Question

A sample of 50 elements from a population with a standard deviation of 20 is selected. The sample mean is 150. The 90% confidence interval for is: a.165.0 to 185.0. b.146.4 to 153.6. c.145.3 to 154.7. d.171.8 to 188.2.

Answer 1

Answer:

c.145.3 to 154.7.

Step-by-step explanation:

We have 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 Z-table as such z has a p-value of $1 - \alpha$.

That is z with a p-value of $1 - 0.05 = 0.95$, so Z = 1.645.

Now, find the margin of error 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.

$M = 1.645\frac{20}{\sqrt{50}} = 4.7$

The lower end of the interval is the sample mean subtracted by M. So it is 150 - 4.7 = 145.3.

The upper end of the interval is the sample mean added to M. So it is 150 + 4.7 = 154.7.

Thus the correct answer is given by option c.