Question

a chemist examines 17 seawater samples for iron concentration for the sample data is 0.704 cc/cubic meter with a standard deviation of 0.0142. determine the 99% confidence interval for the population mean from concentration. Assume the population is approx. normal.

Step 1. Find the critical value that should be used in constructing the confidence interval ( round your answer to 3 decimal places)

Step 2. Construct the 99% confidence interval (Round answer to 3 decimal places)

Answer 1

Answer:

Critical value: $z= 2.575$

99% confidence interval: (0.695 cc/cubic meter, 0.713 cc/cubic meter).

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 Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.005 = 0.995$, so $z = 2.575$ is the critical value

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}} = 2.575*\frac{0.0142}{\sqrt{17}} = 0.0089$

The lower end of the interval is the mean subtracted by M. So it is 0.704 - 0.0089 = 0.695 cc/cubic meter.

The upper end of the interval is the mean added to M. So it is 0.704 + 0.0089 = 0.713 cc/cubic meter.

So

99% confidence interval: (0.695 cc/cubic meter, 0.713 cc/cubic meter).