Homework 1

Homework 1 - STAT 380, Fall 2010 HW 1 Solutions 1. Solve :...

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Unformatted text preview: STAT 380, Fall 2010 HW 1 Solutions 1. Solve : a) Ex 1.3 in the book. 219 214 .. 205 =209.9 psi 10 Mean_noaging=222.1 psi There are 10 observations, so the median is the average of the 5th and 6th observation in the ordered arrangement. So Median_age=(209+211)/2=210 psi Similarly, Median_nonaging=221.5 psi Median and mean are similar in both the groups. mean_aging= b) Ex 1.9 in the book. s.d_aging= (219 209.9) 2 ... (205 209.9) 2 =6.49 psi 9 s.d._noaging=4.86 psi. 2. In a study to test whether directed reading activities in the classroom help elementary school students improve aspects of their reading ability, a treatment class of 21 third-grade students participated in these activities for eight weeks, and a control class of 23 third-graders followed the same curriculum without the activities. After the eight-week period, students in both classes took a Degree of Reading Power (DRP) test which measures the aspects of reading ability that the treatment is designed to improve. The DRP scores are listed below: (Source: http://lib.stat.cmu.edu/DASL/Datafiles/DRPScores.html) Treatment: 24 43 58 71 43 49 61 44 67 49 53 56 59 52 62 54 57 33 46 43 57 Control: 42 43 55 26 62 37 33 41 19 54 20 85 46 10 17 60 53 42 37 42 55 28 48 Use summary statistics like mean, median, range, standard deviation to compare the two sets of scores. Comment on the presence of outliers, if any. Based on the data, does it seem that directed reading activities in the classroom improve reading ability? Summary Statistics Mean Median Range Standard Deviation Treatment 51.48 53 47 11.01 Control 41.52 42 75 17.15 Based on the dataset, it seems that directed reading activities in classrooms do have an effect on reading abilities since we observe that average DRP scores for control group is around 10 points less than that of treatment group. Variability in the control group is more than that of the treatment group since there appears to be both very high and low scores in the control group. 3. Solve 1.16 in the book. Show that (x x ) 0 i 1 i n ( xi x ) xi nx xi xi 0 ( since x i 1 i 1 i 1 i 1 n n n n 1n xi ) n i 1 4. Suppose there are three children in a family. Your goal is to count the number of boys or girls in the family. Denote by ‘G’ if it is a girl and by ‘B’ if it is a boy. a. Write out the sample space for this experiment. (Do not ignore the order in which the children are born) Sample space S={BBB,BBG,GBB,BGB,BGG,GBG,GGB,GGG} b. Give an example of an event (not one listed below). Example of an event A: there are exactly two boys in the family={BBG,BGB,GBB} c. Let A be the event that there is at least one boy in the family, B be the event that there is at least one girl and C be the event that there are exactly two girls. List the sample points in AUB, Ac∩Bc, AcUBc, (A∩B)c, BUC, AUC, AUBUC. Draw a Venn diagram for AUBUC for this problem. Here A={BBB,BBG,GBB,BGB,BGG,GBG,GGB} B={BBG,GBB,BGB,BGG,GBG,GGB,GGG} C={BGG,GBG,GGB} Then, AUB={BBB,BBG,GBB,BGB,BGG,GBG,GGB,GGG} Ac∩Bc= AcUBc={GGG,BBB} (A∩B)c={GGG,BBB} BUC={BBG,GBB,BGB,BGG,GBG,GGB,GGG}(Note this is same as B since C is a subset of B, that is entirely contained in B) AUC={ BBB,BBG,GBB,BGB,BGG,GBG,GGB } (Note this is same as A since C is also a subset of A) AUBUC= {BBB,BBG,GBB,BGB,BGG,GBG,GGB,GGG}=S 5. By shading appropriate regions of a Venn diagram, for any three events A,B and C, verify (A∪B) ∩C=(A∩C) ∪ B∩C) 6. Four out of 20 tires in storage are defective. If 4 tires are randomly chosen for inspection (each tire has the same chance of being selected), what is the probability that only one of the defective tires will be included in the selected four? 20 4 tires can be chosen out of 20 tires in =4845. You can choose 1 defective out of 4 4 16 4 defective items in = 4 ways and 3 non-defective items in 560 ways. Hence 1 3 the total no of ways=4x560=2240 (by multiplicative rule). So, P( 1 defective is chosen)=2240/4845=0.462 7. Samples of a cast aluminum part are classified on the basis of surface finish (in microinches) and edge finish. The results of 100 parts are summarized as follows: Edge finish excellent good Surface excellent 80 2 Finish good 10 8 Let A denote the event that a part has excellent surface finish, and let B denote the event that a part has excellent edge finish. a. If a part is selected at random (all the parts have equal chance of being selected), determine the probabilities i) P(A) =82/100=0.82 ii) P(B)=90/100=0.9 iii) P(Ac)=0.18 iv) P(A∩B) =0.80 v) P(A∪B) =0.92 vi) P(A|B)= P(A∩B) /P(B)=0.8/0.9=0.889 vii) P(Ac|Bc) = P(Ac∩Bc)/P(Bc)=0.08/0.1=0.8 viii)P(Ac|B)=0.1/0.9=0.111 b. Are A and B independent events? Justify. P(A∩B)=0.8 not equal to P(A)P(B)=0.82 x 0.9 So A and B are not independent. ...
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This note was uploaded on 01/25/2011 for the course MATH 380 taught by Professor Staff during the Fall '08 term at UNL.

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