Document 271039

Sample Examination Three-Answers
Section
I
l.
(E) The sum of the probabiiities must be one'
0.1 + 0.2+ 0.3+0'4+ 0 5 = 1'5
z.
(c)
ln the case of (E) the sttm is
as for each increase in 1 second
0.8 minutes. The slope of 0. I 8 can be interpreted
0.18 rninutes interval ot-time
in duratior-r of an eruption. there rviil be an increase of
in duration would mean an
the next eruptron. Therefbre, an increase of 60 seconds
1
until
estimated increase of
a
-') .
(c)
60(0'18)= 10'8 minutes'
62%.',rhe percent Variation in
.r,
explained by the least square regression of
,v
on x
isr2. Thevalueof rrcanbefoundbyusingtheformula'.b--rfi's"tuittgforrweget:
,=b r,\ s,=0.307 .sr=0. 73andbrepresentstheslopeofttreregressionlinewhichis
given as: 0.90. Pluggirrr into tlte fortnula rve get: r'=
we square 0.79. r U 79 tr = 6lo
4. (C) 0.4402This
/ n+g\
0'96\i.r,r,
= O'79To obtain r2'
o
observations anci each
represents a binomial setting with independent
observation having the sanre probability of success.
rhe Tl-83 calculator. plug in
I
-
N= 4O,p= 0.6 and k> 25, Using
binomialcdf (40, 0.6.24')which is
1
-
Probabiiity of 24
or less will prefer the brand ilame'
5'(B)Thesamplemeanrisanunbiasedestimatorofthepopulationmeanp.
6. (D) 0.783. This a binomial distribution
because observations are independent and the
observation' N= 15' P= 0'4 and
probability of ou'ning a cell phone is 40% for each
k=5.6.7,.."15.UsingtheTi-83:i-binomialcdf(15'0'4'4)=0'Zlg:
'7.
"\1
|
fn.
Et f#
llo
probability
of at least
probability of having no girls,tt
(+)
=
one girl" is 1-(probabilityof nogirls)' The
fr'
so' t
-#=ffi
OR
TI-83:
1
-
binomialcdf (7' 0.5' 0)
up to one, P(y= 1) must be 0'5
0.1. Since a probability distribution must add
at'rd y= 1)=P(r= 2)xP(l= 1)since
because0.l +0.5+ 0.4= 1. FurlhermoreP(r =2
s. (A)
.r and ) are
independent'
t'et the P(x=2)=xz'P(y=1)=O'S and the
P(x=2,!='s')=0'2.Then(.x:X0.5)=0.2'SolvingforX2'wegetxz=0.4.Thesumof
32
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&-
Section
Sample Examination Three-Answers
I
33
probabilities in
probability distribution must be l. To find xr or P (r =.- 2), add all the
the distribution ofx and subtract from 1.
0.1 + 0.3 + 0. I + 0.4 = 0.9. Then P (x = -2)=1-0.9=.1
a
9.
(D)83'9.Theexpectedcellcountcanbecalculated,,ineffi.
instrument is Rorv 1'
For the cell representing right-handed girls who play a musical
Column l. the formula becornes
{?$9
= 83'9
10. (B) 75. The formula which provides us with the smallest sample
maximurn error, and
{
n
=l
a
is: '=(##)
confidence level
required given a
,size
' This gives us
(i.qo \(2.2)\'
\i-]}.1/
| =1+.Zl.We round this up to 75.
[''')
11. (C) 14. Creating
a sample space for the 36 possible sums:
n
0
0
6
6
o
6
12
12
11
IL
12
l8
18
6
6
t2
la
LL
6
ta
IL
t2 t8 l8
12
6
6
lz
ta
l/-
6
6
la
IL
1")
12
ta
IL
t8
18
t2
1a
18
r8 18
IL
18
18
4i
t8 24 z+
aA
24
We find the probability distribution for the sum to be:
Sum
-Probability
1
0
-Jt)
6
6
-JO
1a
t2
;.10
l8
;JO
12
.A
LA
4
36
the multiplication rule for
(These probabilities could have also been obtained by using
events") The expected
independent events and the addition rule for mutually exclusive
varue is
zx r(x)=
o(+).r(*)+ r2(*)-' 't(*)+z+(*)='n
or
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34
Sample Examination Three-Answers
Section
I
t2. (E) As one increases speed, the time required to travel a given distance decreases. This
indicates a negative correlation.
ta
lJ.
