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

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Answer:

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

Step-by-step explanation:

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

[tex]\alpha = \frac{1-0.95}{2} = 0.025[/tex]

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

So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]

Now, find M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

[tex]M = 1.96*\frac{10.51}{\sqrt{30}} = 3.76[/tex]

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.

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