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normal distribution question watch

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    hi, i have a lengthy normal dist question which i need help with

    The following is a random sample of size 9 from a normal population with mean µ and variance σ²
    2.8, 3.0, 3.0, 3.8, 2.4, 3.1, 3.3, 2.8, 4.0

    (a) estimate the parameters µ and σ from these data
    (b) using Students t distribution, determine the 95% and 99% confidence intervals for µ
    (c) test the null hypothesis µ=2.6=:µ0 against the alternative µ≠2.6 using significance levels 0.05 and 0.01
    (d) explain in the above context the null hypothesis and the alternative hypthesis
    (e) using this example, does any correspondence between hypothesis testing and confidence intervals occur to you?

    thanks for whoever helps me, hopefully the saint johnny w!
    attached are the sheets that came with this. im not sure really how to do this Q or how to extract the info from the tables when required so please be as patronising as you like :tsr:
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  1. File Type: pdf 2.PDF (96.5 KB, 90 views)
  2. File Type: pdf 3.PDF (62.8 KB, 200 views)
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  4. File Type: zip 1.zip (55.4 KB, 102 views)
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    (Original post by manps)
    The following is a random sample of size 9 from a normal population with mean µ and variance σ²
    2.8, 3.0, 3.0, 3.8, 2.4, 3.1, 3.3, 2.8, 4.0

    (a) estimate the parameters µ and σ from these data
    Surely you can atleast do this bit, Not sure about my statistics its been a long time.

    I think µ ≈ 3.13 and σ ≈ 1.54
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    i agree with your answer to µ but i get 0.473755... for σ and that also makes more sense than your answer :rolleyes:
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    (Original post by manps)
    i agree with your answer to µ but i get 0.473755... for σ and that also makes more sense than your answer :rolleyes:
    Maybe, I'm not sure
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    s. d.=+sqrt[(x^2)bar-((x bar)^2)]
    x bar=(1/9)(2.8+3.0+3.0+3.8+2.4+3.1+3.3+2 .8+4.0)=(1/9)(28.2)=3.133

    =>((x bar)^2)=9.818

    (x^2)bar=(1/9)(7.84+9+9+14.44+5.76+9.61+10.8 9+7.84+16)=(1/9)(90.38)=10.042

    =>s. d.=+sqrt[0.224]=0.473

    Newton.
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    can anyone help me with the rest of the question?
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    (a)
    Estimate of \mu
    = \overline{x}
    = 3.1333

    Now we use the standard formula that gives, based on a sample, an unbiased estimator of the population variance (the "sample variance" formula, which has a 1/(n - 1) factor).

    Estimate of \sigma^2
    (1/8)\sum_{i=1}^9 (x(i) - \overline{x})^2
    = (1/8) \big( \sum_{i=1}^9 x(i)^2 - 9*3.1333^2 \big)
    = 0.2525

    \hat{\sigma}
    = Estimate of \sigma
    = sqrt(0.2525)
    = 0.5025

    (b)
    From 3.pdf, the Student's t-value for a 95% confidence interval with 8 degrees of freedom is 2.306. (DoF = n - 1 = 9 - 1 = 8. Memory aid: "n = 1" must be a degenerate case, since you can't estimate variance from a single observation.)

    95% CI
    = [\overline{x} - 2.306\hat{\sigma}, \overline{x} + 2.306\hat{\sigma}]
    = [1.975, 4.292]

    From 3.pdf again, the Student's t-value for a 99% confidence interval with 8 degrees of freedom is 3.355.

    99% CI
    = [\overline{x} - 3.355\hat{\sigma}, \overline{x} + 3.355\hat{\sigma}]
    = [1.447, 4.819]
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    (c)
    For a level-0.05 test, we reject H0 iff

    | (\overline{x} - 2.6)/\hat{\sigma} | >= 2.306

    Since | (3.1333 - 2.6)/0.5025 | = 1.061, we accept H0.

    For a level-0.01 test, we reject H0 iff

    | (\overline{x} - 2.6)/\hat{\sigma} | >= 3.355

    Since | (3.1333 - 2.6)/0.5025 | = 1.061, we accept H0.

    (Are you sure that "2.6" is right? It seems surprising that the question setter would make the two tests produce the same result.)

    (d)
    I'm not sure what they want here.

    (e)
    Let p be a parameter (eg, the mean). Suppose that we have a 95% confidence interval I for p. Let k be a proposed value of p (eg, 2.6). Then the following procedure gives a 0.05-level test of H0: p = k against H1: p \neq k.

    Procedure
    - Accept H0 if k \in I.
    - Reject H0 if k \not\in I.

    The procedure works because if H0 is true then, by definition of a 95% confidence interval, P(k \in I) = 0.95.

    Example
    We can save effort in (c) by using the confidence intervals from (b). Both CIs contain 2.6, so we accept H0 in each case.
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    Did you know you can check your answers by using the TI-83 calculator? *Great way to check your work

    But use "statistics", plug in the data into the list, and it can calculate the mean, variance, etc.

    You can also test hypotheses (z tests) and calculate the p-value if you plug in the data!

    You can also do t-tests and X^2 tests as well.
 
 
 
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