Question

2. (8 marks in total) Suppose that in an experiment, two methods A and B are under investigation. Two independent samples of sizes n1 = n2 = 10 are observed with response values as follows: A 85.43 84.90 83.10 85.55 86.24 87.81 81.99 82.41 88.50 88.36 B 88.51 86.89 89.71 89.42 88.94 88.96 87.75 89.27 88.72 85.84 (a) (2 marks) Test the hypothesis whether or not the two methods have the same mean response. (b) (2 marks) Test the hypothesis whether or not Method B has higher mean response. (c) (2 marks) Construct a 95% confidence interval of the mean differ- ence: PB - MA (d) (2 marks) Use R to carry out the randomization test in (a) and (b).

Answer 1

Answer: ---- Given values can yield calculations tabulated as: Date: ---4/10/2019 Method A (x - mean(x)) (x - mean(x))^2 Method By ly - mean(y))^2 x 85.43 ly-meanly) 0.109 0.001 0.000001 88.51 0.011881 84.9 -0.529 0.279841 86.89 -1.511 2.283121 83.1 -2.329 5.424241 89.71 1.309 1.713481 85.55 0.121 0.014641 89.42 1.019 1.038361 86.24 0.811 0.657721 88.94 0.290521 87.81 2.381 5.669161 88.96 0.539 0.559 -0.651 0.312481 81.99 -3.439 11.826721 87.75 0.423801 8241 -3.019 9.114361 89.27 0.869 0.755161 88.5 3.071 9.431041 88.72 0.319 0.101761 88.36 2.931 8.590761 85.84 -2.561 6.558721 Sum = 854.29 51.0085 Sum = 884.01 13.5 (a) Since, the sample sizes are 10 (< 30) here. We can take these as small samples and go for t-test.

Let X and Y are two random variates representing the respective values got from the two methods A and B. Let Hy and Hy are the respective means. Null Hypothesis Alternate Hypothesis Ho: Llx = hly H : Hx * uy (Two tailed test) Here, x = 85.429, y = 88.40, n = n2 = 10, Variance of the difference of means is given by $a = oth-((x - 5)2 + 2(y - 3)2) = () (51.0085 + 13.5) = 3.58 Under null hypothesis: 3-5 -2.972 t = -3.512 Tabulated to.os for (10+10-2) = 18 degrees of freedom is 2.10 Since, |t| = 3.512 > 2.10, hence the null hypothesis is rejected under 5% level of significance. Hence, we can conclude that the two methods DO NOT have same mean response. (b) In this case, Null Hypothesis Alternate Hypothesis HoHxHLY Hu: Mx >= Hly (One tailed test) From previous results, under null hypothesis, t = -3.512 < 2.10 = to.os (18 degrees of freedom) Since t<toos, the null hypothesis is accepted under 5% level of significance. Hence, we can conclude that the method B mean response is higher than method A mean response.

(c) The 95% confidence interval for difference in means is given by (Hy-Hz)-5-) <2.10 2.10 0.846- i.e.-2.10 < (HyHx) -2.972 i.e. 2.972 - 2.10x0.846 < (Hy - ux) <2.972 +2.10x0.846 i.e. 1.1954 < (Hy – Hx) < 4.749 Hence the 95% confidence interval is [1.195, 4.749]. (d) For randomization test an R-Script can work here as: ------------ ## Assign number of replication repl <- 10000 d<- (x-y) m0 <- mean(d) # perform a one-sample randomization test on d # for the null hypothesis HO: mu_d = 0 vs H1 mu_d != 0 (i.e. two tailed) #here the test statistic is the mean and using normal approximation to binomial proportion rndmdist <- replicate(repl, mean((rbinom(length(d), 1,.5)*2 - 1)*d))

# two tailed p-value: sum(abs(rndmdist) >= abs(mo))/length(rndmdist) The p-value can show that in (1 - p-value)*100 = 98.52% (approx) cases method B mean response will become higher than method A.