p.4 (1.1) 1,2,6
1-"No car" vs. "I did not have a car or way to leave"
2-the table wording may be more accurate, but the graph is easier to quickly comprehend
6-the bar heights are generally similar (and the rankorder is similar) for men and women
p.14 (1.2) 4,7,13,14,19,50,51(population cannot be determined),53,54,55,64
4-for a qualitative variable, the description of an element is a quality or property, for a quantitative variable, the description of an element is a number
14-(female,66,freshman,3,550)
50-the population is all veterans, the sample is the 20 she examined
51-the sample is the students who were querried, the populaton could be all students at PCC, all community college students, all college students, or many other possibilities.
54-statistical inference is used to infer behavior of all drivers from the 100 surveyed.
64-a quantiy calculated from the data is a statistic; The average lifetime if new lightbulbs is 2000 hours, based on a sample of 100 bulbs.
p.45 (2.1) 1,2,15,16,17,18,39
2-A frequency distribution gives the actual number of each category of object, a relative frequency distribution is technically any numbers proportional to the raw cout, but in general usage percentages which sum to 100 percent
16-A: 11/25; B: 3/25; AB: 1/25; O: 10/25
18-(since this is ascii, I shall not draw the pie chart)
p.62 (2.2) 1,2,22,23,24,25,27,32,33,34,35,36
2-fewerclasses may be easier to comprehend, but you lose some information
22-You could treat this as discrete data, but one possibility is
60-69: 4
70-79: 9
80-89: 4
90_99: 3
24-(again, this is not something it is easy to do in ascii)
32-a) 10, they should use a broken zigzag to indicate the graph does not go down to zero maintaining scale. I would probably go down to zero at scale because the gap is so small (in which case I maight have 12 classes). b)2.5, which is the same for all classes as it should be. c) frequency, since it uses actual counts.
34-a) If he is using left endpoint inclusion, you cannot tell, either 0,1, or 2; b) divide your answer to a) by 19; c) 5 (if you are using left endpoint inclusion,; d) 5/19
36-skewed to the left
p.77 (2.4) 6,7
p.92 (3.1) 3,15,23(also midrange(from lecture)),31,47,56; look at 64,65,66
23-midrange is (65+100)/2=82.5
56-a) themean is less than the median; b) the mean is greater than the median; c) you cannot tell, the mean and median are close.
64-65-66- I want you to look at these in order to be aware that there are other averages which are appropriate for specific problems.
p.109 (3.2) 1,9,11,23,29,35,calculate the sample variance and standard deviation for that data set,51,52,53,54,89,90
35-the sample variance (divide by n-1) is 116.67 , the sample standard deviation is 10.80
52-two standard deviation units above the mean means .5(1-.95) = .025 (or .0228)
54-within 2 standard deviation units, but below the mean .5(.95 = .475 (or .4772)
p.118 (3.3) 1,5,9,11(also median(from lecture))
p.132 (3.4) 1,3,15,17,19,25,31
p.142 (3.5) 3,19,20,21,24,25
20-this will depend on you definition of quartiles, but I would use (85.5-68)=17.5
24-again, difficult using ascii

[First Test]

p.205 (5.1) 3,12,16,21,25,26,27,28,71,73,75
12-no, there is a negative 'probability',and the 'probabilities' do not sum to 1
16-yes, the probabiities are non-negative and sum to 1.
26-0, a die shows only a single number in a single roll
28-observing a 3 is a subset of observing a 3 or a 5, hence the latter will always occur if the former does.

p.214 (5.2) 5,41,57,58,59,60,61,62,66,73,75,77
58-.5=1/2, half (26) of the cards are red, half (26) are not
60-9/52; of the 12 face cards (JQK), 3 are diamonds, hence 9 are not
62-3/12, 3 of the face cards are diamonds
66-5/8, they mean exactly two tails, your tree should have 8 tips, HTT,THT,TTH are the three ways to get 2 tails

p.230 (5.3) 3,63,79,81,82,83,84,89,90,98,102,103,104,105
82-0.4, by the definition of independence
84-0.7=.05+.04-(0.5 × 0.4)
90-no, P(P and C) > 0
98- a) 16/49 = (4/7) × (4/7); b) 2/7 = 12/42 = (4/7) × (3/6)
102- a) P(F)=314/618; b) P(M)=304/618; c) 412/618
104- a) P(More|F)=212/314; P(More|F)=(200/304)

p.245 (5.4) 3,8,17,19,25,27,29,31,33,37,39,43,45,46,51,54,60
8-9=3 ×3 (assuming one scoop of ice creamand one topping)
46- ABC,ABD, ACD, BCD; 4 (one needs to figure out a systematic way to list all the possibilities)
54-120=5!
60-455=(15 × 14 × 13 )/(3 × 2 × 1)

p.263 (6.1) 2,7,9,19,21,22,55
2-yes, your height varies during the day, so if it is measured at a random time it is a random variable; even if you consider your height constant, if you assign a random time to measure it is technically a random variable, but most people would not consider your height as printed onyour driver's license a random variable
22-yes, the probabilities (are no negative and) sum to 1.

