$95\%$ confidence interval with unknown standard deviation

Question

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?

Answer 1

If the standard deviation is not known, the pivotal quantity to be used is the following:

$$t=\frac{\overline{X}_n-\mu}{S}\sqrt{n -1}$$

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 $\pm 1.96$