Question

A manufacturer produces piston rings for an automobile engine. It is known that ring diameter is normally distributed with σ=0.001 millimeters. A random sample of 15 rings has a mean diameter of x¯=74.021. Construct a 99% two-sided confidence interval on the true mean piston diameter and a 95% lower confidence bound on the true mean piston diameter. Round your answers to 3 decimal places.

Answer 1

Answer with explanation:

The confidence interval for population mean is given by :-

$\overline{x}\pm z^* SE$                   (1)

, where $\overline{x}$ = sample mean

z* = critical value.

SE = standard error

and $SE=\dfrac{\sigma}{\sqrt{n}}$ , $\sigma$ = population standard deviation.

n= sample size.

As per given , we have

$\overline{x}=74.021$

$\sigma=0.001$

n= 15

It is known that ring diameter is normally distributed.

$SE=\dfrac{0.001}{\sqrt{15}}=0.000258198889747\approx0.000258199$

By z-table ,

The critical value for 95% confidence  = z*= 1.96

A 99% two-sided confidence interval on the true mean piston diameter :

$74.021\pm (2.576) (0.000258199)$     (using (1))

$74.021\pm 0.000665120624$

$74.021\pm 0.000665120624\\\\=(74.021- 0.000665120624,\ 74.021+ 0.000665120624)\\\\=(74.0203348794,\ 74.0216651206)\approx(74.020,\ 74.022)$  [Rounded to three decimal places]

∴  A 99% two-sided confidence interval on the true mean piston diameter  = (74.020, 74.022)

By z-table ,

The critical value for 95% confidence  = z*= 1.96

A 95% lower confidence bound on the true mean piston diameter:

$74.021- (1.96) (0.000258199)$    (using (1))

$74.021- 0.00050607004=74.02049393\approx74.020$ [Rounded to three decimal places]

∴  A 95% lower confidence bound on the true mean piston diameter= 74.020