Question

HELP. How do I calculate the confidence interval 90%, 95%, and 99% for a sample from a normal population when X=40, o=7, and n=13?

Answer 1

Solution

We can assume a normal distribution since the problem states that the samples came from this distributoon

And for this case we can use the following formula:

$\operatorname{mean}\pm Z_{\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}}$

For this case mean = 40 , sigma = 7, n=13 and z_alpha/2 represent a quantile from the normal standard distribution, for the confidence levels 90%, 95% and 99% the z values are: 1.64, 1.96 and 2.58 and replacing we have:

90% confidence

$40\pm1.64\cdot\frac{7}{\sqrt[]{13}}=(36.816;43.184)$

95% confidence

$40\pm1.96\cdot\frac{7}{\sqrt[]{13}}=(36.195;43.805)$

99% confidence

$40\pm2.58\cdot\frac{7}{\sqrt[]{13}}=(34.991;45.009)$