In: Statistics and Probability
Two catalysts are being analyzed to determine how they affect the mean yield of a chemical process. Specifically, catalyst 1 is currently in use, but catalyst 2 is acceptable. Since catalyst 2 is cheaper, it should be adopted, providing it does not change the process yield. An experiment is run in the pilot plant and results in the data shown. (a) Is there any difference between the mean yields? Use α = 0.05 and assume equal variances. (b) Find the 100(1-α) % confidence interval for (µ1 - µ2). Use α = 0.05.
Observation number | Catalyst 1 | Catalyst 2 | |
1 | 91.5 |
89.19 |
|
2 |
94.18 | 90.95 | |
3 | 82.18 | 90.46 | |
4 | 95.39 | 93.21 | |
5 | 91.79 | 97.19 | |
6 | 89.07 | 97.04 | |
7 | 94.72 | 91.07 | |
8 | 89.21 | 92.75 |
(a)
:Mean yield with catalyst 1
: Mean yield with catalyst 1
Null hypothesis : Ho :
Alternate Hypothesis : Ha : (Two tailed test)
Test Statistic when equal variances are assumed:
Sample mean :
Sample standard deviation :
Sample 1 : Catalyst 1 :
Observation number | x1: Catalyst 1 | (x1 - ) | (x1 - )2 |
1 | 91.5 | 0.495 | 0.245025 |
2 | 94.18 | 3.175 | 10.080625 |
3 | 82.18 | -8.825 | 77.880625 |
4 | 95.39 | 4.385 | 19.228225 |
5 | 91.79 | 0.785 | 0.616225 |
6 | 89.07 | -1.935 | 3.744225 |
7 | 94.72 | 3.715 | 13.801225 |
8 | 89.21 | -1.795 | 3.222025 |
Total | 728.04 | 128.8182 | |
Mean: | 91.005 |
n1 : sample size = 8
Sample mean :
Sample standard deviation :
Sample 2 : catalyst 2
Observation number | Catalyst 2 | (x2-) | x2-)2 |
1 | 89.19 | -3.5425 | 12.54931 |
2 | 90.95 | -1.7825 | 3.177306 |
3 | 90.46 | -2.2725 | 5.164256 |
4 | 93.21 | 0.4775 | 0.228006 |
5 | 97.19 | 4.4575 | 19.86931 |
6 | 97.04 | 4.3075 | 18.55456 |
7 | 91.07 | -1.6625 | 2.763906 |
8 | 92.75 | 0.0175 | 0.000306 |
Total | 741.86 | 62.30695 | |
Mean: | 92.7325 |
n2 : sample size of sample 2 = 8
Sample mean :
Sample standard deviation :
As P-Value i.e. is greater than Level of significance i.e
(P-value:0.3656 > 0.05:Level of significance); Fail to Reject
Null Hypothesis
Not enough evidence to conclude that there is any difference
between the mean yields
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(b)
Formula for Confidence Interval for Difference in two Population means
100(1-) = 100(1-0.05) = 95%
95% Confidence Interval for Difference in two Population means
95% Confidence Interval for Difference in two Population means = (-5.6898,2.2348)