(B) 0.07. This isageometricdistribution,where p=0.47 andl - p=0.53 andx=4. The
probability is calculated by multiplying: 0.533 0.47 = 0.07. This is the probability of
getting three girls and then one boy in that order.
14. (C) H,,:
15.
H= 36. The null hypothesis is: Hu: F= hypotheticalmean.
(B) 0.29. Standard deviation for difference between two means, standard deviations
nr-=
F;i- ,;lf
* =V -l3O- * :16f
unknown is: u.fr
i6. (B) Since
fi
P(A) P(B)=
= 0.29
(O.S+)(0.20)= 0.108 = P(AandB);the events are independent.
17. (B) Because of differences in reading levels between urban, suburban and rural schools,
a stratified sampiing metliod is most appropriate in the proportion which exists in the
population.
18.
19.
(D) I
-p
by definitron of Pouer. Pon'er is one minus the probability of a Type II error.
or r = 577.2.
for .r. *e set: 84 =
(B) 580 Solve the equation: : = +
To get 0.84 for z, we
Rounded to the nearest multrple of 10 rre get 580 *'hich is B.=#
to
the
1eft
of
the
cune
under
z. The closest value
for
the
area
looked up 1- 0.2 or 0.8
using the table is 0.7995 *hich gires 1'ou a:-score of 0.84. This may also be solved
using your graphing calculator b1 using 2nd DIST 3: invNorm (area to left of z, mean,
standard deviation). Forourproblem rve rvould do invNonn(O.8,510,80). The answer
from the calculator will vary slightll' frorn the table answer. Either way rounding to the
nearest multiple of ten gives you the same answer.
^I
where I'20. (A) (35.5, 44.5) The formula to use is: p + ,iry.
for a95o/o confidence interval, for this example it would be:
n, Since z = 1.96
04T051
/a5O\'-?5oi
-o4+
1.96\ -i53r=(355..1+5)
--Zs3-v.*r4s0-..,-\
t80
+,
o^
each population are independent. Two sample /-tests are used
when the population standard deviations are not known. Two sample l-tests are used if
samples are independent. If samples are dependent, then a matched pair procedure
should be used.
21. (B) The samples from
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Section I
Sample Examination Three-Answers
35
and III only. The third quartile is 59.8 so try definition of the third quar1ile,25Yo
of the values lie above 59.8. 31.2 and 59.8 are the first and third quarliles and again by
definition, 50o/o of the values lie in this region. Selection I says the mean is less than
44.5 which is the median. This would be true if the graph was lefl skewed. In drawing
the box plot there is no evidence to believe this.
22. (D) II
t0
60
c
50
40
30
20
L).
(E)1, Ilandlli.Oneproperll ofresidualsislre.siductls=0.Sincethemeanresiduai is
Lresiduols
. this ri,ould har,'e to be equal to 0. A curved pattern in a residual plot does
,
show that the relationship betr.veen the two variables can best be modeled by a nonlinear model. The residual plot can be a plot of residuals against the x values, as well
as the residuals against the fttted y values.
/-+.
(A) I only. The normal curve is symmetric about the mean. If a normal curve does not
have a mean of 0, it is not symmetric about 0. The empirical rule states that
approximately 680/o (not 50%) of the values lie within 1 standard deviation of the mean.
25.
(A) A continuous variable takes all values on the number line, Shoe sizes may come in
6 or 6.5 but cannot take on a value such as 6.23384 and therefore are not continuous.
26.
