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    I AM ASKED TO CALCULATE THE CONFIDENCE INTERVAL, THE SAMPLE SIZE IS 7 BUT THE POPULATION SIZE IS IN THE THOUSANDS. I AM NOT GIVEN ANY STANDARD DEVIATION ALL I HAVE IS n, N and TO CALCULATE 99% CONFIDENCE INTERVAL. DO I USE Z OR T? I AM SO CONFUSED
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    I am rusty in stats but if I remember well you use t-distribution if population is normal, sample size is small and variance of population is unknown (has to be estimated from the sample)
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    (Original post by TeeEm)
    I am rusty in stats but if I remember well you use t-distribution if population is normal, sample size is small and variance of population is unknown (has to be estimated from the sample)
    thankyouuuu
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    no worries
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    (Original post by TeeEm)
    I am rusty in stats but if I remember well you use t-distribution if population is normal, sample size is small and variance of population is unknown (has to be estimated from the sample)
    This is correct.
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    (Original post by rayquaza17)
    This is correct.
    what if i am not given the significance
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    Test at 5%. If you do not reject here, then you must not reject at a lower percentage. If you do reject, repeat with 1% then 0.1%


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    (Original post by rayquaza17)
    test at 5%. If you do not reject here, then you must not reject at a lower percentage. If you do reject, repeat with 1% then 0.1%


    posted from tsr mobile
    you are the best

    so it does not matter that i was not given any percentage?
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    (Original post by rayquaza17)
    Test at 5%. If you do not reject here, then you must not reject at a lower percentage. If you do reject, repeat with 1% then 0.1%


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    when u say 5% do you mean siginificance level of 0.05? my problem is q=1-significance level. when looking at the t table it seems that my hypothesis 0 can be rejected at a lower significance level
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    Yes 5% is the same as 0.05.

    If it can be rejected at a lower significance level, then test it at that level level as well.
 
 
 
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