# $95\%$ confidence interval with unknown standard deviation

## The Question

We have the results for the heart beats per minute of a sample of $$15$$ people: $$54, 63, 58, 72, 49, 92, 70, 73, 69, 104, 48, 66, 80, 64, 77$$ Give a $$95\%$$ confidence interval for the mean number of heart beat per minute

i. if the standard deviation is known and equal to $$\sigma=15$$

ii. if the standard deviation is not known.

(Assume normal distribution.)

## My Understanding

The mean value is $$\overline{x}\approx 69.27$$.

In case of known standard deviation the interval is $$\left (\overline{x}-z\cdot \frac{\sigma}{\sqrt{n}}, \overline{x}+z\cdot \frac{\sigma}{\sqrt{n}}\right )=\left (69.27-1.960\cdot \sqrt{15}, 69.27+1.960\cdot \sqrt{15}\right )$$

In case of unknown standard deviation we calculate the estimated from the given data, $$s\approx 15.17$$. Is the formula again tha same? Or is the $$z$$ value different?

$$t=\frac{\overline{X}_n-\mu}{S}\sqrt{n}$$
which follows a Student T distribution with $$n-1$$ degrees of freedom
Thus the Confidence interval is the same but the quantiles are the two ones taken from a Student T with 14 dof: $$\pm2.14$$ instead of $$\pm 1.96$$
• Therefore the formula is the same but when we have known standard deviation we have $z$ and when we have unknown we have $t$ (from t-distribution), right? Feb 25 '21 at 9:20