Question

Q4 (please also show the steps):

Q4 Consider a problem of comparing the means of two population means. (If you are using a calculator to obtain answers directly, please write down the steps involved.) For sample 1: *1 = 10.9, 51 = 5.4, n = 15. For sample 2: 72 = 12.3, 52 = 4.8, n2 = 13. Compute a 95% confidence interval for Mi - 12 if 1. both sample 1 and sample 2 are normally distributed; 01 = 02 nevertheless unknown. 2. both sample 1 and sample 2 are normally distributed and we are not sure if 01 = 02. 3. both sample 1 and sample 2 are normally distributed and it is given that a = 02 = 5. 4. both sample 1 and sample 2 are not normally distributed. (Can we actually do it?) MacBook

Answer 1

1 Answey Given that Sample 1 a = 10.9, 3, -5.4., 9, = 15 Samples 12.3 6, -4.8 n2=13. - Pos 954. Confidence Ynterval 0 Both Simple and Carpte 2 Ore normally distributed 6:02. (un ucun) Estiralean Variance is given by s': (1,-1) 81 +(1:1) n+ 2 - 2 (15-1) (5 4 + (13-1) 483" 15+13-2 14 X 99.16 + 12 X 93.04 26 408-24 +276.48 2-6 684.72 26 = 26.335 S? SE 126.335 SE5132 957 c. I for M, M₂ 1 + X (3173) * tops, 15 + 1 ((109 - 13.3) 7 42.056,8 5,132 t-table. 2.056 From Since to.025,26

61.4 F 10.551392 70.14359 = (-14 I 3.9983) (-5.3983, 2.5983) are normally distributed and 2) Both Sample i and Sample 2. sure it ad que au net 95% confidence Pietaval for + 17 no (1-4) 7 toos, Motos (110.9 - 12.3) 7 2056 5.41 15 + 4.82 13 (10.9 - 12-3) † 2:056 29.16 +23.04 15 13 ) 1944 +1.772 (10.9 - 12.3 I 2.056 C10.9-12.3 F 2.056 3.716 ) 6-1.4 F 2.056 (1927)) = (-1.4 3.92.0) . = GIl +3.920, -1.4-3.920) = (2.52, -5.32) = (-5.32, 2.52). Both Samples normally distributed and 61=62-5 95% Confidence interval for M, M2 is.

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