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Margaret [11]
3 years ago
5

A manufacturer produces piston rings for an automobile engine. It is known that ring diameter is normally distributed with σ=0.0

01 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.
Mathematics
1 answer:
oksian1 [2.3K]3 years ago
7 0
<h2>Answer with explanation:</h2>

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

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