Question

The pulse rate (in bpm) of a random sample of 30 Peruvian Indians was collected. The mean pulse rate of the sample is 70.2, with a sample standard deviation of 10.51. Compute a 95% confidence interval for the population mean. Assume the pulse rate is normally distributed

Answer 1

The 95% confidence interval for the population mean is between 66.44 bpm and 73.96 bpm.

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1-0.95)/(2) = 0.025`

Now, we have to find z in the Ztable as such z has a pvalue of
`1-\alpha`.

So it is z with a pvalue of
`1-0.025 = 0.975`, so
`z = 1.96`

Now, find M as such

`M = z*(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

`M = 1.96*(10.51)/(√(30)) = 3.76`

The lower end of the interval is the mean subtracted by M. So it is 70.2 - 3.76 = 66.44 bpm

The upper end of the interval is the mean added to M. So it is 70.2 + 3.76 = 73.96 bpm

The 95% confidence interval for the population mean is between 66.44 bpm and 73.96 bpm.