Question

The number of hours that a nine month old baby sleeps at night are normally distributed with a population standard deviation of 1.5 hours and an unknown population mean. A random sample of 22 nine month old babies is taken and results in a sample mean of 12 hours. Find the margin of error for a confidence interval for the population mean with a 90% confidence level.

Answer 1

0.526

Explanation:

To find the margin of error we need to identify three things: the z-score, σ, and n.

1.Find zα2 using invNorm. The invNormfunction has one input: probability.

Here, α=1−0.90=0.10. Probability is then 1−0.102=0.95. To find our z-score, we select invNorm after pressing 2nd then VARS. Type invNorm(0.95). The output Is 1.6448. This is the z-score.

2. σ=1.5.

3. n=22.

4. We type 1.6448×1.522√ on the calculator. The output is 0.526, when rounded to three decimal places. This is the margin of error.

Answer 2

The margin of error is of 0.7123 hours.

Explanation:

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

`\alpha = (1-0.9)/(2) = 0.05`

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
`1-0.05 = 0.95`, so
`z = 1.645`

Now, find the margin of error 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.

Find the margin of error for a confidence interval for the population mean with a 90% confidence level.

We have that
`\sigma = 1.5, n = 12`. So

`M = 1.645*(1.5)/(√(12)) = 0.7123`

The margin of error is of 0.7123 hours.