Question

In: Statistics and Probability

Two catalysts are being analyzed to determine how they affect the mean yield of a chemical...

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

Solutions

Expert Solution

(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)


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