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Thread: 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:

    $\displaystyle
    t=\sqrt{N}\frac{\bar{x}-\mu}{s}
    $

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

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

    $\displaystyle
    t=\sqrt{10} \frac{9.4-10}{1.8}\approx -1.054
    $

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

    $\displaystyle p(t \le -1.054)=0.1597$,

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

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

    $\displaystyle
    z=\sqrt{10} \frac{\bar{x}-\mu}{s}
    $

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

    $\displaystyle p(z \le -1.054)=0.1459$,

    and so the same conclusion would be arrived at.
    Last edited by CaptainBlack; Jul 8th 2006 at 03:15 AM.
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  3. #3
    Grand Panjandrum
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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:

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

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

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

    $\displaystyle 2.09 \le x^2 \le 21.7$,

    or:

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

    so:

    $\displaystyle
    \frac{(N-1) s^2}{2.09} \ge \sigma^2 \ge \frac{(N-1) s^2}{21.7}
    $

    substituting the values for $\displaystyle s$ and $\displaystyle N$:

    $\displaystyle
    10.12 \ge \sigma^2 \ge 0.975
    $

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

    RonL
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  4. #4
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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; Jul 8th 2006 at 03:24 AM.
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  6. #6
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    Re: p value and confidence interval

    "The data do not present sufficient evidence to indicate that the mean weight per can is less than 10 ounces..."

    the confidence interval for 98% for population mean:

    is it CI= 7.7943, 11.0057 OR CI= 8.2843, 10.5157???
    Last edited by BreeLee; Nov 8th 2015 at 07:20 PM.
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