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    Hi everyone,

    I wonder if someone could help me please or at least point me in the right direction.

    I have a large amount of data, 500 pieces, which I've narrowed down to 495 using outliers.

    What I have to do now is hypothesise that the data is normally distributed by carrying out a chi squared test. The data has been sorted into 10 classes but the question is this

    How do I calculate the expected frequencies of each and how can I calculate the degree of freedom?

    Thanks in advance

    Ady
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    Since you're estimating 2 population parameters Mu and Sigma from the data the degs of freedom = 10-1-2=7 (assuming no cell amalgamation?)

    Could you give the classes so as to explain the expected freqs? I guess you're using software for all of this?
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    Since you're estimating 2 population parameters Mu and Sigma from the data the degs of freedom = 10-1-2=7 (assuming no cell amalgamation?)

    Could you give the classes so as to explain the expected freqs? I guess you're using software for all of this?

    Hi,

    The classes I have got are as follows

    Class Frequency

    5.40 – 5.46 4
    5.47 – 5.53 19
    5.54 – 5.60 51
    5.61 – 5.67 108
    5.68 – 5.74 123
    5.75 – 5.81 106
    5.82 – 5.88 62
    5.89 – 5.95 14
    5.96 – 6.02 6
    6.03 – 6.09 2

    I'm not using any software to do this, it's all being done manually. If there is a useful piece of software, Excel or SPSS, that could do this I would be glad to know of it.

    Thanks

    Andy
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    Ok, I've adjusted the bins slightly:

    (<5.47) 4
    (5.47, 5.54) 19
    (5.54, 5.61) 51
    (5.61, 5.68) 108
    (5.68, 5.75) 123
    (5.75, 5.82) 106
    (5.82, 5.89) 62
    (5.89, 5.96) 14
    (5.96, 6.03) 6
    (>6.03) 2

    For each bin you need to work out it's normal probability, e.g. for (5.82, 5.89) work out P(5.82<X<5.89) under the null hypothesis with Mu and Sigma estimated from the data. To get the expected freqs multiply each normal probability by n=495.

    If any of the expected freqs are <5 then a cell amalgamation process is often done.
 
 
 
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