Question

A geologist examines 14 geological samples for iron concentration. The mean iron concentration for the sample data is 0.181 cc/cubic meter with a standard deviation of 0.0318. Determine the 90% confidence interval for the population mean iron concentration. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Answer 1

Answer:

The 90% confidence interval for the population mean iron concentration is between 0.167 cc/m³ and 0.195 cc/m³.

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:

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

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.

$M = 1.645*\frac{0.0318}{\sqrt{14}} = 0.0140$

The lower end of the interval is the sample mean subtracted by M. So it is 0.181 - 0.0140 = 0.167 cc/m³.

The upper end of the interval is the sample mean added to M. So it is 0.181 + 0.0140 = 0.195 cc/m³.

The 90% confidence interval for the population mean iron concentration is between 0.167 cc/m³ and 0.195 cc/m³.