ii) Show using the data provided that SST is 184. (3 marks) iii) At the 5% level of significance, can the researcher conclude the interaction of sintering time on the compressive strength of the three different metals? (4 marks) iv) Should the researcher continue the study without considering the interaction between sintering times and metal types? Justify your answer. (2 marks)QUESTION 1 Analysis of Variance a) A computer manufacturing company conducts the acceptance sampling for incoming Source DF Adj ss Adj MS F-Value computer chips. The probability of conforming chips is 80%. If three or fewer non- Factor V1 SSA MSA conforming chips are found, the entire lot is accepted without inspecting the remaining Error 498 . 667 MSE chips in the lot. However, if four or more chips are nonconforming, every chip in the entire Total 10 1140.545 lot is carefully inspected at the supplier's expense. Assume all assumptions of ANOVA are met, answer the following questions. i) Find the probability of more than 12 chips in the sample of 20 are the conforming chips. State TWO (2) important assumptions that must be held true if the ANOVA test was to (4 marks) be used. (2 marks) ii) Find the mean and standard deviation of the random variable X, as described in (i). i) Find the missing values of SSA, MSE and F in the ANOVA table above. (3 marks) (2 marks) i) Is there enough evidence to indicate that at least one of the treatment methods iii) After receiving a huge shipment of computer chips, the company randomly selects 800 produced a mean student response which was different from the other methods? Test chips. Assume that the probability of non-conforming computer chips being supplied is using a = 0.05. (4 marks) 0.001. What is the probability that the lot will be accepted? Use a suitable approximation. (4 marks) b) An experiment was conducted to determine the effects of sintering time on the compressive strength of three different types of metal. The researcher measured the compressive strength of three specimens of each metal type at each sintering time; 100 minutes, 150 b) Extruded plastics rods are automatically cut into nominal lengths of 120 cm. Actual lengths minutes, and 200 minutes. The result is shown in the following table. are normally distributed with a mean of 120 cm and a variance of 100 cm. Sintering Time Metal Type i) Find the probability that a strip selected at random has length within 5 cm of the mean. 2 3 100 23, 20, 21 22, 19, 20 19, 18, 21 (3 marks) 150 22, 20, 19 24, 25, 22 20, 19, 22 200 18, 18, 16 21, 23, 29 20, 22, 24 ii) Strips that are shorter than _ cm are rejected. Determine the value of _ if 5% are rejected. The Minitab output for the above data is as follows. (3 marks) Analysis of Variance ii) Suppose there are 20 samples of the extruded plastics rods selected, calculate the probability that the sample mean has a length over 114 cm. Source DE Adj SS Adj MS F-Value P-Value (4 marks) Time V3 6.222 3.111 F 0. 480 Metal VA 46.222 23. 111 FB 0 . 012 Time*Metal VS J 14.556 FAXB 0.026 Error V6 73.333 4.074 QUESTION 2 Total 26 SST a) A clinical psychologist wishes to compare the treatment methods used for reducing hostility Assume all the assumptions of ANOVA are met, answer the following questions levels among university students. Three types of treatment methods were proposed and a psychological test (HLT) was used to measure the degree of hostility. An analysis of i) Identify the experimental design. State the response variable in the experiment. variance (ANOVA) test was performed, and the output is as shown below. (2 marks)KEY FORMULAS Binomial probability formula P(X = x) =(1-pyx; x=0,12 ....n Poisson probability formula P(X = x)_ex -; x = 0,1, 2, ... x! CONFIDENCE INTERVALS Parameter & description Two-sided (1 - @)100% confidence interval Mean, u, variance, 6 unknown, small samples of =n-1 Difference in means, #1 -#2 , 1+1 variances of =02" and unknown of = 1 + my -2, 5p = (n, - 1)s,+ (12 - 1)s 2 My + 12 - 2 Difference in means, Hy - H2. variances of #62 and unknown di = - (s,2/n, + 5 2 /n2 P m -1 12 - 1 Mean difference for paired samples, My of =n-1 where n is no. of pairs Variance, 62 (n -1)s' (n-1)s' of =n-1 Ratio of the variances ,?/62 1 $7 Fa/2: VIV2 V1 = 0 -1, v2 = 02 -1