 # 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.
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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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