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.
0

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