 # 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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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 $$\displaystyle{\left({86.8341},\ {107.1659}\right)}.$$