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    Hi, can someone check for my answer? Because my answers do not agree with the answers given. So I'm not sure whether what I did is right or not. Thanks.

    1) The numbers 24.4, 18.9, 12.8, 20.5, 19.1, 15.2, 21.7, 14.6 from a random sample of values of a normally distributed r.v. Find a 98% C.I. for the mean mu of X.

    So the sample mean is 147.2/8 = 18.4
    s² = [2816.56 - (147.2)²/8] / 7 = 15.44

    So the 98% C.I.
    = [18.4 ± 2.326sqrt(15.44/8) ]
    = [15.17, 21.63]

    But the answer given is [14.23, 22.57]

    2) The mature weight of a certain breed of dog has a normal distribution. A sample of these dogs was taken and the following results found:
    Sample size = 10
    Sample mean = 3.84
    Sample standard deviation = 0.4673
    Construct a 95% C.I. for the mean mature weight of this type of dogs.

    So 95% C.I.
    = [3.84 ± 1.96*0.4673/sqrt(10) ]
    = [3.55, 4.13]

    But the answer given is [3.506, 4.174]
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    You don't know the true variance so you should be looking at the t distribution tables.
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    (Original post by SsEe)
    You don't know the true variance so you should be looking at the t distribution tables.
    But I thought if the distribution is normal then we don't use the t-distribution? I'm confused.
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    If the distribution is normal then you use the normal tables if you know the true variance and the t tables if you only know the sample variance.
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    (Original post by SsEe)
    If the distribution is normal then you use the normal tables if you know the true variance and the t tables if you only know the sample variance.
    Oh I see. thank you.
 
 
 
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