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Question 5 A chemical engineer carried out several batches of experiments to evaluate two types of catalysts. The objective is to determine whether catalyst B produces higher mean process yield than catalyst A. In this case, a random sample of 12 batches were prepared using catalyst A, resulting in an average yield of 86 % and sample standard deviation of 3. Another random sample of 15 batches were prepared using catalyst B, resulting in an average yield of 89% with a sample standard deviation of 2. Assume that the population yield measurements from the two types of catalysts are approximately normally distributed with the same population variance. (a) Formulate the null and alternative hypotheses. (2 marks) (b) (c) Compute the pooled variance. (3 marks) At 5 % level of significance, perform a two-sample t-test to determine whether there is evidence to support the claim that catalyst B produces a higher mean yield than catalyst A. (10 marks) Construct a 99% confidence interval on the difference in mean yields that can be used to test the claim in part (c) above. (5 marks)

Answer 1

Data Summary

	n	Mean (M)	Variance (SS)	Standard Deviation (SD)
X1	12   (n1)	86 (M1)	9   (SS1)	3
X2	15    (n2)	89 (M2)	4   (SS2)	2

a) Objective is to test whether catalyst B produces higher mean process yield than catalyst A

Hence, the null and alternative hypotheses are :

Ho : μ1 = μ2        where μ1, μ2 are the population means for   
Ha : μ1 < μ2           for the mean process yields of catalyst A and catalyst B respectively

b)

First find the degrees of freedom

Degrees of freedom are calculated as

df1 = n1 - 1   df2 = n2 - 1   df = n1 + n2 - 2  

Degrees of Freedom  
df1 = 11   df2 = 14
df = 25  

Pooled variance is calculated using the formula

dfiSS1 + df2SS2 dfi + df2

Pooled Variance Sp²  
Sp² = 6.2  

c)

Mean Squared Error is calculated using the following formula

п1 п

Mean Squared Error S_(M1-M2)                    
S_(M1-M2) = 0.9644                      
                      
t-statistic is calculated using following formula

t-statistic = -3.1109                     
For t = -3.1109 df = 25 we find the Left Tailed p-value using Excel function t.dist                      
p-value = t.dist(-3.1109, 25, TRUE)                      
p-value = 0.0023                      
                      
Decision                      
0.0023 < 0.05                      
that is p-value <= α                      
Hence we REJECT Ho                      
                      
Conclusion                      
There exists enough statistical evidence at α = 0.05 to show that catalyst B produces higher mean process yield than catalyst A            
                      

en-this (271—16) – (?W-W)

d)

For 99% confidence interval, margin of error (ME) is given by                  

ME = z ME = 2 (951), (952 (SS2)2 n2 ni

For 99%, α = 0.01, α/2 = 0.005                  
From the z-tables, or Excel function NORM.S.INV(α/2)                  
z' = NORM.S.INV(0.005) = 2.576           (We take the positive value for calculations)      

ME = 2.576 (cena

ME = 7.2021                     
99% Confidence interval is given by                   

M1 - M2 + Margin of Error

= (86-89) ± 7.2021                  
= (-10.2021, 4.2021)                  
                  
99% confidence interval for the difference in mean yields is (-10.2021, 4.2021)          

Since 0 is part of the interval, we can say that mean difference is 0 that is we do not reject Ho.

       

Note: The test in (c ) is based on 5% level of significance (95% confidence), while here we have calculated

99% confidence interval. Hence, the difference in the two decisions. Therefore, the two (c and d) cannot be compared.