Y-55 (0,1) We accept Ho when the test statistics T(a)=(Y-55)/ 36 36 lying in the in- terval Co=[-1.96, 1.96 under the size of the test a=0.05 We now have 532 gi=532 and y =48.4 Then 48.4-55 1.01 V36.36 which is in the accept region. Therefore we accept the null hypothesis that Ho xaMl Assume a random sample of size 11 is drawn from a normal distribution N(u, a2) In particular,=62,y=52,y3=68,4=23,y5=34,y6=45,yr=27,y= 42, y9=83, 910=56 and y11 =40. Test the null hypothesis that Ho: A= 55 versus h1:≠55 Since a is unknown, the sample mean distributed as Y therefore under Ho: A=55 however it is not a pivotal test statistics since an unknown parameters o2or Y¯ − 55 √ 36.36 ∼ N(0, 1). We accept H0 when the test statistics τ (x) = (Y¯ − 55)/ √ 36.36 lying in the in￾terval C0 = [−1.96, 1.96] under the size of the test α = 0.05. We now have X 11 i=1 yi = 532 and y¯ = 532 11 = 48.4. Then 48.4 − 55 √ 36.36 = −1.01 which is in the accept region. Therefore we accept the null hypothesis that H0 : µ = 55. Example: Assume a random sample of size 11 is drawn from a normal distribution N(µ, σ 2 ). In particular, y1 = 62, y2 = 52, y3 = 68, y4 = 23, y5 = 34, y6 = 45, y7 = 27, y8 = 42, y9 = 83, y10 = 56 and y11 = 40. Test the null hypothesis that H0 : µ = 55 versus H1 : µ 6= 55. Since σ 2 is unknown, the sample mean distributed as Y¯ ∼ N(µ, σ 2 /n), therefore under H0 : µ = 55 Y¯ ∼ N(55, σ 2 /n) or Y¯ − 55 p σ 2/n ∼ N(0, 1), however it is not a pivotal test statistics since an unknown parameters σ 2 . 10