# QUIZ

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```					22S:8 MIDT-A 2 April 2007                     print NAME___________________

1)   Find the SD of the following data set:     12 18 18 24 28
a) 4.52        b) 5.68         c) 6.16            d) 7.08           e) 8.60

82  2 2  2 2  4 2  82
mean = 20.                                6.16
4

(.05)
(.03)
2)   Find the height of the missing rectangle
in the histogram →                                                    ?

0              10                25   30

a) .02         b) .03          c) .05             d) .06            e) .08

other rectangles have areas (10)(.03) = .30 & (5)(.05) = .25 for a total = .55
Thus, missing area = 1 - .55 = .45 so that the missing height is .45/15 = .03

SD
3)   The coefficient of variation of a data set is defined as CV 
mean

Suppose the mean and SD of a data set are 50 & 12 respectively. The data
is transformed as follows: new value = 5(old value) + 6.
Find the new CV.

a) .23         b) .28          c) .35             d) .40            e) .44

new mean = 256 & new SD = 60 so that new CV = 60/256 = .234

4)   Find the correlation between the x & y variables
x      3       4       7      7      7       8             SD = 2            mean = 6
y      6       4       12     12     14      12            SD = 4            mean = 10

a) .75         b) .80          c) .84             d) .90            e) .95
(1.5)(1)  (1)(1.5)  (.5)(.5)  (.5)(.5)  (.5)(1)  (1)(.5)
r                                                                       .90
5

5)   Which statement is true regarding data sets A & B ?

a)   median(A) > median(B)                        A
b)   maximum(A) < 3rd quartile(B)
c)   IQR(A) > IQR(B)
d)   1st quartile(A) > median(B)                  B
e)   none of the above are true                   __________________________
0     10    20    30    40

6)   At right is the stemplot indicating        1      6   6   8   9
the daily profits (in thousands of         2      0   0   1   2 2 3 6 6 6
dollars) at XYZ Corporation                3      0   0   2   4 7
over a 22 day period in March.             4      1   5   8   8
Find the IQR of this data.

a) 12           b) 14          c) 15          d) 16                e) 18
quartiles are colored purple in the stemplot. 34 – 20 = 14

7)   Which one of the following statements is true about the correlation coefficient, r ?

a)   r is affected by the units of measurement (inches, feet, meters, etc.) naaaa
b)   r is not affected by the presence of outliers naaaa
c)   r is always between 0 and 1 naaaa
d)   a high ( + or - ) r is proof of a cause & effect relationship between x & y naaaa
e)   none of the above statements are true

8)   The daily numbers of visitors to the Tate Modern Gallery is approximately
normally distributed with mean = 6320 & SD = 288. What proportion of
days does the Tate receive more than 6500 visitors ?

a) .27           b) .34           c) .40          d) .46           e) .52

(6500 – 6320) / 288 = .62         table look-up

9)   In (8), on the busiest 10% of days, the Tate receives about __?__ or more
visitors.

a) 6420          b) 6496          c) 6564         d) 6688          e) 6824
1.28 has 90% area to the left (& thus 10% to the right) under the std normal curve.
So, 6320 + 1.28(288) = 6688

10)   The lifetimes of ACME batteries (when run 24-7) are normally distributed with
mean = 62 hours. Suppose 20% of these batteries last more than 70 hours.
What is the SD of these lifetimes?

a) 7.78        b) 8.26         c) 9.52         d) 10.28        e) 12.50

.84 has 80% area to its left (& thus 20% to its right) under the std normal curve.
70  62
Solve the equation             .84 to get x = 9.52
x

11)   Suppose that 5% of the suits filed with Superior Court are declared frivolous.
Find the probability that at least one of the next five suits filed are frivolous.

a) .118        b) .226         c) .284         d) .325         e) .388

1 – (.95)5 = .266

12)   In problem (11), find the probability that 2 of the next 10 suits filed are frivolous.

a) .075        b) .125         c) .156         d) .180         e) .196

This one’s a binomial:    (10C2)(.05)2(.95)8 = .075

13)   In problem (11) find the probability that the first frivolous suit filed this month is
the 10th suit that is filed.

