A polymer is manufactured in a batch chemical process. Viscosity measurements show that it is approximately normally distributed with a sta

Question

A polymer is manufactured in a batch chemical process. Viscosity measurements show that it is approximately normally distributed with a standard deviation of 20. A random sample of 42 batches has a mean viscosity of 759. Construct a 99% confidence interval around the true population mean viscosity.

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Lyla 1 week 2021-10-10T23:04:42+00:00 1 Answer 0

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    2021-10-10T23:06:15+00:00

    Answer: (751.05, 766.95)

    Step-by-step explanation:

    We know that the confidence interval for population mean is given by :-

    \overline{x}\pm z*\dfrac{\sigma}{\sqrt{n}},

    where \sigma =population standard deviation.

    \overline{x}= sample mean

    n= sample size

    z* = Two-tailed critical z-value.

    Given :  \sigma= 20

    n= 42

    \overline{x}=759

    We know that from z-table , the two-tailed critical value for 99% confidence interval : z* =2.576

    Now, the 99% confidence interval around the true population mean viscosity :-

    759\pm (2.5760)\dfrac{20}{\sqrt{42}}\\\\=759\pm (2.5760)(3.086067)\\\\=759\pm7.9497=(759-7.9497,\ 759+7.9497)\]\\=(751.0503,\ 766.9497)\approx(751.05,\ 766.95)

    ∴ A 99% confidence interval around the true population mean viscosity : (751.05, 766.95)

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