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Math Help - p value and confidence interval

  1. #1
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    p value and confidence interval

    In a random sample of ten cans of corn from supplier B, the average weight per can of corn was sample mean or x bar = 9.4 ounces with standard deviation s = 1.8 ounces. Does this sample contain sufficient evidence to indicate the mean weight is less than 10 ounces? Use the p-value approach. Find a 98% confidence interval for population variance
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  2. #2
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    Quote Originally Posted by bombo31
    In a random sample of ten cans of corn from supplier B, the average weight per can of corn was sample mean or x bar = 9.4 ounces with standard deviation s = 1.8 ounces. Does this sample contain sufficient evidence to indicate the mean weight is less than 10 ounces? Use the p-value approach.
    Assuming that the distribution of can weights is normal, the random variable:

    <br />
t=\sqrt{N}\frac{\bar{x}-\mu}{s}<br />

    where \bar{x} is the sample mean, \mu is the population mean,
    s is the sample standard deviation (the one where division is by
    N-1 rather than N as we need the unbiased estimator of the
    population variance here), and N is the sample size, has a
    Student's t-distribution with N-1 degrees of freedom.

    The null hypothesis is that \mu=10, and then we have:

    <br />
t=\sqrt{10} \frac{9.4-10}{1.8}\approx -1.054<br />

    Now looking this up in an appropriate table for 9 degrees of freedom gives:

    p(t \le -1.054)=0.1597,

    or \sim 16 \%, and with this level of probability we would not reject the
    hypothesis that \mu=10.

    Now we could have tried using the large sample assumption, which would then
    justify treating:

    <br />
z=\sqrt{10} \frac{\bar{x}-\mu}{s}<br />

    as being normally distributed, then looking this up in a normal table we would
    find that:

    p(z \le -1.054)=0.1459,

    and so the same conclusion would be arrived at.
    Last edited by CaptainBlack; July 8th 2006 at 03:15 AM.
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  3. #3
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    Quote Originally Posted by bombo31
    In a random sample of ten cans of corn from supplier B, the average weight per can of corn was sample mean or x bar = 9.4 ounces with standard deviation s = 1.8 ounces. ..


    .. Find a 98% confidence interval for population variance
    If we assume that the weight of cans has a normal distribution then:

    x^2=\frac{(N-1)s^2}{\sigma^2} \sim \chi^2_{N-1}

    That is x^2 has a chi-squared distribution with N-1 degrees of freedom.

    Now we want a 98% confidence region so we need the critical values
    of \chi^2_{9} for 1% and 99% (as the area between will constitute 98%.
    These critical values are: 21.7 and 2.09, so for
    our confidence region:

    2.09 \le x^2 \le 21.7,

    or:

    2.09 \le \frac{(N-1)s^2}{\sigma^2} \le 21.7,

    so:

    <br />
\frac{(N-1) s^2}{2.09} \ge \sigma^2 \ge \frac{(N-1) s^2}{21.7}<br />

    substituting the values for s and N:

    <br />
10.12 \ge \sigma^2 \ge 0.975<br />

    Hence the 98% confidence interval for \sigma^2 (variance) is:
    [0.975,10.12].

    RonL
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    Thanks

    Thanks much.I think there is a miscalculation in the original statistic. I got
    -1.053. Thanks a lot. You are the best.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by bombo31
    Thanks much.I think there is a miscalculation in the original statistic. I got
    -1.053. Thanks a lot. You are the best.
    Typo that propagated through the working

    I will correct it.

    Note same error crept into your other question as well, both now corrected

    Thanks

    RonL
    Last edited by CaptainBlack; July 8th 2006 at 03:24 AM.
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