a) .01         b) .03          c) .05          d) .08          e) .10

P(1st nine are not frivolous & the 10th is frivolous) = (.95)9(.05) = .031

14)   Two cards are chosen without replacement from the box A A B B B C
Find the probability that the cards are DIFFERENT.
(hint: think “outside the box” i.e., think “complement”)

a) .56         b) .64          c) .68          d) .70          e) .73
1 – P(same) = 1 – P(both AA) – P(both B) = 1 – (2/6)(1/5) – (3/6)(2/5) = .73

15)   Suppose 72% of Iowa adults drink tea or coffee, 48% drink coffee, and 28%
drink tea and coffee. Find the probability that a randomly chosen Iowa adult
drinks tea.

a) .14         b) .30         c) .40          d) .52         e) .64

P(coffee or tea) = P(coffee) + P(tea) – P(coffee and tea)
.72 = .48 + P(tea) - .28               So, P(tea) = .52

16)   In problem (15), find the probability that an Iowa adult drinks coffee, given that
s/he drinks tea.

a) .538        b) .625        c) .688         d0 .725        e) .745

Draw a diagram with 100 Iowans, 28 of them drinking coffee and tea, etc.
From the diagram we have P(coffee | tea) = 28/52 = .538

17)   Three independent observations are made from a standard normal distribution.
Find the probability that ALL THREE are less than 1.5

a) .64         b) .70         c) .76          d) .81         e) .87

area to the left of 1.5 under the std normal curve = .9332
.93323 = .81

18)   Suppose P(L > 5) = .75, and P(L > 6) = .60      Find P(L > 6 | L > 5).

a) .50         b) .70         c) .80          d) .90         e) .95

P(both) / P(given) = P(L > 6 and L > 5) / P(L > 5) = P(L > 6 ) / P(L > 5) = .80

19)   A gambler is equally likely to select a card from box A or B:

A    1 2 3 4 5                         B    1 2 3 4 5 6

Use Bayes’ Theorem to find the probability that the gambler selected from box A,
given that s/he chose a number less than 4.

a) .545        b) .612        c) .688         d) .741        e) .768
P ( 4 | A) P ( A)                     (3 / 5)(1 / 2)
P ( A |  4)                                                                             .545
P ( 4 | A) P ( A)  P ( 4 | B) P ( B ) (3 / 5)(1 / 2)  (3 / 6)(1 / 2)

20)   Use the following information to calculate the regression equation that predicts
the monthly number of TV sets sold at Crazy Eddie’s Electronics Emporium
from the monthly number of DVD players sold there.

Avr. number of TV’s sold  = 288                       SD = 28
Avr. number of DVD’s sold = 200                       SD = 25          r = .80

Now use your equation to predict the number of TV’s sold during a month when
240 DVD players were sold.

a) 312            b) 320            c) 324            d) 330           e) 336

28(.8)
slope          .896         y  .896x  b                     288  .896(200)  b
25
b = 108.8             y  .896x  108.8

predicted value:            y  .896(240)  108.8  323.84

21)   In (20), find the residual associated with June 2007 when 380 TV’s and
290 DVD players were sold.

a) 6.8            b) 8.5            c) 11.4           d) 15.8          e) 18.2

y  .896(290)  108.8  368.64               380 – 368.64 = 11.36

22)   In problem (21), Regression SS = 840 and Residual SS = 290. Find R2.

a) 67%            b) 70%            c) 74%            d) 80%           e) 86%

840 / (840 + 290) = 74.3%

23)   A computer fits the equation y = 32 + 2.4x1 + 3.5x2 to salary data at
XYZ corporation. y = salary, x1 = years of service and x2 = 1 for male
employees or 0 for female employees. Last Tuesday I caught up to Molly
Stevenson. She told me she’s been with XYZ for 10 years and earns a
salary of 60. Find her residual.
a) - 2         b) -1         c) 0            d) +2            e) +4

y = 32 + 2.4(10) + 3.5(0) = 56       residual = 60 – 56 = 4

24)    Four couples got married over the weekend:

husband’s age                22      23      25      26
wife’s age                   18      23      23      28
predicted husband's age                                       ← fill in this line first

The regression equation is husband’s age = .4 wife’s age + 14.8
Find the Regression SS.

a) 4           b) 5          c) 6            d) 8             e) 10

25)    A regression model relates salary to years of service, age, gender (male or
performance (excellent, good, fair & poor). Including the “loose constant” term,
how many terms are in this model ? (hint: this problem’s a no-brainer)

a) 157         b) π          c) 7.13         d) 9             e) Detroit, Michigan

bonus ☺: In problem (11), find the probability that the 3rd fraudulent claim filed
this month is the 10th claim filed.

