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Math Help - Question 1

  1. #1
    randomer86
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    Question 1

    Hi guys, i need help.

    A 90% confidence interval for μ based on a sample of size 17 from a Normal distribution has upper end-point equal to 525.9. Given that the sample standard deviation is s = 583, calculate the lower end-point of the interval, correct to 1 decimal place.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by randomer86 View Post
    Hi guys, i need help.

    A 90% confidence interval for μ based on a sample of size 17 from a Normal distribution has upper end-point equal to 525.9. Given that the sample standard deviation is s = 583, calculate the lower end-point of the interval, correct to 1 decimal place.
    Here we are dealing with the t-distribution.

    t=(mu-m)/(s/sqrt(n))

    has a t-distribution with n-1 degrees of freedom. The 90% confidence region
    for t with 16 degrees of freedom is:

    (-1.75, 1/75),

    or for mu:

    (-1.75 s/sqrt(n) + m, 1.75 s/sqrt(n) + m) = (-1.75x583/sqrt(17)+m, 1.75x583/sqrt(17)+m)
    ...................= (-247.446+m, 247.466+m)

    But we are told that the upper end point is 525.9, so:

    525.9=247.466 + m,

    so m=278.434, and so the lower end of the interval is: 30.968, or rounding
    to 1 decimal place: 31.0.

    RonL
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