DATA A Carbon Registry provides approved measurement methodologies for energy co
ID: 3290331 • Letter: D
Question
DATA
A Carbon Registry provides approved measurement methodologies for energy companies who wish to show greenhouse gas emission reduction through retrofitting high bleed pneumatic controllers with low bleed controllers. Sample emission data from two different controller manufacturers were analyzed to determine if there was a difference between the average emission levels for the two manufacturers.
1. State null and alternative
2. Is there a difference in emission levels between the two manufacturers? The data is provided above. Use = .10 and assume unequal variances. Report the p-value in conclusion. If you used excel, write excel function used.
3. Which type of error (Type I or Type II) could have been made?
Mftr 1 655 Mftr 2 367 Mftr 1 1052.2 Mftr 2 432 Mftr 1 597.6 Mftr 2 528 Mftr 1 559.2 Mftr 2 732 Mftr 1 415.2 Mftr 2 463.2 Mftr 1 636 Mftr 2 576 Mftr 1 744 Mftr 2 696 Mftr 1 540 Mftr 2 535.2 Mftr 1 285 Mftr 2 726.7 Mftr 1 984 Mftr 2 851.5 Mftr 1 1034 Mftr 2 401.3 Mftr 1 712 Mftr 2 386 Mftr 1 657 Mftr 2 441.9 Mftr 1 703 Mftr 2 355.6 Mftr 1 324 Mftr 2 401.3 Mftr 1 518.1 Mftr 2 447 Mftr 1 274.3 Mftr 2 441.9 Mftr 2 406.4Explanation / Answer
Solution:-
1) State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1 - 2 = 0
Alternative hypothesis: 1 - 2 0
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.
2) Formulate an analysis plan. For this analysis, the significance level is 0.10. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 67.32
DF = 33
t = [ (x1 - x2) - d ] / SE
t = 1.76
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.
Since we have a two-tailed test, the P-value is the probability that a t statistic having 33 degrees of freedom is more extreme than -1.76; that is, less than -1.76 or greater than 1.76.
Thus, the P-value = 0.088
Interpret results. Since the P-value (0.088) is less than the significance level (0.10), we cannot accept the null hypothesis.
3) We could have made Type I error.
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