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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?

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

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

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 $\pm 1.96$

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  • $\begingroup$ 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? $\endgroup$
    – pingu
    Feb 25 '21 at 9:20
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    $\begingroup$ @pingu. Exaclty. If my answer has been useful you can mark it as accepted $\endgroup$
    – tommik
    Feb 25 '21 at 9:21
  • $\begingroup$ Your answer has been very helpful! Thank you! $\endgroup$
    – pingu
    Feb 25 '21 at 9:24

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