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Math Help - Confidence interval problems?

  1. #1
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    Confidence interval problems?

    Hi all,

    I really hate Statistics since the problems are very difficult





    For the 9th problem:

    The sample mean is (20+27+15+20+16+22+13)/7=19

    The sample variance = E[X] - 19

    = (400 + 729 + 225 + 400 + 256 + 484 + 169) / 7 - 361

    = 19.428.

    Bessel's correction >> 19,428 . 7/6 = 22.666

    standard deviation = 4.76

    ... but how to go on?




    For 8th problem:

    Distribution binomial


    mean = np = 13

    standard deviation = [np (1-p)] (square root) = 3.486

    ... but how to go on?



    I could not solve the 7th one.
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  2. #2
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    Re: Confidence interval problems?

    You are taking statistics- don't you know what a normal distribution [b]is[b]. First consult a table of the standard normal distribution to see how many standard deviations from the mean lie within 95% or 99%. Then use the mean and standard deviation you have calculated to find the numbers you want. The table at Standard Normal Distribution Table tells me that 95% of the standard normal distribution is within 1.91 standard deviations of the mean. Since the standard deviation, in the last problem, is 4.76, 95% will be within (1.91)(4.76), or approximately 9, of the mean. Since the mean is 19, that is from 19- 9= 10 to 19+ 9= 28.
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  3. #3
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    Re: Confidence interval problems?

    Quote Originally Posted by HallsofIvy View Post
    You are taking statistics- don't you know what a normal distribution [b]is[b]. First consult a table of the standard normal distribution to see how many standard deviations from the mean lie within 95% or 99%. Then use the mean and standard deviation you have calculated to find the numbers you want. The table at Standard Normal Distribution Table tells me that 95% of the standard normal distribution is within 1.91 standard deviations of the mean. Since the standard deviation, in the last problem, is 4.76, 95% will be within (1.91)(4.76), or approximately 9, of the mean. Since the mean is 19, that is from 19- 9= 10 to 19+ 9= 28.
    Problems 7 and 9 are small sample questions so the critical values from the t-distribution should be used with $\nu=n-1$, also the sd should be estimated with the $\frac{1}{n-1}$ formula not the $\frac{1}{n}$ formula (which is what Bessel's correction does).

    .
    Last edited by zzephod; June 6th 2014 at 09:17 PM.
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  4. #4
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    Re: Confidence interval problems?

    7) For a sampling distribution the mean is the same and the standard deviation becomes 64/sqrt(15)
    a 99% CI for the mean would be 912+/-2.9768*64/sqrt(15)=
    912+/-49.1908=(862.81, 961.19) (2.9768 is t_alpha/2 with 14 degrees of freedom)

    8)13/200=.065
    a 95% CI for the proportion of defective light bulbs would be
    .065+/-1.96*sqrt(.065(1-.065)/200)=
    .065+/-.034167=
    (.0308, .0992)

    9) Assuming your standard deviation and mean are correct, 2.4469 is the t value corresponding to alpha/2 with 6 degrees of freedom
    since the sample size is small and the population standard deviation is unknown, we must use the t distribution (even though the sample came from a normal distribution).
    19+/-2.4469*4.76/sqrt(7)=
    19+/-4.4022=
    (14.60, 23.40)
    Last edited by RShnay; June 8th 2014 at 07:37 AM.
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