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Math Help - Determining the confidence interval and claim justification

  1. #1
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    [SOLVED]Determining the confidence interval and claim justification

    The following mesaurements were recorded for the drying time, in hours of a certain brand of latex paint.

    3.4, 2.5, 4.8, 2.9, 3.6, 2.8, 3.3, 5.6, 3.7, 2.8, 4.4, 4.0, 5.2, 3.0, 4.8

    Assuming that the measurements represent a random sample from a normal population,
    a) find the 95% confidence interval for the means
    b) find the 95% confidence interval for the variances
    c) with 0.01 level of significance is it justifiable to claim that there the mean is less than 3.75 hours?
    d) is there a reason to believe that the variance is equal to 1?

    I tried solving this a little while ago but I just can't understand it. My professor teaches us just too fast and gives us tons of notes to copy that's why I can't keep up with what he is teaching. I based my computations on my notes but still I could not do it. I can't think of the correct null hypothesis and alternative hypothesis for this. Any help will be appreciated, thank you very much.
    Last edited by kusama; March 17th 2009 at 01:39 PM.
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  2. #2
    MHF Contributor matheagle's Avatar
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    You need to compute the sample mean \bar x and the sample variance s^2.

    a) find the 95% confidence interval for the mean.

    The interval here is (\bar x -ts/\sqrt{n}, \bar x +ts/\sqrt{n})

    where n is the number of observations and t is a percentile point from your t table.
    The degrees of freedom is n-1 (which is your row) and the column is \alpha=.025.

    b) find the 95% confidence interval for the variances

    THERE is only one variance here. The confidence interval is ...

    \bigg({(n-1)S^2\over \chi_1^2} , {(n-1)S^2\over \chi_2^2}\biggr)

    where \chi_1^2 has n-1 degrees of freedom and \alpha=.025.

    while \chi_2^2 has n-1 degrees of freedom and \alpha=.975.

    c) with 0.01 level of significance is it justifiable to claim that there the mean is less than 3.75 hours?

    Most likely, your instructor wants you to just examine whether or not 3.75 is in the interval from part (a)
    That's not really right, since that was a two sided CI and this is a one sided test.

    d) is there a reason to believe that the variance is equal to 1?

    This sounds like a two sided test. So once you obtain the interval, all you need to do is examine whether or not it contains the value 1.
    Last edited by matheagle; March 14th 2009 at 02:49 PM.
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  3. #3
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    Quote Originally Posted by matheagle View Post
    You need to compute the sample mean \bar x and the sample variance s^2.

    a) find the 95% confidence interval for the mean.

    The interval here is (\bar x -ts/\sqrt{n}, \bar x +ts/\sqrt{n})

    where n is the number of observations and t is a percentile point from your t table.
    The degrees of freedom is n-1 (which is your row) and the column is \alpha=.025.

    b) find the 95% confidence interval for the variances

    THERE is only one variance here. The confidence interval is ...

    \bigg({(n-1)S^2\over \chi_1^2} , {(n-1)S^2\over \chi_2^2}\biggr)

    where \chi_1^2 has n-1 degrees of freedom and \alpha=.025.

    while \chi_2^2 has n-1 degrees of freedom and \alpha=.975.

    c) with 0.01 level of significance is it justifiable to claim that there the mean is less than 3.75 hours?

    Most likely, your instructor wants you to just examine whether or not 3.75 is in the interval from part (a)
    That's not really right, since that was a two sided CI and this is a one sided test.

    d) is there a reason to believe that the variance is equal to 1?

    This sounds like a two sided test. So once you obtain the interval, all you need to do is examine whether or not it contains the value 1.
    Thank you very much for your help! I was able to understand letters a and b but I was a little bit confused about letters c and d, I couldn't understand what you meant..I'm sorry for being so slow.
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  4. #4
    MHF Contributor matheagle's Avatar
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    Parts c and d should be done by hypotheses testing.
    I'm not sure you've covered that.
    I'm guessing that all your teacher wants you to do, is look at the interval and see if those numbers are in there.

    ACTUALLY, the level from part a to part c is different.
    SO in part c you need to do a hypothesis test, where your test is...

    H_0:\mu=3.75 vs. H_a:\mu<3.75

    BUT in part d I do not see an \alpha, so maybe all you need to do is obtain the CI from part b and see if the number 1 is in that interval.
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  5. #5
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    Quote Originally Posted by matheagle View Post
    Parts c and d should be done by hypotheses testing.
    I'm not sure you've covered that.
    I'm guessing that all your teacher wants you to do, is look at the interval and see if those numbers are in there.

    ACTUALLY, the level from part a to part c is different.
    SO in part c you need to do a hypothesis test, where your test is...

    H_0:\mu=3.75 vs. H_a:\mu<3.75

    BUT in part d I do not see an \alpha, so maybe all you need to do is obtain the CI from part b and see if the number 1 is in that interval.
    We have done some hypothesis testing, that's the one with the Ho and H1 right sir? and how could I check if the interval is in the interval? would I have to subtract the interval to 1 and if yields to a positive answer it is within 1 and if it is outside it is not?
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  6. #6
    MHF Contributor matheagle's Avatar
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    Get the interval in part b. These are reasonable values for your variance.
    Look at it. If the number 1 is in there, then it is acceptable to assume that
    \sigma^2 can take on the value 1.
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  7. #7
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    Quote Originally Posted by matheagle View Post
    Get the interval in part b. These are reasonable values for your variance.
    Look at it. If the number 1 is in there, then it is acceptable to assume that
    \sigma^2 can take on the value 1.
    do you mean in the table sir? I'm so confused I'm sorry.

    EDIT: I get this part now sir, thank you very much for your help! I'll lock this now...if I can haha thanks again!
    Last edited by kusama; March 17th 2009 at 01:38 PM.
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