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  1. #1
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    Probability questions

    If standard deviation is 80 and mean is 10:

    P ( X < 100)

    How do you do that?

    ...

    And, confidence intervals are very hard. Can someone break how to do a 95% interval down. Say 1000 people were interviewed about smoking and 600 said they did. How would you do that interval? Can do you do it on a calculator?
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  2. #2
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    And, confidence intervals are very hard. Can someone break how to do a 95% interval down. Say 1000 people were interviewed about smoking and 600 said they did. How would you do that interval? Can do you do it on a calculator?
    Confidence intervals aren't difficult. Since you have a proportion, Just use the formula E=z\sqrt{\frac{pq}{n}}. This is the margin of error.

    p=0.60, q=1-p, n=1000. z that corresponds to 95% CI is 1.96

    E=1.96\sqrt{\frac{(.60)(.40)}{1000}}=0.0304

    0.60+0.0304=0.6304; 0.60-0.0304=0.5696

    So, you can say with 95% confidence that the proportion of people who say they smoke is between 56.96% and 63.04%.
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  3. #3
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    how do you determine the margin of error for a dataset? is it fixed?
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  4. #4
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    Quote Originally Posted by GoneFishing View Post
    If standard deviation is 80 and mean is 10:

    P ( X < 100)

    How do you do that?
    let z=(x-10)/80, then if x is normally distributed with mean 10 and sd 80,
    then z has a standard normal distribution and:

    p(X<100)=p(z<1.125)

    and this can be looked up in a table of the standard normal distribution,
    and is ~0.87.

    RonL
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by galactus View Post
    Confidence intervals aren't difficult. Since you have a proportion, Just use the formula E=z\sqrt{\frac{pq}{n}}. This is the margin of error.

    p=0.60, q=1-p, n=1000. z that corresponds to 95% CI is 1.96

    E=1.96\sqrt{\frac{(.60)(.40)}{1000}}=0.0304

    0.60+0.0304=0.6304; 0.60-0.0304=0.5696

    So, you can say with 95% confidence that the proportion of people who say they smoke is between 56.96% and 63.04%.
    You need to say that you are using the normal approximation to the
    binomial distribution here. The 1.96 only applies to the normal distribution
    which is justified here as we have a large population and we are not
    skulking in the far skirts of the distribution.

    RonL
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