(D) 1.699. To find the value of / we
us the area under one
t-value of |.6997
27.
L+9q
,ur"*[email protected]=
tail. Looking up 0.05 for 30 -
1
= 0.05. This gives
degrees of freedom, one finds
a
.
(B) A type I error is reject Hu. but Hu is true, a Type II error is when we fail to reject
Hu, but Hu is not true. Since we are failing to reject the null hypothesis, this is an
example of a type II error
78. (C) The interquartile range (lQR) is obtained by subtracting Qr from p:, therefore
IQR= (49 - 27) = 22. To determine the existence of an outlier, compute 1.5 x IQR add
this value to Q: and subtract it from Qt lf any piece of data falls outside the range of
these two values, it is an outlier.
(0,) + 1.5(22) = 49 + 33 = 82
(0,1
- t.s (22) = 27 - 33 = - 6
Since all the values in (C) fall above 82, they are outliers.
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36
Sample Examination Three-Answers
Section I
29' (B) The distribution of a sample
mean when o is not known can be represented
by a
/-distribution, provided the sample is a simple random
sample. and the population from
which the sample is taken is normal. The level of significance
does not play a factor in
this decision at ail. nor does the value of the sample-mean.
30. (D) ijsing IJ.:trt = 15 and H^:,u<
15,r
=t!?,-^t)
=- 1.40, which results
.43t
in ap_value
/36
of betrveen 0.05 and 0.10. usingthe r-score char1. (r-teston-fl-g3
yierds p= 0.0g59)
at
Jt.
(E)
i.
2'
3'
-fhe
basic principles of statistical design of experiments are:
response, most easily by
comparing several treatments.
control of the effects of rurking variables onthe
Randomizatio,n, the use of impersonal chance to assign
subjects to treatments.
Replication oithe experiment on many subjects to reJuce
chance variation in
the results.
-12.
(B) A census is a complete count of an entire population.
The nervspaper is polling the
entire population of their readers. Although the sampling
design may not be random,
this sample survei- is intended to make inferences about
all the paper,s
readers.
33.
(B) Since an outlier can stronsly' affect the ralLre of the mean (1t),
it is non-resistant.
since both the standard cleviation (o) and ccrrelarion coefficilnt
(r) depend on the
mean' they are also non-resistant. The meclian i-s not affected
by an outlier, so it is
resistant.
a1
J{.
(C) To select a stratifled random sample. first divide the population
into groups of
similar individuals, called strata. l-hen choose a separate SRS
in each stratum and
combine these SRS's to form the full sample.
35.
(D) Increasing each grade by 5 points ',vill not affect the distance
each grade is from the
nlean. therefbre the standard deviation will not change.
However, measures of center
like the mean and rnedian rvill increase by 5.
Jo.
(A) Substituting the value x= 3 into the equation of the least-squares
regression line
results in i= i.3 +0.21(3)= 1.3 + 0.81=2.i1. The value
of the residual is observed
predicted
-17.
-
: 2 - 2.1 1 =- 0.1 i.
(B) fhe distribution is skewed to the reft, w'hich impries thar the mean
is less than
the
median, since the mean is non-resistant. The mode must be
90 and the median is 85. so
the mean is less than the mode as r.vell.
38' (D) The fact that a distribution
has a larger mean does not have any bearing on how
tightly the data is clustered around the mean.
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Sample Examination Three-Answers
Section I
39. (B) The pooled sample proportion for a 2-sample test of hypothesis is gi'ren
p=
#&rr=2#ffi=
40. (E) All of
JI
by:
0.3025
tl-re statements are true regarding the effects
of lurking variables.
trn fact,
lurking variable can falsely suggest a strong relationship between the variables. or
cari hide a relationshrp that really exists.
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38
Sample Examination Three-Answers
Section
Section
II
II
Question One
l.
Let P be the event an employee passes the dexterity test, then P' is the event an
employee does not pass. Let W be the event that the employee is a \voman, then W' is
the event the employee is a man.
