up the standard normal probability table for alpha = 0.1, you will see a value like z0.1=1.6. I am making this up as an example, you should look at the table yourself. But usually the table is for a two-tailed test. To use a two-tail table for a one-tail test, you want to look up alpha = 0.2. Let's say you did this and came up with a critical value z0.2. Because this is a lower-tail test, you want the negative value of z0.2, that is -z0.2. Now,
    [tex]-z_{0.2}=-\frac{x_c-\mu_o}{\sigma}.[/tex]
    Clearly if you knew xc you could determine muo. You have to use the alternative dist. to derive xc.
    To do this, look up z0.02. Then solve xc from
    [tex]z_{0.02}=\frac{x_c-\mu_a}{\sigma}.[/tex]
    Now substitute xc into the previous equality and solve for muo.
     
    Last edited: Jul 27, 2005
  11. Jul 27, 2005 #10
    oh.....ok i think i have an example like that from my class notes....thanks alot.


    Anyone have ideas as to how to correctly do the first two i posted?
     
  12. Jul 27, 2005 #11

    EnumaElish

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    For the first 2 questions you have to assume that the 120 random var.s are independent of each other. That's a standard assumption but was not stated explicitly in this problem.
    "At least 20 will receive an A" = (NOT "exactly 19 receive A") AND (NOT "exactly 18 receive A") AND ... AND (NOT "exactly 1 receives A") AND (NOT "exactly 0 receives A"). Now use the binomial formula to calculate each of the terms in quotes (e.g. Prob{"exactly 19 receive A"}), then take the complement ("NOT") to get each paranthesis term (e.g. Prob{NOT "exactly 19 receive A"} = 1 - Prob{"exactly 19 receive A"}), then multiply the complements. Formally this is equivalent to deriving the prob. distribution of the 20th highest order statistic out of a sample of 120 normal random variables. Typically this is a non-Normal distribution, closer (pehaps identical?) to a Log-normal.
    Need to derive the probability distribution of the sample average for a sample of 120. That distribution will also be Normal, you just need to figure out what its mean and the standard dsitribution are. Show that cumulative distribution as F. Then Prob(Average > 80) = 1 - F(80).
     
    Last edited: Jul 27, 2005
  13. Jul 27, 2005 #12
    How do I derive the probability distribution?
     
  14. Jul 27, 2005 #13
    Could one use the central limit theorem to say that the mean of the sample is approximately equal to the mean of the population?
     
  15. Jul 27, 2005 #14

    EnumaElish

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    Well, the sample average (Xbar) is a linear combination of 120 normal random var's (RVs). Each RV ~ N(mu,sigma) and they are independent. A linear combo of several Normal RV's is itself a Normal RV, and all you need to derive is its mean and std. dev. For the mean, you can use an expected value theorem that says expectation is a linear operator, hence:

    E(Xbar) = E(Sum[Xi/n]) = E(Sum[Xi]/n) = E(Sum[Xi])/n = Sum[E(Xi)]/n = Sum[mu]/n = n mu/n = mu.

    So the mean of the sample average is identical to any one of the RV's mean (all of which are equal to mu = 70). No big surprise here.

    Next, you need to derive the std. dev. of the sample mean. It should be equal to sigma/sqrt(n), where sigma is the std. dev. given in the problem. But you should check this formula. I'll think about how it is derived and come back if I can remember it.
     
    Last edited: Jul 27, 2005
  16. Jul 27, 2005 #15

    EnumaElish

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    This theorem (like the linear expectation theorem) is from Mood, Graybill, Boes' Introduction to Probability and Statistics textbook (I am sure about the author names but slightly less so about the title):

    For RV's X, Y and constants a, b, Var(aX+bY) = a2Var(X) + b2Var(Y) + abCov(X,Y). In your case the X is identical to Y, there are 120 RV's, and all Cov terms = 0 (because of independence). "Sum" denotes summation over i = 1 through 120. Therefore: Var(Xbar) = Var(Sum[Xi/n]) = Var(Sum[(1/n)Xi]) = (1/n)2Sum[Var(Xi)] = (1/n)2Sum[sigma] = (1/n)2n sigma2 = sigma2/n. Since std. dev. = sqrt(Var), it follows that Std. Dev. of the sample average = sigma/sqrt(n).

    We have proven that Xbar is ~ N(mu, sigma/sqrt(n)) where mu = 70, sigma = 10, n = 120. This is your F distribution.
     
    Last edited: Jul 27, 2005
  17. Jul 27, 2005 #16
    ok thanks for your help everyone. I think i understand how to do it.
     
  18. Aug 15, 2005 #17
    59 years since Cox, and people still insist on the tail area tests?
     
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