A sample standard deviation of 10 weights of packages of grass seed distributed by a certain company was calculated to be 0.286. Assume that

Question

A sample standard deviation of 10 weights of packages of grass seed distributed by a certain company was calculated to be 0.286. Assume that weights are normally distributed and find a 95% confidence interval for the standard deviation of all such packages of grass seed

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Ariana 1 week 2021-09-13T15:36:11+00:00 1 Answer 0

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    2021-09-13T15:37:11+00:00

    Answer:

    \frac{(9)(0.286)^2}{19.02} \leq \sigma^2 leq \frac{(9)(0.286)^2}{2.70}

     0.0387 \leq \sigma^2 \leq 0.2727

    Now we just take square root on both sides of the interval and we got:

     0.197 \leq \sigma \leq 0.522

    Step-by-step explanation:

    Data given and notation

    s=0.286 represent the sample standard deviation

    \bar x represent the sample mean

    n=10 the sample size

    Confidence=95% or 0.95

    A confidence interval is “a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval”.  

    The margin of error is the range of values below and above the sample statistic in a confidence interval.

    The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

    Calculating the confidence interval

    The confidence interval for the population variance is given by the following formula:

    \frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

    The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

    df=n-1=10-1=9

    Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical values.  

    The excel commands would be: “=CHISQ.INV(0.025,9)” “=CHISQ.INV(0.975,9)”. so for this case the critical values are:

    \chi^2_{\alpha/2}=19.02

    \chi^2_{1- \alpha/2}=2.70

    And replacing into the formula for the interval we got:

    \frac{(9)(0.286)^2}{19.02} \leq \sigma^2 leq \frac{(9)(0.286)^2}{2.70}

     0.0387 \leq \sigma^2 \leq 0.2727

    Now we just take square root on both sides of the interval and we got:

     0.197 \leq \sigma \leq 0.522

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