Question

The data below are daily carbon monoxide emissions, in parts per million (ppm).

45, 30, 38, 42, 63, 43, 102, 86, 99, 63, 58, 34, 37, 55, 58, 153, 75, 58, 36, 59, 43, 102, 52, 30, 21, 40, 141, 85, 161, 86, 71, 12.5, 20, 4, 20, 25, 170, 15, 20, 15.

Data from DASL http://lib.stat.cmu.edu/DASL/Datafiles/Refinery.html.

We wish to know if these data suggest that the overall average daily emission is more than 45 ppm.

(a) Report an appropriate hypothesis test with α=0.01.Write out the full procedure taught in this class. (Hint: do 1-Var Stats first to get Sx, and use Sx for the test, not σx.)

(b) If we were making an error here, which type would it be? Write out in words, in context, what making such an error would mean

Answer 1

n = 40

Mean of sample: x = sum/n = 2367.5/40 = 59.188

Standard deviation of sample: s (Using Exce; function STDEV.S)= 41.620

Population mean: μ = 45

α=0.01

a) H0: μ = 45

H1: μ > 45

Test statistic: t = (x-μ)/(s/n^0.5) = (59.188-45)/(41.62/20^0.5) = 1.525

Degrees of freedom: df = n-1=40-1=39

Critical t (Using Excel function (T.INV(probability,df)) = ABS(T.INV.2T(0.01.39)) = 2.43

SInce test statistic is less than critical value, we do not reject the null hypothesis and conclude that μ = 45.

So, overall average daily emission is equal to 45 ppm.

b) Type 1 error could have been made here.

It means rejecting a null hypothesis that is true. Had the error been made, we would have concluded that μ > 45, instead of μ = 45.