The concentration of carbon monoxide (CO) in a gas sample is measured by a spectrophotometer and found to be 85 ppm. Through long experience

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

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Edana Edana 1 day 2021-07-22T06:51:53+00:00 1 Answers 0 views 0

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    2021-07-22T06:53:17+00:00

    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.

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