A statistics practitlioner took a randoml sample of 54 observations form a popu

Sinead Mcgee 2021-10-02 Answered
A statistics practitlioner took a randoml sample of 54 observations form a population whose standard deviation is 29 and computed the sample mean to be 97.
Note: For each confidence interval, enter your answer in the forem (LCL, UCL). You must include the parentheses and the comma between the confidence limits.
A) Estimate the population mean with \(\displaystyle{95}\%\) confidence.
B) Estimate the population mean with \(\displaystyle{90}\%\) confidence.
C) Estimate the population mean with \(\displaystyle{99}\%\) confidence.

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Expert Answer

krolaniaN
Answered 2021-10-03 Author has 12799 answers
Setp 1
Given:
\(\displaystyle\sigma={29}\)
\(\displaystyle{n}={54}\)

\(\displaystyle\overline{{{x}}}={97}\)
a) From the Standard Normal Table, the value of \(\displaystyle{z}\cdot\) for \(\displaystyle{95}\%\) level is 1.96
The \(\displaystyle{95}\%\) confidence interval for the population mean is obtained as below:
Sample statistic \(\displaystyle\pm{z}\cdot{S}{E}=\overline{{{x}}}\pm{z}\cdot{\frac{{\sigma}}{{\sqrt{{{n}}}}}}\)
\(\displaystyle={97}\pm{\left({1.96}\times{\frac{{{29}}}{{\sqrt{{{54}}}}}}\right)}\)
\(\displaystyle{97}\pm{7.7349}\)
\(\displaystyle={\left({89.2651},\ {104.7349}\right)}\)
Thus, the \(\displaystyle{95}\%\) confidence interval for the population mean is \(\displaystyle{\left({89.2651},\ {104.7349}\right)}.\)
Step 2
b) From the Standard Normal Table, the value of \(\displaystyle{z}\cdot\) for \(\displaystyle{90}\%\) level is 1.645.
The \(\displaystyle{90}\%\) confidence interval for the population mean is obtained as below:
Sample statistic \(\displaystyle\pm{z}\cdot{S}{E}=\overline{{{x}}}\pm{z}\cdot{\frac{{\sigma}}{{\sqrt{{{n}}}}}}\)
\(\displaystyle={97}\pm{\left({1.645}\times{\frac{{{29}}}{{\sqrt{{{54}}}}}}\right)}\)
\(\displaystyle={97}\pm{6.4918}\)
\(\displaystyle={\left({90.5082},\ {103.4918}\right)}\)
Thus, the \(\displaystyle{90}\%\) confidence interval for the population mean is \(\displaystyle{\left({90.5082},\ {103.4918}\right)}.\)
Step 3
c) From the Standard Normal Table, the value of \(\displaystyle{z}\cdot\) for \(\displaystyle{99}\%\) level is 2.576
The \(\displaystyle{99}\%\) confidence interval for the population mean is obtained as below:
Sample statistic \(\displaystyle\pm{z}\cdot{S}{E}=\overline{{{x}}}\pm{z}\cdot{\frac{{\sigma}}{{\sqrt{{{n}}}}}}\)
\(\displaystyle={97}\pm{\left({2.576}\times{\frac{{{29}}}{{\sqrt{{{54}}}}}}\right)}\)
\(\displaystyle={97}\pm{10.1659}\)
\(\displaystyle={\left({86.8341},\ {107.1659}\right)}\)
Thus, the \(\displaystyle{99}\%\) confidence interval for the population mean is \(\displaystyle{\left({86.8341},\ {107.1659}\right)}.\)
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