Find the p-value of the appropriate test . Also find 90% & 95% CI's for the parameter.

1.       Researchers are interested in the effectiveness of a new allergy drug. In particular, they
are interested in knowing whether most allergy patients would benefit from taking it. 432 of 800
randomly chosen allergy patients benefited by taking the new drug.

hyp:    50% of all allergy patients would benefit from taking this drug    vs.
alt:    more than 50% would benefit                                        p

432
p
ˆ         .54
800
.54  .50
T                   2.26
(.50)(.50)
800
p  value  .012
(.54)(.46)
95% conf int  .54  1.96
800
replace 1.96 by 1.65 to get 90% confidence
________________________________________________________________________
2.      Average life of 400 ACME car batteries is 2004 days with SD = 35.5 days.

hyp:    average life of ALL bateries = 2000 days          vs.
alt:    average life of ALL bateries > 2000 days                   µ (large sample)

2004  2000
T                   2.25
35.5 / 400
p  value  .012
35.5
95%  2004  1.96
400
_______________________________________________________________________
3.      54 of 225 randomly chosen boron absorbing rods have cracks.

hyp:    percentage of ALL rods with cracks = 20%          vs.
alt:    percentage of ALL rods with cracks > 20%                           p
54
p
ˆ         .24
225
.24  .20
T                1.50
(.20)(.80)
225
p  value  .0668
(.24)(.76)
95% conf int  .24  1.96
225
________________________________________________________________________
4.     Average caloric content of 6 fast food burgers at a local chain = 768, SD = 171.4

hyp:       average caloric content of ALL burgers = 900   vs.
alt:       average caloric content of ALL burgers < 900          µ (small sample)

This is a small sample problem (use t distribution)

768  .900
T                 1.89      p-value = area to the LEFT of -1.89 under the t5 curve
171 .4 / 6

ALWAYS CHECK THAT THE DATA IS IN THE DIRECTION OF THE
ALTERNATIVE, & THEN TAKE THE SMALLER AREA (in this case the area to the
left - - not to the right - - of -1.89) FOR YOUR p-value

768 < 900 IS IN THE DIRECTION OF THE ALTERNATIVE.

AREA LEFT OF -1.89 = AREA RIGHT OF 1.89                          .05 < p-val < .10

___________________________________________________________________
5.     15 of 60 randomly chosen Iowans are worried about the economy. 24 of 80 randomly
chosen Californians are worried about the economy.

hyp: same percentage of worry in each state vs.
alt: higher percentage in CA                                     p1 – p2

39
p1  .25
ˆ                  p 2  .30
ˆ             pooled         .278
140
.30  .25
T                             
(.278)(.722) (.278)(.722)

60           80
________________________________________________________________________
6.     50 Friday customers at Backwater Mike's:          avr. tip = \$2.40 SD = .60
80 Saturday customers at Backwater Mike's:        avr. tip = \$2.82 SD = .80
hyp:     Friday & Saturday tips the same., on average   vs.
hyp:     average tip higher on Saturday                 µ1 - µ2 (large sample)

2.82  2.40
T                     3.41      p-val = .0003
.60 2 .80 2

50    80

________________________________________________________________________

8.      108 of 144 randomly chosen cars in Iowa are found to be in violation of U.S. emission
standards. Let p = the proportion of cars in Iowa in violation. DOT wishes to test
Ho: p = .8 vs. H1: p < .8                       p

108
p
ˆ          .75
144
.54  .50
T                 1.20
(.80)(.20)
144
p  value  .1151
(.75)(.26)
95% conf int  .75  1.96
144
________________________________________________________________________
9.      In pblm (8), let µ denote the average odometer reading of observed cars. DOT also
wishes to test Ho: = 45,000 vs H1 > 45,000              Avr reading of the cars is
47,186 miles, SD = 12,024 miles                                 µ (large sample)

47,186  45,000
T                       2.18
12,024 / 144
p  value  .0146
12,024
95%  47,186  1.96
144

________________________________________________________________________
10.     The 5 readings (in meters) given by a rangefinder in repeat measurements of the distance
between an observer and a bridge are independent n(, 4) random variables. Suppose
average reading is 1802.6 m. and SD = 1.88 (note: this is a different SD than the one I
used in class). We wish to test Ho: µ = 1800 vs. H1: µ > 1800
where µ denotes the true distance.                              µ (small sample)
1802 .6  1800
T                      3.09     From t4 curve .01 < p-val < .02
1.88 / 5

