The standard deviation for a population is \sigma=14.8. A sample of 21 obs

Khaleesi Herbert 2021-10-02 Answered
The standard deviation for a population is \(\displaystyle\sigma={14.8}\). A sample of 21 observations selected from this population gave a mean equal to 139.05. The population is known to have a normal distribution.
Round your answers to two decimal places.
a) Make a \(\displaystyle{99}\%\) confidence interval for \(\displaystyle\mu\). ( Enter your answer; \(\displaystyle{99}\%\) confidence interval, lower bound ,Enter your answer; \(\displaystyle{99}\%\) confidence interval, upper bound )
b) Construct a \(\displaystyle{97}\%\) confidence interval for \(\displaystyle\mu\). ( Enter your answer; \(\displaystyle{97}\%\) confidence interval, lower bound ,Enter your answer; \(\displaystyle{97}\%\) confidence interval, upper bound )
c) Determine a \(\displaystyle{95}\%\) confidence interval for \(\displaystyle\mu\). ( Enter your answer; \(\displaystyle{95}\%\) confidence interval, lower bound ,Enter your answer; \(\displaystyle{95}\%\) confidence interval, upper bound )

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Expert Answer

lobeflepnoumni
Answered 2021-10-03 Author has 16565 answers

Step 1
From the provided information,
Population standard deviation \(\displaystyle{\left(\sigma\right)}={14.8}\)
Sample size \(\displaystyle{\left({n}\right)}={21}\)
Sample mean \(\displaystyle{\left(\overline{{{x}}}\right)}={139.05}\)
a) Since, population standard deviation is known therefore z distribution will be used.
The z value at \(\displaystyle{99}\%\) confidence level from the standard normal table is 2.58.
The required \(\displaystyle{99}\%\) confidence interval for µ can be obtained as:
\(\displaystyle{C}{I}=\overline{{{x}}}\pm{z}_{{\frac{\alpha}{{2}}}}{\frac{{\sigma}}{{\sqrt{{{n}}}}}}\)
\(\displaystyle={139.05}\pm{\left({2.58}\right)}{\frac{{{14.8}}}{{\sqrt{{{21}}}}}}\)
\(\displaystyle={139.05}\pm{8.33}\)
\(\displaystyle={\left({130.72},\ {147.38}\right)}\)
Thus, the lower bound of \(\displaystyle{99}\%\) confidence interval is \(\displaystyle{130.72}\) and the upper bound is 147.38.
Step 2
b) The z value at \(\displaystyle{97}\%\) confidence level from the standard normal table is 2.17.
The required \(\displaystyle{97}\%\) confidence interval for \(\displaystyle\mu\) can be obtained as:
\(CI=\bar{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\)
\(\displaystyle{139.05}\pm{\left({2.17}\right)}{\frac{{{14.8}}}{{\sqrt{{{21}}}}}}\)
\(\displaystyle={139.50}\pm{7.01}\)
\(\displaystyle{\left({132.04},\ {146.06}\right)}\)
Thus, the lower bound of \(\displaystyle{97}\%\) confidence interval is 132.04 an the upper bound is 146.06
Step 3
c) The z value at \(\displaystyle{95}\%\) confidence level from the standard normal table is 1.96.
The required \(\displaystyle{95}\%\) confidence interval for \(\displaystyle\mu\) can be obtained as:
\(\displaystyle{C}{I}=\overline{{{x}}}\pm{z}_{{\frac{\alpha}{{2}}}}{\frac{{\sigma}}{{\sqrt{{{n}}}}}}\)
\(\displaystyle{139.05}\pm{\left({1.96}\right)}{\frac{{{14.8}}}{{\sqrt{{{21}}}}}}\)
\(\displaystyle={139.05}\pm{6.33}\)
\(\displaystyle={\left({132.72},\ {145.38}\right)}\)
Thus, the lower bound of \(\displaystyle{95}\%\) confidence interval is 132.72 and the upper bound is 145.38

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