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# You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the po

Confidence intervals
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asked 2020-12-15
You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. From a random sample of 58 ​dates, the mean record high daily temperature in a certain city has a mean of $$\displaystyle{83.43}^{{\circ}}{F}$$. Assume the population standard deviation is $$\displaystyle{14.02}^{{\circ}}{F}$$. $$\displaystyle{90}\%=$$ $$\displaystyle{95}\%=$$ Which interval is wider?

## Answers (1)

2020-12-16

Step 1 Given Data: The sample size is: $$\displaystyle{n}={58}$$ The mean temperature of the city is: $$\displaystyle\mu={83.43}^{{\circ}}{F}$$ The standard deviation of the temperature is: $$\sigma=14.02^{\circ}F$$ 95% Population: The expression calculate the 95% of the Population is, $$\displaystyle{1}-{2}\alpha={0.95}$$
$$\displaystyle\alpha={\frac{{{1}-{0.95}}}{{{2}}}}$$
$$\displaystyle={0.025}$$ Corresponds to probability 0.025, the z value from the right tail end of the z-table 1.96. Step 2 The expression to calculate the interval of the population is, $$\displaystyle{C}{I}=\overline{{{X}}}\pm{M}.{E}$$
$$\displaystyle=\overline{{{X}}}\pm{Z}\times{\frac{{\sigma}}{{\sqrt{{{n}}}}}}$$ Substitute values in the above expression. $$\displaystyle{C}{I}={83.43}\pm{1.96}\times{\frac{{{14.02}}}{{\sqrt{{{58}}}}}}$$
$$\displaystyle={83.43}\pm{3.60}$$
$$\displaystyle={\left({83.43}-{3.60},{83.43}+{3.60}\right)}$$
$$\displaystyle{\left({79.83},{87.03}\right)}$$ Thus, the interval width of the 95% level is 3.6. and the interval is (79.83, 87.03). Step 3 90% Population: The expression calculate the 95% of the Population is, $$\displaystyle{1}-{2}\alpha={0.90}$$
$$\displaystyle\alpha={\frac{{{1}-{0.90}}}{{{2}}}}$$
$$\displaystyle={0.050}$$ Corresponds to probability 0.050, the z value from the right tail end of the z-table is 1.64. The expression to calculate the interval of the Population is, $$\displaystyle{C}{I}=\overline{{{X}}}\pm{M}.{E}$$
$$\displaystyle=\overline{{{X}}}\pm{Z}\times{\frac{{\sigma}}{{\sqrt{{{n}}}}}}$$ Substitute values in the above expression. $$\displaystyle{C}{I}={83.43}\pm{1.96}\times{\frac{{{14.02}}}{{\sqrt{{{58}}}}}}$$
$$\displaystyle={83.43}\pm{3.02}$$
$$\displaystyle={\left({83.43}-{3.60},{83.43}+{3.02}\right)}$$
$$\displaystyle={\left({80.41},{86.45}\right)}$$ Thus, the interval width of 90% population level is 3.02, and the interval is (80.41, 59.45). Thus, 95% population has higher interval width.

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