Question

Presume we know that sample sizes are n = n = 14,= 7.7 & X = 4.6, and si=5&s= 7.25 a) Consider that of = c. and that the information is drawn from normal distributions. (a = 0.05) i) Test the prediction H: U = U, against Hall 7 . ii) Construct 95% confidence interval for 4 - uz to support your answer (0) and that the information is pulled from normal distributions. b) Consider that (a = 0.05) iii) Test the prediction Ho: 4, - uz = 2 against Hai u - uz <2. iii) Construct 95% confidence interval for u-u, to support your answer (iii).

Answer 1

a) Assuming ,

q; = q;

i)

Hypotheses : To test null hypothesis H H1 - H2 = 0 against alternative hypothesis H1 M1-462 70 This is a two tailed test as the alternative hypothesis contains ' sign. Here, 2i = 7.70, si = 5.0000, n1 = 14, 2 = 4.60, să = 7.2500, n2 = 14 The test statistic can be written as: t=- (61 - 72) SpV + which under Ho follows a t distribution with 26 df. where pooled variance (n1 - 1)s +(n2 - 1)s ni + n2 - 2 = 6.125000 and standard error for the difference between means 1 1 1 SE=18*1 + == V nin2 SE = v=VA B = V6.1250w • * * –2.315082 6.125000 * 1 = = 2.315032 Decision rule / Rejection region : We reject H, at 0.05 level of significance if P-value < 0.05 or if tstat > t0.025,71 +11-2 = t0.025,26 = 2.055529 0.4- 20.3- 20.2- 0.0- 1 -4 - 2 0 Now, Value of the test statistic : The value of the test statistic is (7.70 – 4.60) Istat = = 3.314039 3.314 V6.125000 * V + associated degrees of freedom = nu + n2 - 2 = 14 + 14 -2 = 26 The critical value = tcritical = to.025,26 =+ 2.055529 2.056 P-value = P(|t74 +12-21 > tstat) = P(|t26 > tstat) = 2* P(t26 > 3.314039) = 2 * (1 - P(t26 < 3.314039)) = 2 * (1 - 0.998644) = 0.002712 -0.0027 Conclusion: Since p-value < 0.05 and Itstat > tcritical = 2.056, so we reject H, at 0.05 level of significance Hence, we can conclude that there is significant difference between two population means.

ii)

Here, či = 7.700, $i = 2.236, ni = 14, 22 = 4.600, S2 = 2.693, n2 = 14 and associated degrees of freedom df = n1 +12 – 2 = 14+14 – 2 = 26 and pooled variance (n1 – 1)sí +(n2 - 1)s º2 = 6.125000 ni +n2 - 2 and pooled standard deviation 16.125000 = 2.474874 and standard error for the difference between means $P = Vet-V I 4 = V0 12:50 • Vein + = 2.315082 SE = V8 11 -= V6.125000 * V ni n2 = 2.315032 14 . 14 At 95 percent level of confidence, the margin of error is 1 1 1 ME = t 1-0.95,72 +12-2*8p* V ni 'n2 = + 0.28 * 2.4749 + V ia tida = 2.055529 + 2.4749* Vis + V 14 = 1.922772

b) i)

a 95 percent confidence interval for the differnce in population means (Mi - M2 ) is 1 = 21 - 12 sp1 — + V nin — *t 1-0.950 26 2 ,26 = 21 - 22 + ME = (7.700 – 4.600) +1.922772 = 3.100 + 1.922772 = (1.177228,5.022772) ~ (1.18,5.02)

Hypotheses : To test null hypothesis H0:41 - U2 = 0 against alternative hypothesis H1:41-42 +0 This is a two tailed test as the alternative hypothesis contains '#' sign. Here, ci = 7.70, $1 = 2.24, ni = 14,12 = 4.60, S2 = 2.69, n2 = 14 The test statistic can be written as: t = (x1 – 2) – 0 which under H, follows a t distribution with 25 df. where 2 6922 + )2 (2,24 + 2.00%)? ~ 25 df= no-1 and standard error for the difference between means

Decision rule / Rejection region : We reject Ho at 0.05 level of significance if P-value < 0.05 or if tstat > t0.025,df = to.025.25 = 2.059539 0.4- 0.3- density 0.1 - Nowo- Value of the test statistic : The value of the test statistic is Estat =- (7.70 – 4.60) V 2.24 + 2.622 = 3.314039 3.314 associated degrees of freedom = 25 The critical value = t critical = t0.025.25 = 2.059539 2.060 P-value = P(|t25/> tobs) = 2 * P(t25 > 3.314039) = 2 * (1 – 0.998597) = 0.002806 ~ 0.0028 Conclusion: Since p-value < 0.05 and tobs > tcritical = 2.060, so we reject Ho at 0.05 level of significance Hence, we can conclude that there is significant difference between two population means.

ii)

Here, #1 = 7.700, 1 = 2.236, n1 = 14, +2 = 4.600, S2 = 2.693, n2 = 14 and associated degrees of freedom ar G+ )? - = 25 The Standard error is The critical value is tcritical = t 1-0.95 25 = t0.025,25 = 2.059539 ~ 2.060 The margin of error is 2.242 2.692 - 2.692 ME= t 1-0.95,25 * V ni 2.242 14 =t0.025,25 * 1 / = 2.060 * 1 + = 1.927000 ~ 1.927 n2 14 14 14

a 95 percent confidence interval for the differnce in population means (M1 - U2 ) is SA = či - 2 + 1 = + = *t 1-0.950 2 n2 = 21 - 22 ME = (7.700 – 4.600) = 1.927 = 3.100 + 1.927 = (1.173, 5.027) ~(1.17,5.03)