If the inspection division of a county weights and measures department wants to estimate the mean amount of soft-drink fill in 2-liter bottl

Question

If the inspection division of a county weights and measures department wants to estimate the mean amount of soft-drink fill in 2-liter bottles to within LaTeX: \pm± 0.01 liter with 95% confidence and also assumes that the standard deviation is 0.05 liter, what sample size is needed?

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Amaya 2 weeks 2021-10-12T10:20:13+00:00 1 Answer 0

Answers ( )

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    2021-10-12T10:21:47+00:00

    Answer:

    n=97

    Step-by-step explanation:

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

    Normal distribution, is a “probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean”.

    Assuming the X follows a normal distribution

    X \sim N(\mu, \sigma=0.01)

    We know that the margin of error for a confidence interval is given by:

    Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

    The next step would be find the value of \z_{\alpha/2}, \alpha=1-0.95=.05 and \alpha/2=0.025

    Using the normal standard table, excel or a calculator we see that:

    z_{\alpha/2}=1.96

    If we solve for n from formula (1) we got:

    \sqrt{n}=\frac{z_{\alpha/2} \sigma}{Me}

    n=(\frac{z_{\alpha/2} \sigma}{Me})^2

    And we have everything to replace into the formula:

    n=(\frac{1.96(0.05)}{0.01})^2 =96.04

    And if we round up the answer we see that the value of n to ensure the margin of error required \pm=0.01 Liters is n=97.

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