Step 1

Given:

Sample size, \(\displaystyle{n}={13}\)

Sample mean, \(\displaystyle\overline{{x}}={31}\)

Sample standard deviation, \(\displaystyle{s}={17}\)

The degree of confidence, \(\displaystyle{c}={99}\%{\quad\text{or}\quad}{0.99}\)

Level of significance,

\(\displaystyle\alpha={1}-{c}\)

\(\displaystyle={1}-{0.99}\)

\(\displaystyle={0.01}\)

Let the population mean to be estimated using confidence intervals be \(\displaystyle\mu\)

Step 2

The \(99\%\) confidence interval for the population mean is:

\(\displaystyle\mu=\overline{{x}}\pm{t}_{{\frac{a}{{2}},{d}{f}}}\frac{s}{\sqrt{{n}}}\)

\(\displaystyle={31}\pm{t}_{{\frac{0.01}{{2}},{12}}}\frac{17}{\sqrt{{13}}}\)

\(\displaystyle={31}\pm{3.055}\times\frac{17}{\sqrt{{13}}}\)

\(\displaystyle={31}\pm{14.4}\)

\(\displaystyle={\left({16.6},{45.4}\right)}\)

Thus, the required \(99\%\) confidence interval for the population mean is:

\(\displaystyle{16.6}<\mu<{45.4}\)</span>

Given:

Sample size, \(\displaystyle{n}={13}\)

Sample mean, \(\displaystyle\overline{{x}}={31}\)

Sample standard deviation, \(\displaystyle{s}={17}\)

The degree of confidence, \(\displaystyle{c}={99}\%{\quad\text{or}\quad}{0.99}\)

Level of significance,

\(\displaystyle\alpha={1}-{c}\)

\(\displaystyle={1}-{0.99}\)

\(\displaystyle={0.01}\)

Let the population mean to be estimated using confidence intervals be \(\displaystyle\mu\)

Step 2

The \(99\%\) confidence interval for the population mean is:

\(\displaystyle\mu=\overline{{x}}\pm{t}_{{\frac{a}{{2}},{d}{f}}}\frac{s}{\sqrt{{n}}}\)

\(\displaystyle={31}\pm{t}_{{\frac{0.01}{{2}},{12}}}\frac{17}{\sqrt{{13}}}\)

\(\displaystyle={31}\pm{3.055}\times\frac{17}{\sqrt{{13}}}\)

\(\displaystyle={31}\pm{14.4}\)

\(\displaystyle={\left({16.6},{45.4}\right)}\)

Thus, the required \(99\%\) confidence interval for the population mean is:

\(\displaystyle{16.6}<\mu<{45.4}\)</span>