Question

(b) (6 points) Let to be a particular value of t. Use t-distribution table to find to values such that the following statements are true. (Given that df=10.) (i) P(-to <t<to) = 0.90. (ii) P(t <-to or t > to) = 0.05. (iii) P(t <to) = 0.025. (c) (4 points) The following random sample was selected from a normal distribution: 11 4,6,3,5, 9, 3, 4, 4, 7, 8,5. Hint: xi = 58; x = 346. i=1 (i) Construct a 90% confidence interval for the population . (ii) Construct a 95% confidence interval for the population u.

Answer 1

not to be a particular value of t. Gliven da ello) ☺ Then? P(-20 < £<to) = 0.90 Hese to tasto 10125 90% 1(10) -1.8125 1.8125 Hence the value of to as to = 1.8125

PC e sto or f>lo) = 0.05 > |- P(- lost <2o) 2005 3 pto = tato) = 1-0.05 = P(to = £ <fo) <tstol = 0.95 to 5 16.975, 10) = 2.2202 ello) 95% 2.2282 -9.2282 0 Hancer lle value of to as to = 2.8282

P( t <to) -0.025 Then, to 20.025,10) = -2.2282 = 25% f/10) -2.22020 Hence the vallee of to as a do=-2.2282

Ex=50 © Given n = 11 and X = 346 니 Eni x = li so 17 = 5.2727 Them = and 5? 26? - 11x²) 346 - (2797 = To 40.185 10 =4.0185 and 2 X= 5.2727 s²=4.0185 Wl know theid, X- slun f(n-1) ti

Then 100(1-00% confidence inlesual for He as) ta < t < lo pl-tor spl - to s ta < x-ce S/N) =14 ta) = 1 3p(- do so as t-pes to an 1-4 3 P(x- tos spes set to sn) - 19 Thus, 100(1-23% cófidence hulipal for it as zon X + to X-tos a Jh

0 Let a= 0.10 then a 90% confiawa lulaval for pe as, [x-Cosio in ă-Poslo) in 20046 =(5.9727-46Oslo) It (5.2727 + $0.05, 10) Tu 2.0046 = [4.1778,6.3604] Hence a 90% confidence lulepal for ce asi (4:1772, 63682]

0 Let X=0.05 then, couef dwee mboval for fear 4 95% [x- 46025,10 Ju x + 86.035, 1/5n =>537 foc, es va 2.0046 5.2727- fo.vasio) TUI 20046 5[3.9259, 66194 (tenen a 95% confidence interval for Man 6.6194 [3.9259