Question

The concentration of carbon monoxide (CO) in a gas sample is measured by a spectrophotometer and found to be 85 ppm. Through long experience with this instrument, it is believed that its measurements are unbiased and normally distributed, with an uncertainty (standard deviation) of 9 ppm. Find a 95% confidence interval for the concentration of CO in this sample. Round the answers to two decimal places. The 95% confidence interval is

Answer 1

Answer:

The confidence interval is $(85 - \frac{14.81}{\sqrt{n}},85 + \frac{14.81}{\sqrt{n}})$, in which n is the size of the sample.

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:

$\alpha = \frac{1 - 0.95}{2} = 0.025$

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 $1 - 0.025 = 0.975$, so Z = 1.96.

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.

$M = 1.645\frac{9}{\sqrt{n}} = \frac{14.81}{\sqrt{n}}$

The lower end of the interval is the sample mean subtracted by M. So it is $85 - \frac{14.81}{\sqrt{n}}$

The upper end of the interval is the sample mean added to M. So it is $85 + \frac{14.81}{\sqrt{n}}$

The confidence interval is $(85 - \frac{14.81}{\sqrt{n}},85 + \frac{14.81}{\sqrt{n}})$, in which n is the size of the sample.