Listed below are measured amounts of lend (in micrograms per cubic meter or mu g
ID: 3291563 • Letter: L
Question
Listed below are measured amounts of lend (in micrograms per cubic meter or mu g/m^3) in the ar. The EPA has established an air quality standard for lead of 1 5 mu g/m^3. The measurements shown below were recorded at a building on differed days. Use the given values to construct a 95% confidence interval estimate of the mean amount of lead in the air. Is there anything about this data set suggesting that the confidence interval might not be very good? 5.40 1.300 30 0.76 0.66 1.00 Click here to view a t distribution table. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. What is the confidence interval for the population mean mu? mu g/m^3Explanation / Answer
mean value of measurements (x_bar) is (5.40 + 1.30 + 0.39 + 0.76 + 0.66 + 1.00)/6 = 9.51/6 = 1.58 ug/m^3
standard deviation of measurements (sigma) is sqrt[((5.40 - 1.58)^2 + (1.30 - 1.58)^2 + (0.39 - 1.58)^2 + (0.76 - 1.58)^2 + (0.66 - 1.58)^2 + (1.00 - 1.58)^2)/(6 - 1)] = sqrt(17.9421/5) = sqrt(3.5884) = 1.90
Because there are only six measurements and the problem does not state whether the measurements are normally distributed, the most conservative answer uses the Student's t-distribution. It may be possible to use the normal distribution if other information is available to you.
The critical t-value for a sample size of six (five degrees of freedom) at the 95% confidence level is 2.571.
The confidence level inequality starts out looking like
-2.571 < (x_bar - mu)/(sigma/sqrt(n)) < 2.571
manipulating so mu is alone in the center gives
x_bar - 2.571(sigma/sqrt(n)) < u < x_bar + 2.571(sigma/sqrt(n))
1.58 - 2.571(1.90/sqrt(6)) < u < 1.58 + 2.571(1.90/sqrt(6))
1.58 - 2.57(0.7757) < u < 1.58 + 2.57(0.7757)
-0.41 < u < 3.57
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