Download Stat/For/Hort 571 Midterm Solutions: Probability Distributions and Hypothesis Testing and more Exams Data Analysis & Statistical Methods in PDF only on Docsity! Stat/For/Hort 571 -Midterm I, Fall 99- Brief Solutions Using Table A w-e find P(Z ~1.96) = .025 so we set u. -.u .03 -' -1 ~ = ~ 1'--= 1.96 or"Vn=(1.96x"V.025)/.03 = 10.33. O"I"'n ",.0251 ",n Then n= 107.I. (a) The stem and leaf display is useful for a small data set such as this. 8.130 9.191 10.19953 11.1569 12.1816 13.12 4. First consider the entire sample space (1 = in, O =out). Remember that Butch will never be in the room with a cat. Here y is the number of animals in the living room. M I l 1 O I 0 0 O probability 0 .6 x.4= .24 0 0 .6 x (1-.4) = .36 (1-.6) x .4 = .16 .2 1 -(.24 + .36 + .16 + .2) = .04 The display appears to be roughly symmetrical, centered at about II. (b) The median is the middle value (8d1) of the 15 ordered observations. Median = 10.9 Mean=tyjln= 163.6/15= 10.91 Variance = [11(n -1)][ty? -(1:yjrln] = (1/14) [1819.34- (163.6)z/lS] = 2.501 (a) Because the cats behave independently, the probability that Frisky is in the room is not affected by Mittens' behavior, and the probability is 0.4. (b) This probability is O because Butch and Frisky are never in the living room together. (c) From the table above, the probability that only Frisky is in is 0.16. (d) Here P(O)=0.04, P(I) = 0.72, P(2) = 0.24, p(J) =0. Then }:x.p(x) = OxO.04 + Ix.72 + 2x.24 + JxO = 1.2. (c) We want P( 81. < 2.501). We must convert this to a probability statement about y1. = (n -1)81./a'- because our tables are in terms ofY1.. Thus, {V2 <7)={v2 <~)=pr2 <8.75) We compute 1- p(V1. > 8.75) where y1. has 14 df. Then .10 < P(81. < 2.5) < .25. [Draw a picture!] 2. Let ~ be the mean height of the pin oak seedlings. Then 5. (a) ~ = up = 12x.2 = 2.4 and ~ = uP(l-p) = 12x.2x.8 = 1.92 Ho: 1-1= 1.3 m and HA: I-1F 1.3 m We find the mean height is 8.7/8= 1.0875 and we are given that <i1 = 0.10. Thus z=1:.:::.E-= 1.0875-1.3 =-1901 O"/.r;; .10/.J8 . (b) From (a), 11- a = 2.4- ..JI.92 = 1.014. We want P(Y < 1.014), which for a discrete random variable is P(y=OorY= I)=P(Y=O)+ P(y= I). Then 12! o 12 P(Y=0)=-x.2 x.8 ="069and The p-value is 2P(Z ~ -1.901) = 2p(Z ~ 1.901) = 2 x .0287 = .0574. Thus, the results ~ significant at 10% but not at 5% or 1%. 12.~ P(Y = I) = ~.21.811 = .206 and I In! P(Y=O)+P(Y= 1)=.275. (c) Because np = 20 > 5 and n(l-p) = 80 > 5, we can apply the nonnal approximation to the binomial distribution. Then P(YNA < 1.1- a) is just the probability that a normal RV is below a point I standard deviation below the mean, so P(YNA < 1.1- a) = P(Z < -I) = .1587. Grade Distribution 100:2 90-99:27 80-89:35 70-79:36 60-69:33 50-59:9 <50:13 median = 76 3. (a) We want to find Y. such that P(Y<y.)=0.8when Y- N(3.62, .U2s/1.8). Note that P(Y < Y. ) = .8 implies P(Y>y.)=.2. From Table A we see that P(Z >.84) =.2. Thus ;ji = .84 .Solving for Y. , we fmd Y. = .84x (.{025/ .Jii)+3.62 = 3.651. [Draw a picture!] (b) We want P(l. ~ Y ~ u.) = .95. Converting to z-scores { I. -Jl u. -Jl ) .~ ~ Z ~ ~ = .95 .At the upper end of the mterval we have { Z ~ ~ ) = .025 and we know u. -Jl = .03 (half of the interval). ,02.$' .0~> -{.9(,4 1.9(03.(.,}. - 'I' 0~ r. F I I O I O I O O B I O 1 I O O I O y 3 2 2 2 1 1 1 O
