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Math Help - confidence intervals

  1. #1
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    Smile confidence intervals

    When Gregor Mendel (born 1822, died 1884) conducted his famous genetics experiment with peas, one sample of offspring consisted of 428 green peas and 152 yellow peas. in Mendel's theory such experiments yield a characteristic distribution of colors, but each individual plant is an independent trial that may have one color or the other.

    a) calculate a 95% confidence interval for the population proportion of yellow offspring. (5 points)

    b) according to Mendel's theory the proportion of yellow offspring in this experiment should be 25%. Based on the confidence interval calculated in part a) assess whether the results of this experiment are consistent with Mendel's theory. Justify your answer. (3 points)

    c) suppose someone says "well, the experimental proportion was over 26% and you can't go against the evidence. it looks like Mendel should have adjusted his theory" provide an appropriate answer. (3 points)
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  2. #2
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    Quote Originally Posted by sanja85 View Post
    When Gregor Mendel (born 1822, died 1884) conducted his famous genetics experiment with peas, one sample of offspring consisted of 428 green peas and 152 yellow peas. in Mendel's theory such experiments yield a characteristic distribution of colors, but each individual plant is an independent trial that may have one color or the other.

    a) calculate a 95% confidence interval for the population proportion of yellow offspring. (5 points)

    b) according to Mendel's theory the proportion of yellow offspring in this experiment should be 25%. Based on the confidence interval calculated in part a) assess whether the results of this experiment are consistent with Mendel's theory. Justify your answer. (3 points)

    c) suppose someone says "well, the experimental proportion was over 26% and you can't go against the evidence. it looks like Mendel should have adjusted his theory" provide an appropriate answer. (3 points)
    See the Wikipedia page, you probably want the normal approximation interval.

    CB
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  3. #3
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    so how would you answer that question?
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  4. #4
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    Quote Originally Posted by sanja85 View Post
    so how would you answer that question?
    a) exactly like it says in the Wikipedia article the required interval is:

    \widehat{p}\pm z_{97.5} \sqrt{\frac{\widehat{p}(1-\widehat{p}}{n}}

    where \widehat{p} is the estimate of the proportion derived from your sample, z_{97.5} is the 97.5 percentile of the standard normal distribution and n is the sample size.

    Also this looks like assessed work so I am closing the thread.

    CB
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