A machine produces parts with lengths that are normally distributed with displaystylesigma={0.66}. A sample of 20 parts has a mean length of 75.98. a) Give a point estimate for displaystylemu. (Give your answer correct to two decimal places.) b) Find the 95% confidence maximum error of estimate for displaystylemu. (Give your answer correct to two decimal places.) c) Find the 95% confidence interval for displaystylemu. (Give your answer correct to two decimal places.) Lower limit displaystyle=? Upper limit displaystyle=?

A machine produces parts with lengths that are normally distributed with displaystylesigma={0.66}. A sample of 20 parts has a mean length of 75.98. a) Give a point estimate for displaystylemu. (Give your answer correct to two decimal places.) b) Find the 95% confidence maximum error of estimate for displaystylemu. (Give your answer correct to two decimal places.) c) Find the 95% confidence interval for displaystylemu. (Give your answer correct to two decimal places.) Lower limit displaystyle=? Upper limit displaystyle=?

Question
Confidence intervals
asked 2020-12-12
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.
a)
Give a point estimate for \(\displaystyle\mu\). (Give your answer correct to two decimal places.)
b)
Find the \(95\%\) confidence maximum error of estimate for \(\displaystyle\mu\). (Give your answer correct to two decimal places.)
c)
Find the \(95\%\) confidence interval for \(\displaystyle\mu\). (Give your answer correct to two decimal places.)
Lower limit \(\displaystyle=?\)
Upper limit \(\displaystyle=?\)

Answers (1)

2020-12-13
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.,
\(\displaystyle={75.98}+{0.289260}\)
\(\displaystyle={76.26926}\)
0

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