Question

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Two catalysts may be used in a batch chemical process. Twelve batches were prepared using catalyst 1, resulting in an average yield of 84 and a sample standard deviation of 3. Fifteen batches were prepared using catalyst 2, and they resulted in an average yield of 92 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 a =0.01. Yes. (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 places (e.g. 98.76). Hi-H2 !

Answer 1

a)Given that sample size ni = 12, n = 15 For catalyst 1 71 = 84 and standard deviation S, = 3 For catalyst 2 1 = 92, and standard deviation Sy = 2 Null Hypothesis Ho : There is no difference (4 = ) Alternative Hypothesis Hy : Caatalyst 2 produces higher mean, 4 st (Left tail test) Level of significance a:=1% = 0.01 Degrees of freedom=n, +n -2= 12 +15-2 = 25 T-R Test statistic: 1 = - + ni 12 84-92 = -7.9342 32 22 + 12.15 toy for 1% LOS with 25 d. f is – 2.49 (For left tail, from t - tables) Criterion: If tea <tipall, we reject the null hypothesis Otherwise we accept the null hypothesis. Conclusion: Since tu = -7.934 <-2.49 = fra Hence we reject the null hypothesis (ie)YES, there is an evidence to support that 2nd catalyst mean yield is more than the mean yield of 1st catalyst with 99 confidence. b) 99% confidence interval is $* (7-7)-t. an n, at 25 d.) with 1% level of significance is 2.79 (52 - Ts) tahun + Here taha = = 0.01 (two Since P() >ca) = [ (64–92)-(2.79 Ft) (84-92)+(2.79), 32 22 + 12 15 3 = (-8-2.81, -8+2.81) = (-10.81, - 5.19) =-10.8154-4, 5-5.19