Question

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
ANSWERED
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.

0
 
Best answer

expert advice

Have a similar question?
We can deal with it in 3 hours

Relevant Questions

asked 2021-01-05
You are given the sample mean and standard deviation of the population. Use this information to construct the​ \(\displaystyle{90}\%{\quad\text{and}\quad}​{95}\%\) confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals.
From a random sample of 66 ​dates, the mean record high daily temperature in a certain city has a mean of \(\displaystyle{85.69}^{\circ}{F}\). Assume the population standard deviation is \(\displaystyle{13.60}^{\circ}{F}.\)
The​ \(90\%\) confidence interval is
The​ \(95\%\) confidence interval is
Which interval is​ wider?
Interpret the results.
asked 2021-06-16
Let p be the population proportion for the situation. (a) Find point estimates of p and q, (b) construct 90% and 95% confidence intervals for p, and (c) interpret the results of part (b) and compare the widths of the confidence intervals. In a survey of 2202 U.S. adults, 1167 think antibiotics are effective against viral infections.
asked 2021-06-02
Let p be the population proportion for the situation. (a) Find point estimates of p and q, (b) construct 90% and 95% confidence intervals for p, and (c) interpret the results of part (b) and compare the widths of the confidence intervals. In a survey of 2223 U.S. adults, 1334 say an occupation as an athlete is prestigious.
asked 2021-06-23
Let p be the population proportion for the situation. (a) Find point estimates of p and q, (b) construct 90% and 95% confidence intervals for p, and (c) interpret the results of part (b) and compare the widths of the confidence intervals. In a survey of 1035 U.S. adults, 745 say they want the U.S. to play a leading or major role in global affairs.
...