Question

Question 02: (20 pts) 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. A test is run in the pilot plant and results in the data shown in the table below Observation Number | Catalyst 1 (%) Catalyst 2 (%) 91.50 94.18 92.18 95.39 91.79 89.07 94.72 89.21 89.19 90.95 90.46 93.21 97.19 97.04 91.07 92.75 2 4 7 a) Is there any difference between the mean yields? Use a 0.05 and assume equal variances.(10 pts) se that if catalyst 2 produces a mean yield that differs from the mean yield of catalyst l by 40% we would like to reject the null hypothesis with probability at least 085. What sample size is required? (1o pt)

Answer 1

At first put all the data in an Excel spreadsheet and calculate the mean and standard deviation using the AVERAGEO, STDEV.S() formula respectively. Observation Number Catalyst 1(%) | Catalyst 2(%) 91.5 89.19 94.18 90.95 92.18 90.46 95.39 93.21 91.79 97.19 89.07 97.04 94.72 91.07 89.21 92.75 Or calculate it by using below given formulas: The sample mean of catalyst 1(%): =91.50+94.18+...... +89.21 8 738.04 8 = 92.26 The sample mean of catalyst 2 (%): - 89.19 +90.95 +.....+92.75 _741.86 8 = 92.73 The sample standard deviation of catalyst 1(%): (91.50 -92.26) +(94.18-92.26) +...+(89.21-92.26)? S = 8-1 10.57 +3.71+...+9.27 EV 39.82 = 5.69 = 2.39

The sample standard deviation of catalyst 2(%): |(89.19-92.73) +(90.95-92.73)? +...+(92.75-92.73)2 8-1 12.55+3.18 +...+0.0003 S21 162.31 = 78.90 = 2.98 (a) Let the null and the alternative hypothesis be: HO:/4 - Il = 0 Hall-I = 0 Here, 14. Al are the population means of catalyst 1 and catalyst 2 respectfully. The level of significance a =0.05 The population variances are unknown an equal. Hence, use t-test here. - SIT (n-1) sí +(12-1)s V (n +n, -2) (8-1)2.392 +(8-1)2.982 V (8+8-2) 102.1475 V 14 =V7.29625 = 2.7012 92.26-92.73 t=- -0.47 1.3506 = -0.348

The degrees of freedom are: = n +12 - 2 = 8+8-2 = 14 At 14 degrees of freedom and 0.05 level of significance, the critical value of t-test = +2.144787 (By using T.INV.2T (0.05, 14) function in Excel.) Conclusion: The calculated value of t-test is lying in the acceptance region. -2.144787<-0.348<2.144787 Therefore, do not reject the null hypothesis. Hence, at 0.05 level of significance there is no significant evidence to conclude that there is any difference between the mean yields. (6) Now, d-141 2x2.7012 4 5.4024 = 0.7404 The type II error is: B=1-0.85 = 0.15 n* = x100 0.15 x100 0.7404 = 20.259 - 20 (rounded to the nearest integer) Therefore, the required sample size is: 20+1 = 10.5 Hence, the required sample size is 11. As per the HOMEWORKLIB POLICY, kindly ask one question at a time. Thank you!