Step 1

Given:

A machine produces parts with lengths that are normally distributed with \(\displaystyle\sigma={0.66}\). A sample of 20 parts has a mean length of 75.98.

To Determine:

a) A point estimate for \(\displaystyle\mu.\)

b) The \(95\%\) confidence maximum error of estimate for \(\displaystyle\mu.\)

c) The \(95\%\) confidence interval for \(\displaystyle\mu\)

Step 2

Solution:

A machine produces parts with lengths that are normally distributed with the given values,

\(\displaystyle\sigma={0.66},{n}={20},\overline{{x}}={75.98}\)

a) A point estimate for \(\displaystyle\mu\) is

\(\displaystyle\overline{{\mu}}=\overline{{x}}={75.98}\)

b) For \(95\%\) confidence intervals,

Z - critical value is 1.96.

So, The maximum confidence error of estimate for \(\displaystyle\mu\) is,

\(\displaystyle{M}.{E}.=\frac{{{Z}\times\sigma}}{\sqrt{{{n}}}}\)

\(\displaystyle{M}.{E}.=\frac{{{1.96}\times{0.66}}}{\sqrt{{{20}}}}\)

\(\displaystyle{M}.{E}.=\frac{{{1.96}\times{0.66}}}{{4.4721}}\)

\(\displaystyle{M}.{E}.=\frac{1.2936}{{4.4721}}={0.289260}\)

c) The \(95\%\) confidence interval for \(\displaystyle\mu\),

Lower Limit \(\displaystyle=\overline{{{x}}}-{M}.{E}.\)

L.L.,

\(\displaystyle={75.98}-{0.289260}\)

\(\displaystyle={75.69074}\)

Upper Limit \(\displaystyle=\overline{{{x}}}+{M}.{E}.\)

U.L.,

\(\displaystyle={75.98}+{0.289260}\)

\(\displaystyle={76.26926}\)

Given:

A machine produces parts with lengths that are normally distributed with \(\displaystyle\sigma={0.66}\). A sample of 20 parts has a mean length of 75.98.

To Determine:

a) A point estimate for \(\displaystyle\mu.\)

b) The \(95\%\) confidence maximum error of estimate for \(\displaystyle\mu.\)

c) The \(95\%\) confidence interval for \(\displaystyle\mu\)

Step 2

Solution:

A machine produces parts with lengths that are normally distributed with the given values,

\(\displaystyle\sigma={0.66},{n}={20},\overline{{x}}={75.98}\)

a) A point estimate for \(\displaystyle\mu\) is

\(\displaystyle\overline{{\mu}}=\overline{{x}}={75.98}\)

b) For \(95\%\) confidence intervals,

Z - critical value is 1.96.

So, The maximum confidence error of estimate for \(\displaystyle\mu\) is,

\(\displaystyle{M}.{E}.=\frac{{{Z}\times\sigma}}{\sqrt{{{n}}}}\)

\(\displaystyle{M}.{E}.=\frac{{{1.96}\times{0.66}}}{\sqrt{{{20}}}}\)

\(\displaystyle{M}.{E}.=\frac{{{1.96}\times{0.66}}}{{4.4721}}\)

\(\displaystyle{M}.{E}.=\frac{1.2936}{{4.4721}}={0.289260}\)

c) The \(95\%\) confidence interval for \(\displaystyle\mu\),

Lower Limit \(\displaystyle=\overline{{{x}}}-{M}.{E}.\)

L.L.,

\(\displaystyle={75.98}-{0.289260}\)

\(\displaystyle={75.69074}\)

Upper Limit \(\displaystyle=\overline{{{x}}}+{M}.{E}.\)

U.L.,

\(\displaystyle={75.98}+{0.289260}\)

\(\displaystyle={76.26926}\)