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    Quote Originally Posted by Dr.D View Post
    Listen carefully,

    Based on central limit theorem, 30 or more in sample size allows the sample estimate of sigma to represent the population sigma. The answer is the z-test. Please do not misinform dbardy.

    See
    http://www.sixsigmafirst.com/sampling.htm for a review of this

    lol...that is a general guideline (I tell all my statistics students about people in business and the social sciences how they always assume that if the sample size n>=30 then everything is wonderful and we can always assume normality.)

    This is not the case. The CLT states that in the limit as n->infintiy the distribution will be normally distributed. It tells you nothing about what sample size will guarantee you will have a normal distribution.

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    You are right about the fact that 30 or more does not gurantee normality. However, many authors suggest that 30 or more is recommended by CLT. Given the dbardy is following a convention in his assignment.The z-test should be used.

    I think we should write a book on this (lol). Cause the central limit theorem needs re-developing.

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    Quote Originally Posted by Dr.D View Post
    You are right about the fact that 30 or more does not gurantee normality. However, many authors suggest that 30 or more is recommended by CLT. Given the dbardy is following a convention in his assignment.The z-test should be used.

    I think we should write a book on this (lol). Cause the central limit theorem needs re-developing.
    I'm sure that dbardy will get the "correct" answer according to his/her book I just want dbardy to know that 30 is not a magical threshold where everything turns into a normal distribution no matter what. n=30 is just a guideline to make the problem simpler. In real practice one should test the data to see if it follows a normal distribution before we start using the normal distribution to do statistical inference. I'm sure in most cases it is not...that's a shame.

    A book . I think if we did that you would end up doing most of the writing. I hate to write

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    Don't worry, you will be provide the ideas to my writing. LOL

    Yeah I often urge people to conduct normality tests (like Komologorov-Smirnov, etc) on their data before conducting any parametric tests. Otherwise they should use non-parametric ones.

    The educational system needs to be altered somewhat in this area.

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    Quote Originally Posted by Dr.D View Post
    Don't worry, you will be provide the ideas to my writing. LOL
    That's how I got my last paper published

    Quote Originally Posted by Dr.D View Post
    Yeah I often urge people to conduct normality tests (like Komologorov-Smirnov, etc) on their data before conducting any parametric tests. Otherwise they should use non-parametric ones.
    Good advice

    Quote Originally Posted by Dr.D View Post
    The educational system needs to be altered somewhat in this area.
    I agree.

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    thank you Dr D. Can I just ask you to explain the - 1.96 to 1.96 range? Where does that come from - a chart ?

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    Quote Originally Posted by dbardy View Post
    thank you Dr D. Can I just ask you to explain the - 1.96 to 1.96 range? Where does that come from - a chart ?

    Hi dbardy,

    This comes from the z-distribution curve (normal curve) tables.

    The z-test is based on a two-tailed one since you are looking to see whether there is a difference between the two means (sample vs population).

    The alpha/level of significance is based on .05. So you divide the .05 by 2 (each tail of curve contains the .025 of the alpha). Hence the remaining portion of the curve is 95% (since 5% is taken up already). To find the corresponding z-values, divide the 95% by 2 which is 47.5 (.475). Look in the z-tables body for .475, and you will see the corresponding z-values for that proportion is 1.96. Since it is two-tailed, there is a lower limit (-1.96) and upper limit (+1.96) crossing zero on the curve.

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