________________________________________________________________________
11.     Continuing (8), suppose 82 of 100 randomly chosen Minnesota cars are in violation of the U.S.
standard. DOT wishes to test Ho: The proportions of Iowa and Minn. are the same vs.
H1: the proportion in Minn. is higher.                               p1 – p2

p1  .82
ˆ                   p 2  .75
ˆ             pooled  .779
.30  .25
T                                      .925      p-val = .177
(.779)(.221) (.779)(.221)

144          100

________________________________________________________________________
12.     The mean breaking strength of 120 ACME steel rods is 1026 lbs. with SD = 28 lbs. The mean
breaking strength of 200 GlaxCo steel rods is 1018 lbs. with SD = 16 lbs. Are ACME rods
better, on average, that GlaxCo?                              µ1 - µ2 (large sample)

1026  1018
T                  p-val = area to the right of this number under the std. normal curve
28 2 16 2

120 200
________________________________________________________________________
13. The number of particles of a pollutant in a sample of one cubic meter of air collected near
Schaeffer Hall was measured. Results : 126 154 163 133 118 125 158 143
The observations are assumed to be from a normal distribution whose mean is the actual number
of particles present in the air sample. We wish to test Ho: μ = 150 vs. H1: μ < 150
µ (small sample)

In this problem, YOU have to figure out the sample mean & SD. On the Final, I will
always tell you these numbers, or (at least) tekll you the SD.

140  150
T                 1.67           p-val = area to the RIGHT of +1.57 under the t7 curve
16 .9 / 8
.05 < p-val < .10

________________________________________________________________________
16.     Is "heads" more likely when a coin is spun on a table than when it is flipped in the air ?
100 spins on the table, 62 heads         100 flips in the air, 54 heads p1 – p2
116
p1  .62
ˆ                 p 2  .54
ˆ                pooled         .58
200
.62  .54                                   p-val = .1250
T                                   1.15
(.58)(.42) (.58)(.42)

100        100

note: We could have subtracted .54 - .62 in the numerator of T, & then
too the area to the left of -1.15 but I prefer to keep things on the positive side of
________________________________________________________________________

17.     Are ACME steel rods stronger (on average) than Ming rods ?

100 ACME rods: avr. breaking strength = 1006 lbs. SD = 28 lbs.
80 Ming rods: avr. breaking strength = 1002 lbs. SD = 18 lbs.
µ1 - µ2 (large sample)

1021  1006
T                     4.35                p-val = 0 (appx.)
28 2 18 2

100 80

________________________________________________________________________
18.     Is drug A the same as drug B ? or is A better ? 400 patients take drug A, 224 improve
250 take drug B, 118 improve.                                           p1 – p2

224  118
p1  .56.0
ˆ                     p 2  .472
ˆ               pooled               .526
650
.560  .472                                          p-val = .014
T                                       2.19
(.526)(.474) (.526)(.474)

400          250

19.
Is gas the same price in Iowa (on average) as it is in Minnesota ? Or is it more expensive in
Minnesota ? A random sample of 40 gas prices in Iowa & 50 gas prices in Minnesota resulted
in the observations: Iowa avr. = 1.04, SD = .02 Minn. avr. = 1.06, SD = .025
µ1 - µ2 (large sample)

1.06  1.04
T                       4.21              p-val = 0 (appx)
.02 2 .025 2

40     50
________________________________________________________________________
20.     The scores on an aptitude test are assumed to be normally distributed. Eight prospective
employees are randomly chosen from a large group of applicants. The sample mean =
76 and the sample SD = 4.2 . Ho: μ = 80 vs. Ha: μ < 80.           µ (small sample)

76  80
T           2.69     from t7 curve .01 < p-val < .02
4.2 / 8
________________________________________________________________________
21.     Two types of tire treads are compared. 6 cars are each equipped with a "type A" & a
"type B" tire (randomly placed in the front). The differences in wear between the tires is
noted for each car. We wish to test Ho: A same as B vs. H1: A better than B.

