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# A random sample of size displaystyle{n}={25} from a normal distribution with mean displaystylemu and variance 36 has sample mean displaystyleoverline{{X}}={16.3} a) Calculate confidence intervals for displaystylemu at three levels of confidence: 80%, 90% text{and} 99%. How to the widths of the confidence intervals change? b) How would the CI width change if n is increased to 100? # A random sample of size displaystyle{n}={25} from a normal distribution with mean displaystylemu and variance 36 has sample mean displaystyleoverline{{X}}={16.3} a) Calculate confidence intervals for displaystylemu at three levels of confidence: 80%, 90% text{and} 99%. How to the widths of the confidence intervals change? b) How would the CI width change if n is increased to 100?

Question
Confidence intervals asked 2020-11-29
A random sample of size $$\displaystyle{n}={25}$$ from a normal distribution with mean $$\displaystyle\mu$$ and variance 36 has sample mean $$\displaystyle\overline{{X}}={16.3}$$
a)
Calculate confidence intervals for $$\displaystyle\mu$$ at three levels of confidence: $$80%, 90% \text{and} 99%$$. How to the widths of the confidence intervals change?
b) How would the CI width change if n is increased to 100?

## Answers (1) 2020-11-30
Step 1
a)
Computation of $$80\%$$ confidence interval is obtained below:
It is assumed that the population is approximately normal.
Here, $$\displaystyle\alpha\ {i}{s}\ {0.20}{\left(={1}-{0.80}\right)}$$.
Degrees of freedom is $$\displaystyle{24}={\left({n}-{1}\right)}.$$
Using EXCEL formula $$\displaystyle“={T}.{I}{N}{V}{.2}{T}{\left({0.2},{24}\right)}\text{"}$$, the critical value is obtained as 1.3178. That is, $$\displaystyle{t}_{{\frac{\alpha}{{2}}}}={1.3178}.$$
The $$80\%$$ confidence interval is:
$$\displaystyle{C}{I}={16.3}\pm{\left({1.3178}\right)}\frac{6}{\sqrt{{25}}}$$
$$\displaystyle={\left(\begin{matrix}{16.3}-{1.5814}\\{16.3}+{1.5814}\end{matrix}\right)}$$
$$\displaystyle={\left({14.7186},{17.8814}\right)}$$
Step 2
Computation of $$90\%$$ confidence interval is obtained below:
It is assumed that the population is approximately normal.
Here, $$\displaystyle\alpha\ {i}{s}\ {0.10}{\left(={1}-{0.90}\right)}.$$
Degrees of freedom is $$\displaystyle{24}{\left(={n}-{1}\right)}.$$
Using EXCEL formula $$\displaystyle“={T}.{I}{N}{V}{.2}{T}{\left({0.10},{24}\right)}\text{"}$$, the critical value is obtained as 1.7108. That is, $$\displaystyle{t}_{{\frac{\alpha}{{2}}}}={1.7108}.$$
The $$90\%$$ confidence interval is:
$$\displaystyle{C}{I}={16.3}\pm{\left({1.7108}\right)}\frac{6}{\sqrt{{25}}}$$
$$\displaystyle={\left(\begin{matrix}{16.3}-{1.7108}\\{16.3}+{1.7108}\end{matrix}\right)}$$
$$\displaystyle={\left({14.589},{18.011}\right)}$$
Step 3
Computation of $$99\%$$ confidence interval is obtained below:
It is assumed that the population is approximately normal.
Here, $$\displaystyle\alpha\ {i}{s}\ {0.01}{\left(={1}-{0.99}\right)}.$$
Degrees of freedom is $$\displaystyle{24}{\left(={n}-{1}\right)}$$
Using EXCEL formula $$\displaystyle“={T}.{I}{N}{V}{.2}{T}{\left({0.01},{24}\right)}\text{"}$$, the critical value is obtained as 2.7969. That is, $$\displaystyle{t}_{{\frac{\alpha}{{2}}}}={2.7969}.$$
The $$99\%$$ confidence interval is:
$$\displaystyle{C}{I}={16.3}\pm{\left({1.7108}\right)}\frac{6}{\sqrt{{25}}}$$
$$\displaystyle={\left(\begin{matrix}{16.3}-{2.053}\\{16.3}+{2.053}\end{matrix}\right)}$$
$$\displaystyle={\left({14.247},{18.353}\right)}$$
Step 4
The confidence level is increased, and then the length of the confidence interval gets wider.
Therefore, the confidence interval gets wider.

### Relevant Questions asked 2021-05-05

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Previous studies show that $$\sigma_1 = 19$$.
For Englewood (a suburb of Denver), a random sample of $$n_2 = 12$$ winter days gave a sample mean pollution index of $$x_2 = 37$$.
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