We would like to estimate the true mean number of hours adults sleep at night. Suppose that sleep time is known to follow a Normal distribution with standard deviation 1.5 hours. For a sample of 12, what is the margin of error for a 95% confidence interval

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Answer:

The margin of error for a 95% confidence interval is 0.8487 hours of sleep.

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, we find the margin of error 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.

For a sample of 12, what is the margin of error for a 95% confidence interval

[tex]M = 1.96*\frac{1.5}{\sqrt{12}} = 0.8487[/tex]

The margin of error for a 95% confidence interval is 0.8487 hours of sleep.