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.

Answer 1

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)