Question

Assume that a computer was used to generate the given confidence interval for the population mean. Find the sample mean or margin of error as specified3.23 < < 3.87find the sample meana.3.23 b.3.550 c.3.610 d.3.490

Answer 1

Answer:

The sample mean is

b.3.55

The margin of error is

0.32

Step-by-step explanation:

Deep explanation about a confidence interval

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

$\alpha = \frac{1-0.9}{2} = 0.05$

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 M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

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

$M = 1.645*215 = 353.675$

The lower end of the interval is the mean subtracted by M. So it is 6.4 - 0.3944 = 6.01 hours.

The upper end of the interval is the mean added to M. So it is 6.4 + 0.3944 = 6.74 hours.

In this problem:

The deep explanation is not that important.

We just have to recognize that the interval has a lower end and an upper end. The distance from both the upper and the lower end to the mean is M. This means that the sample mean is the halfway point between the lower end and the upper end.

The margin of error is the distance of these two points(lower and upper end) to the mean.

In our interval

Lower end: 3.23

Upper end: 3.87

Sample mean

$M = \frac{3.23 + 3.87}{2} = 3.55$

So the correct answer is:

b.3.55

The margin of error is

3.87 - 3.55 = 3.55 - 3.23 = 0.32