Question

Two catalysts may be used in a batch chemical process. Twelve batches were prepared using catalyst 1, resulting in an average yield of 85 and a sample standard deviation of 3. Fifteen batches were prepared using catalyst 2, and they resulted in an average yield of 89 with a standard deviation of 2. Assume that yield measurements are approximately normally distributed with the same standard deviation (a) Is there evidence to support the claim that catalyst 2 produces higher mean yield than catalyst 1? Use α-0.01 (b Find a 99% confidence interval on the difference in mean yields that can be used to test the claim in part a Round your answer to two decimal paces e 98.76

Answer 1

To Test :-

H0 :-   $\mu_{1} \geq\mu_{2}$

H1 :-   $\mu_{1} < \mu_{2}$

Test Statistic :-
$t = (\bar{X_{1}} - \bar{X_{2}}) / S_{p}\sqrt{ ( 1 / n1) + (1 / n2)}$
$t = ( 85 - 89) / 2.49 \sqrt{ ( 1 / 12) + (1 / 15 )}$
t = -4.1478

Test Criteria :-
Reject null hypothesis if $t < - t_{\alpha, n1 + n2 - 2}$
$t_{\alpha, n1 + n1 - 2} = t_{ 0.01 , 12 + 15 - 2} = 2.485$
$t < - t_{\alpha, n1 + n2 - 2} = -4.1478 < -2.485$
Result :- Reject Null Hypothesis

Decision based on P value
P - value = P ( t > 4.1478 ) = 0.0002
Reject null hypothesis if P value < $\alpha = 0.01$ level of significance
P - value = 0.0002 < 0.01 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis

There is sufficient evidence to support the claim that catalyst 2 produce higher mean yield than catalyst 1.

Confidence interval is :-
$( \bar{X}_{1} - \bar{X}_{2}) \pm t_{\alpha /2 , n1+n2-2} S_{p}\sqrt{ (1/n1) + (1/n2)}$
$t_{\alpha /2, n1 + n1 - 2} = t_{ 0.01 /2, 12 + 15 - 2} = 2.787$
$( 85 - 89 ) \pm t_{0.01/2 , 12 + 15 -2} 2.49 \sqrt{ (1/12) + (1/15)}$
Lower Limit = $( 85 - 89 ) - t_{0.01/2 , 12 + 15 -2} 2.49 \sqrt{ (1/12) + (1/15)}$
Lower Limit = -6.6881
Upper Limit = $( 85 - 89 ) + t_{0.01/2 , 12 + 15 -2} 2.49 \sqrt{ (1/12) + (1/15)}$
Upper Limit = -1.3119
99% Confidence Interval is ( -6.69 , -1.31 )