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Math Help - Sampling Distribution Problem

  1. #1
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    Sampling Distribution Problem

    I am in a statistics class at Michigan State and am having trouble completing this problem that was assigned for homework. If somebody could give me some advice/pointers I would greatly appreciate it.

    A Gallup Poll finds that 30% of adults visited a casino in the past 12 months, and that 6% bet on college sports. These results come from a random sample of 1011 adults. For an SRS of size n=1011:

    a) What is the probability that the sample proportion P-hat is between .28 and .32 if the population proportion is p=.30?

    b)What is the probability that the sample proportion P-hat is between .04 and .08 if the population proportion is p=.06?

    c) How does the probability that P-hat falls within (+or-)0.02 of the true p change as p gets closer to 0?


    I thought I figured out a), however my answer did not match what was in the back of the book. The method I used was to find the standard deviation (which I computed as .01441) and to solve for each probability based on the Z-score equation (Z= (X-m)/SD). Once i had both probabilities from the z-scores, I subtracted the probability that it was below .28 from the probability that was below .32, leaving me with the probability that it was in the middle of the two. I came up with .7685, however the back of the book has the correct answer as .8354.

    Any help is appreciated, thank you!
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  2. #2
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by carte175 View Post

    A Gallup Poll finds that 30% of adults visited a casino in the past 12 months, and that 6% bet on college sports. These results come from a random sample of 1011 adults. For an SRS of size n=1011:

    a) What is the probability that the sample proportion P-hat is between .28 and .32 if the population proportion is p=.30?

    a) Here p=.3, (1-p)=q=.7, n=1011, and so, \frac{pq}{n}=.000208. Thus, P(.28\leq\hat{p}\leq\\.32)=P(\frac{.28-.30}{\sqrt{.000208}}<br />
\leq\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}}\leq\frac{.28-.30}{\sqrt{.000208}}) =<br />
P(-1.3877\leq\\Z\leq\\1.3877)=.834771.

    Now try to do part (b) using the same techniques. Look up "normal approximation to the binomial distribution" or maybe just the "centeral limit theorem" section in your text for more information on problems of this nature (or just google it).
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  3. #3
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    Thank you, this is precisely what I needed.
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