# You are given the sample mean and standard deviation of the population. Use this information to construct the​ displaystyle{90}%{quadtext{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.

Question
Confidence intervals
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.

2021-01-06
Step 1
Given:
This mean recorded temperature is: $$\displaystyle\overline{{x}}={85.69}{F}^{\circ}$$
The population's standard deviation is: $$\displaystyle\sigma={13.60}{F}^{\circ}$$
The number of random samples is: $$\displaystyle{n}={66}$$ dates.
Confidence level $$\displaystyle-{90}\%$$:
The expression for the confidence level is,
$$\displaystyle{1}-{2}\alpha={0.9}$$
$$\displaystyle\alpha=\frac{{{1}-{0.9}}}{{2}}$$
$$\displaystyle={0.05}$$
Step 2
From the graph of the normal distribution, the confidence level is symmetric about the P axis. The expression to calculate the corresponding value of P is,
$$\displaystyle{P}={1}-\alpha$$
$$\displaystyle={1}-{0.05}$$
$$\displaystyle={0.95}$$
From the z-table corresponding to $$\displaystyle{P}={0.95}$$, the value of z is 1.65
The expression to calculate the interval is,
$$\displaystyle{c}.{I}=\overline{{x}}\pm{z}\frac{\sigma}{\sqrt{{n}}}$$
Substitute the values in the above expression.
$$\displaystyle{c}.{l}={85.69}\pm{1.65}\times\frac{13.60}{\sqrt{{66}}}$$
$$\displaystyle={85.69}\pm{2.762}$$
$$\displaystyle={\left({85.69}-{2.762},{85.69}-{2.762}\right)}$$
$$\displaystyle={\left({82.928},{88.452}\right)}$$
Therefore, the interval for confidence leven $$90\%$$ is from 82.928 to 88.452
Step 3
Confidence level $$- 95\%:$$
The expression for the confidence level is,
$$\displaystyle{1}-{2}\alpha={0.95}$$
$$\displaystyle\alpha=\frac{{{1}-{0.95}}}{{2}}$$
$$\displaystyle={0.025}$$
From the graph of the normal distribution, the confidence level is symmetric about the P axis. The expression to calculate the corresponding value of P is,
$$\displaystyle{P}={1}-\alpha$$
$$\displaystyle={1}-{0.025}$$
$$\displaystyle={0.975}$$
From the z-table corresponding to $$\displaystyle{P}={0.975}$$, the value of z is 1.96.
The expression to calculate the interval is,
$$\displaystyle{c}.{l}=\overline{{x}}\pm{z}\frac{\sigma}{\sqrt{{n}}}$$
Substitute the values in the above expression.
$$\displaystyle{c}.{l}-{85.69}\pm{1.96}\times\frac{13.60}{\sqrt{{66}}}$$
$$\displaystyle={85.69}\pm{3.281}$$
$$\displaystyle={\left({85.69}-{3.281},{85.69}-{3.281}\right)}$$
$$\displaystyle={\left({82.409},{88.971}\right)}$$
Hence, the interval for confidence level $$95\%$$ is from 82.409 to 88.971, and the interval for confidence level $$95\%$$ is the widest one.

### Relevant Questions

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?
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State the null and alternate hypotheses.
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The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations,
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$$\displaystyle{0.125}<{P}-\text{value}<{0},{250}$$
$$\displaystyle{0},{050}<{P}-\text{value}<{0},{125}$$
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P-value $$\displaystyle<{0.005}$$
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P.vaiue Pevgiue
P-value f P-value
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