# If displaystyle{n}={13},{left(overline{{x}}right)}={31},{quadtext{and}quad}{s}={17}, construct a confidence interval at a 99% confidence level. Assume the data came from a normally distributed population. Give your answers to one decimal place.

Question
Confidence intervals
If $$\displaystyle{n}={13},{\left(\overline{{x}}\right)}={31},{\quad\text{and}\quad}{s}={17},$$ construct a confidence interval at a $$99\%$$ confidence level. Assume the data came from a normally distributed population.

2021-03-09
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>

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