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

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Answer:

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

Step-by-step explanation:

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

[tex]\alpha = \frac{1 - 0.95}{2} = 0.025[/tex]

Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].

That is z with a pvalue of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.

Now, find the margin of error M as such

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

[tex]M = 1.645\frac{9}{\sqrt{n}} = \frac{14.81}{\sqrt{n}}[/tex]

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

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

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