NEED HELP /W HW QUESTIONS! HE PROVIDED SOLUTIONS (listed below) BUT THE EXPLANATIONS ARE NOT MY CUP OF TEA. NEED MORE CLARITY AND DETAILS SO PLEASE PROVIDE AS MUCH AS POSSIBLE! THANK YOU

12.21 A law student wants to check on her professor's claim that convicted embezzlers spend on the average 12.3 months in jail. So she decides to test the null hypothesis = 12.3 against the alternative hypothesis a * 12.3 at the 0.05 level of significance, using a random sample of n = 35 such cases from court files. What will she conclude if she gets 7 = 11.5 months and uses the five-step significance test described on pages 294 through 297, knowing that o = 3.8 months?12.23 In a study of new sources of food, it is reported that a pound of a certain kind of fish yields on the average 3.52 ounces of FPC (fish-protein concentrate) used to enrich various food products, with the standard deviation o = 0.07 ounce. To check whether a = 3.52 ounces is correct, a dietician decides to use the alternative hypothesis u # 3.52 ounces, a random sample of size n = 32, and the 0.05 level of significance. What will he conclude if he gets a sample mean of 3.55 ounces of FPC per pound of the fish?12.32 A new tranquilizer given to n = 16 patients reduced their pulse rate on the average by 4.36 beats per minute with a standard deviation of 0.36 beat per minute. Assuming that such data can be looked upon as a random sample from a normal population, use the 0.10 level of significance to test the pharmaceutical company's claim that on the average its new tranquilizer reduces a patient's pulse rate by 4.50 beats per minute.12.34 Five Golden Retrievers weigh 64, 66, 65, 63, and 62 pounds. Show that the mean of this sample differs significantly from / = 60 pounds, the mean of the population sampled, at the 0.05 level of significance.12.36 Random samples showed that 40 executives in the insurance industry claimed on the average 9.4 business lunches as deductible biweekly expenses, while 50 bank executives claimed on the average 7.9 business lunches as deductible biweekly expenses. If, on the basis of collateral information, it can be assumed that of = o = 3.0 for such data, test at the 0.05 level of significance whether the difference between these two sample means is significant.12.38 An investigation of two kinds of photocopying equipment showed that a random sample of 60 failures of one kind of equipment took on the average 84.2 minutes to repair, while a random sample of 60 failures of another kind of equipment took on the average 91.6 minutes to repair. If, on the basis of collateral information, it can be assumed that o = 02 = 19.0 minutes for such data, test at the 0.02 level of significance whether the difference between these two sample means is significant.Question 12.21 We are testing the the null hypothesis u = 12.3 and alternate hypothesis a # 12.3 at 0.05 level of significance. Size of random sample n = 35. We also know ) = 11.5,0 = 3.8 (1) Ho : H = 12.3 (2) Ha: # 12.3 (3) a = 0.05 f -M 11.5 - 12.3 (4) z = =-1.25 a/vn 3.8/ 35 (5) Based on a = 0.05, Zai, = 20025 = 1.96 (6) Since |z| Zay, we reject HoQuestion 12.32 Sample SE22 n = l,samp1 mean J? = 4.36.3ampi'e standard deviation 0.36. We would like to rest the hypothesis thatrhe mean is 4.50 at [LII] level of significance. (1)130 \"4 = 4.50 (2) Hum at 4.50 (3) a: = 0.10 (4} We use t distribution because It 4: 31] and a is unknown. _fp _ 4.354.50 Sh"; [3.353% E = ']._5E1 (5) Based am: = 0.11]. degrees of freedom = 15,. ruff: = tum; = 1.753 (5) Since III .1 Infra, we fail to reject Ha Question 12.34 The 5 golden retrievers weigh 64, 66, 65, 63, and 62. We need to show that mean differs from a = 60 at 0.05 level of significance. (1) Ho : 4 = 60 (2) Ha: u # 60 (3) a = 0.05 (4) We use t distribution because n tay, we reject HoQuestion 12.36 Random sample of 40 executives show that mean is 9.4, while 50 executives show that mean is 7.9. 01 = 02 = 3. We want to test if the difference between sample means is significant at 0.05 level of significance. (1) Ho : M1 = #2 (2) Ha:M # 2 (3) a = 0.05 71 - 12 9.4 -7.9 (4) 2 = = = 2.36 1017 / n + 2" / nz 132 /40 + 32 /50 (5) Based on a = 0.05, Zaj = Zoo25 = 1.96 (6) Since z > Zap. we reject Ho Question 12.38 In a random sample of 60 failures, it took an average of 84.2 minutes, while in another sample, of 60 failures, it took an average of 91.6 minutes. of = 02 = 19. We want to test if the difference between sample means is significant at 0.02 level of significance. (1) Ho : 1 = M2 (2) Ha: * #2 (3) a = 0.02 71 - X2 84.2 - 91.6 (4) 2 = = = -2.13 192/60 + 192/60 (5) Based on a = 0.02, Zaj = Zoo1 = 2.33. (6) Since Iz|