Question

(c) Construct a 99% confidence interval for u if the sample size, n, is 35.

Answer 1

Answer:

The confidence interval is $(\overline{x} - 1.99\frac{\sigma}{\sqrt{35}}, \overline{x} + 1.99\frac{\sigma}{\sqrt{35}})$, in which $\overline{x}$ is the sample mean and $\sigma$ is the standard deviation for the population.

Step-by-step explanation:

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

$\alpha = \frac{1 - 0.99}{2} = 0.005$

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 pvalue of $1 - 0.005 = 0.995$, so Z = 2.575.

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.

In this question:

$M = 1.99\frac{\sigma}{\sqrt{35}}$

The lower end of the interval is the sample mean subtracted by M, while upper end of the interval is the sample mean added to M. Thus, the confidence interval is $(\overline{x} - 1.99\frac{\sigma}{\sqrt{35}}, \overline{x} + 1.99\frac{\sigma}{\sqrt{35}})$, in which $\overline{x}$ is the sample mean and $\sigma$ is the standard deviation for the population.