If n={13}, (x)=31, and s=17, construct a confidence interval at a 99% confidence level.

Question

If $n = 13, (\overset{―}{x}) = 31, and s = 17,$ 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.

Answer 1

Step 1
Given:
Sample size, $n = 13$
Sample mean, $\overset{―}{x} = 31$
Sample standard deviation, $s = 17$
The degree of confidence, $c = 99 % or 0.99$
Level of significance,
$α = 1 - c$
$= 1 - 0.99$
$= 0.01$
Let the population mean to be estimated using confidence intervals be $μ$
Step 2
The $99 %$ confidence interval for the population mean is:
$μ = \overset{―}{x} \pm t_{\frac{a}{2}, d f} \frac{s}{\sqrt{n}}$
$= 31 \pm t_{\frac{0.01}{2}, 12} \frac{17}{\sqrt{13}}$
$= 31 \pm 3.055 \times \frac{17}{\sqrt{13}}$
$= 31 \pm 14.4$
$= (16.6, 45.4)$
Thus, the required $99 %$ confidence interval for the population mean is:
$16.6 < μ < 45.4$

If n={13}, (x)=31, and s=17, construct a confidence interval at a 99% confidence level.

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