CAR      1     2        3       4       5        6
A        48    55       49      51      56       50
B        45    53       50      48      52       46      µ (small sample)
diff

YOU WORK WITH THE DIFFERENCES

3        2     -1       3       4       4

sample mean = 2.5               sample SD = 1.87

2.5  0
T                3.27         from t5 curve         .01 < p-val < .02
1.87 / 6

__________________________________________________________________
22.     A large company believes that more than 20% of its employees arrive late for work. On a
Monday morning in November, 256 randomly chosen employees are observed. Of these,
60 are late.                                                            p

p  .23475
ˆ
.234  .20
T                1.36
(.20)(.80)
256
p  value  .0869
(.234(.766)
95% conf int  .234  1.96
256

_______________________________________________________________________
23.      A cereal company claims that 40% of its boxes contain a prize. A consumer group
doubts this claim. 625 randomly chosen boxes are opened. 228 contain a prize. p
p  .365
ˆ
.365  .40
T                1.79
(.40)(.60)
625
p  value  .0367
(.365)(.635)
95% conf int  .365  1.96
625

Note in this example & similar examples involving p that the hypothesized value is
used in the square root (denominator) of T, whereas the observed sample proportion is
used in the square root of the confidence interval.

________________________________________________________________________
24.     The consumer group in (23) measured the weights of the sampled boxes of cereal. They
found avr. weight = 15.88 ounces, SD = .36 . Ho: µ = 16 oz vs. H1: µ < 16 oz
µ (large sample)

________________________________________________________________________
25.     A quick check of the records shows that the 256 employees of problem (22) averaged
12.4 days absent in 1996, with SD = 2.8 . Ho: µ = 12 vs. H1: µ > 12.
µ (large sample)

12.4  12
T              2.28
2.8 / 256
p  value  .011
2.8
95%  12.4  1.96
256
________________________________________________________________________
26.     Five 1999 BMW's are crashed into a brick wall at 10 mph. Average damage = \$2,162
SD = \$268. Ho: = \$2000 vs. H1: > \$2000.                µ (small sample)

2162  2000
T                  1.35
268 / 5
.10  p  value  .15 from t 4 table
___________________________________________________________________
26.     Five 1999 BMW's are crashed into a brick wall at 10 mph. Average damage = \$2,162
SD = \$268. Ho: = \$2000 vs. H1: > \$2000.                µ (small sample)
2162  2000
T                  1.35
268 / 5
.10  p  value  .15 from t 4 table
___________________________________________________________________
27.     52 of 100 randomly chosen college graduates support O'Brien for Congress.
92 0f 150 randomly chosen non-college graduates support O'Brien for Congress.
Does a higher percentage of non-college grads support O'Brien ?       p1 – p2

pooled extimate of p = (52+92) / 100 + 150) = .576
(. 52  .613 )  0
Area to the right of T                                         under the std normal curve
(. 576 )(. 424 ) (. 576 )(. 424 )

100                150

_____________________________________________________________________
28. 50 Crinkly french fries: Avr calories = 12.4 SD = 2.6
80 Home Style french fries: Avr. calories = 13.2            SD = 1.8

µ1 - µ2 (large sample)

Does the average Home Style FF have more calories than the average Crinkly ?

13.2  12.4
Area to the right of T                          under the std normal curve
2       2
2.6   1.8

50    80

29.     8 patients take a drug. Their before & After blood pressures are measured.

before:         68       62       78     56        90       88       71     70
after           72       60       86     62        88       90       78     78

What can you conclude ?       Does the drug increase BP ?       µ (small sample)

Work with the differences
4     -2     8               6      -2          2        7      8
mean = 3.875                 SD = 4.155

3.875  0
T               2.638
4.155 / 8
.01  p  value  .02 from t 7 curve
________________________________________________________________________
30.     A simple random sample of 4 "one-pound" cans of ACME coffee were weighed. The
measurements (in ounces) were              15.8   15.7      15.3   16.1
Does this data indicate that the true average weight of all "one-pound" cans of ACME
coffee is less than 16 ounces?                                     µ (small sample)

________________________________________________________________________
31.     Eight deluxe burgers purchased at Sky Burger were measured for their caloric content.
The following results were obtained: average caloric content = 1032, SD = 171.4 Is
this significant evidence that the average caloric content of Sky Burger deluxe burgers
exceeds 950? (You may assume that caloric content of Sky Burger deluxe burgers is
normally distributed.)             µ (small sample)

1032  950
T               1.35
171.4 / 8
.10  p  value  .15  from t 7 curve

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