p.276 (6.2) 6,8,35,36,37,38,39,40,50,54,57(the instructor choosing the questions randomly is irrelevant),61
6-no, each toss has more than two possible outcomes which are used to get the final value
8-yes, X=the number of 6's, n=3, p/1/6, q=5/6
36-..0162=4 × (1/6)^3 × (1/6) + (1/6)^4
38-.8681=(5/6)^4 + 4 × (1/6) × (5/6)^3
40-0 (there are at most four doubles)
50- a) &$181; = 4/6, the expected (average) number of doubles that will occur if you repeat the experiment many times; b) 5/9 = 4 × (1/6) × (5/6); c) .7454 = (5/9)^.5
54- (I shall beg off on this because I am working in ascii)

p.285 (6.3) 1,9,25,26,27,29,30,32,34,41
26-0.5 (by symmetry)
30-0.6826 (if you use the empirical rule you will get .68)
32-1 (to as many decimal places as our tables have)
34-B has the greater standard deviation (6) because more data is further from the mean

p.296 (6.4) 3,5,7,15,17,19,21,41,47,56,66,68
56- -.03
66- .67
68- -2.58

p.309 (6.5) 1,9,13,15,23,29,33,35(is this continuous or discrete?),59,61,69,71
p.315 (6.6) 7,14,15,21,26,30,32
14- .5704; E[X]=20, sigma=3.16, z= (17.5-20)/3.16=-.79, z=(22.5-20)/3.16=.79, .7852-.2148=.5704
26- a) .7157 ; ((24.5-.53 × 50)/(50 × .53 &#@15; .47)^.5 = -.57, .7157; b) .0548 ; ((24.5-.38 × 50)/(50 × .38 &#@15; .62)^.5 = 1.60, .0548
30- a) .0003 b) .0951 32- a) .4325 b) .5675 c) .4325 (you will get more accurate answers if you use your calclator instead of tables)

p.329 (7.1) 2,5,11,17,23,29,40,44
2- the distribution of outcomes ensuing from different samples; because we can only take samples, so it describes what we are doing
40- a) 68, 1 = 4/(16)^.5 ; b) .0228 (z= (70-68)/1) ; c) .0013 (z = -3)
44- a) 69.96 inches b) 66.04 inches c) (66.04, 69.96) (actually, symmetric should be modifying 'middle', not 'sample')

p.340 (7.2) 5,11,15,16,17,25,26,31

16- a) 50, b) .75 = 6/8, 64 is large enough for it to be approximately normal
26- 16 is too small to invoke the central limit theorem, but IF prices are mound shapes (approximately normal) $2.95

[Second Test]

p.348 (7.3) 5,7,28,35,39,43
28- .4129
p.366 (8.1) 7,8,25,27,31,39,41,45,47,49,53,61
8- increase the sample size
p.378 (8.2) 35 (use s for sigma)
p.390 (8.3) 9,13,17,19,32,41,59,60,61
32- a) .0588 = 1.96 ×: (.1 × .9 / 100)^.5, b) .00784 = 1.96 ×: (.2 × .8 / 100)^.5, c) .0898 = 1.96 ×: (.3 × .7 / 100)^.5, d) .0960 = 1.96 ×: (.4 × .6 / 100)^.5, e) .0980 = 1.96 ×: (.5 × .5 / 100)^.5,
60- a) increase b) increase c) increase all manifest a larger confidence interval which the true value is more likely to contain the true value.

p.412 (9.1) 9,13,15,20,21
20- a) H0: no change to mortality and morbidity ensuing from accidents, HA change in mortality or morbidity ensuing from accidents b) change when there is change, or no change when there is change (change has been observed, the analysis is predicated on the assumption that the observatons are a random subset of a much larger population) c) Type I: conclude a change,when no change really occurred, d) Type II: conclude no change, when a change had indeed occurred

p.420 (9.2) 2,13,15,21,23,31,39
2- a standardized measure of how far your observation is from what you expect.

p.433 (9.3) 11,13,19,21,29,37
p.460 (9.5) 6,7,11,15,21,23,29,31
6- p is the actual parameter of the populaton,the p-value is a standardized measure of how consistent your observation is with that parameter

p.541 (11.1) 2,7(phraseology is poor),9,11,15,19,21,27,31
2- it tests how consistent your observations are with the hypothsized relative frequencies

p.554 (11.2) 4,5,7,11,12,19,29(make sure you multiply by 1000, the chi-squared value in the back of the text may be wrong),30

4- calculate the marginal relative frequencies, calculate the expected joint relative frequencies assuming indepence, convert to frequencies.
12- a) H0: E and F are independent b) all expected values are greater than 5 c) for 5% significance with 2 degrees of freedom X^2 = 5.991, reject if observed X^2 is larger d) .6340 e)because the X^2 from data is so small, we cannot conclude dependence between the variables.
30- 3.5165 is less than 3.841, so there is no evidence that the proportions differ at the 5% level.

[Third Test]

[notation in the text is confusing, s has two meanings in this secton] p.160 (4.1) 8,9,13,15,18,19,23,35,43,44,55,56
8- no, we are only considering linear relationships
18- a) perhaps parabolic b) the y values go down, then up
44- -.83, y valus tend to decrease as x increases
56- e) .8792 (this is essentially the procedure I use in class
p.174 (4.2) 3,4,7,9,12,13,53
4- y is the original data value, y-hat is the corresponding value on the least squares regression line.
12- near zero

p.185 (4.3) 6,7,11,23,29,37,41

6- r^2 close to one means there is a strong linear association, hte points lie close to a line, if r^2 is close to zero, there is no obvious linear association

[Final]