(a)
The probability that the empioyee passed the test if it is known that the employee
.
P(PnW)-# -48-.,
pfw)
-\ ' =i{=ii=0.64
185
--,-,--.\
isa r.roman is P(PiW)=
(b)
The probability that the employee is a woman. gii en that it is an employee who
passed the test is denoted as
(c)
P(wlo)=
r./
i$*a
p(p)
=
#3
= #= 0.387
L?!-tz4-v.Je
185
Are the events "selecting an employee r,r'ho passed the test" and "selecting an
emplovee rvho is a woman" independent events? Justify' your answer.
Student must perform a chi-square test:
H,,'. Gender and passing lhe test are inclepenclent
Ho: Gender ancl passing the Iest ctre not inc/epertdenl
Assumption that all expeced counts > 5 is true.
Row Tolal x Coluryn Total
Exoectecl _
'
Lrrl]no I OtJl
e.g. Expected courrt fbr "men" uho "passed"
ll0x 124
- -T85-_ 27=?8
JI
z
- 44
expecred -('^-2#)
)t
.,.-- /'
=(observcd-expecred)2
^
|
2A
_1342f
\r*--JTl
----ItT--
-tT
-
(nt-tfl)
l{qq
.2
-
37
2
s
_91 \
"- 3l l
\r'|
915
-iT
=r,..5'-i
d.f.=(row total- 1)x(coiumn total- 1)=
(2-l)(Z-t1=1
p-value =0.4696.
We rvould reject the null hypothesis if xr> 3.84 for a 0.05 level of significance.
Due to large p-value. rve do not reject H".
Conclusion: Insuffrcient evidence to reject the null hypothesis. Our data do not
reject the claim that "selecting an employee rvho passed the test" and "selecting
an employee who is a woman" are independent.
Note: Students may try to use the multiplication rule for determining independence.
This is only partiallv correct because variability is not being taken into consideration.
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Section
Il
Sample Examination Three-Answers
Question Two
Correct answer
#I
Two hundred patients with this type of cancer are randomly selected to get one of the
A, B, and C. Randomizarion could be accomplished by assigning each of
the 200 patients a number between I and 200. Select the first 66 of the patients using a
random number table to get treatment A, the next 66 patients to receive treatment B and the
remaining 67 Io geI treatment C.
three procedures
After a select period of time, results are collected for each group and compared.
Randomly
select ,/
2oo Patientsto
each type
4t
of \
Procedure \
Proceclure
'4
Prttcedure B
Comoare results
Procedure C
Correct answer #2:
If there is reason to believe that t.omen
react differently than men then blocking by
gender is appropriate. The 60 rvomen uould be randomly selected to each of the three
procedures, Number 60 women 1-60. and using a random number table assign the first
randomly selected 20 women to receive procedure A, the next 20 to receive procedure B and
the remaining 20 to receive procedure C.
Follow the same procedure for the 140 men. (i.e. the first 46 randomly selected get
procedure A, the next47 to get procedure B and the last 47 to get procedure C).
Compare results within the blocks.
Procedure A
Randomly assign
the 60 \vomcn
to each procedure
,/
<\
Procedure B
Compare results
Procedure C
Procedure A
Ranclomly
the 140
assign
men +
toeachprocedure
,/
Procedure B
\
Comoare results
Procedure C
Note: In both answers students could add a fourth group as a control group. This group
would receive an old treatment or a placebo. However, a complete answer would address
why or why not a control group would be used in these circumstances mindful of ethics,
severify of disease, etc. Ideally, all experiments have a control group.
copying or reusing
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40
Section
Sample Examination Three-Answers
II
Question Three
Tlie test required fbr this problem is a two sample difference of means I test since the
groups are independent and since sigma is not known. This test requires the two samples to
be independently selected simpie randoni samples and for the two populations to be
normally distributed. It is not verified how these classes are selected, so \\'e will state our
assumption that rve have tu,o independent simple random samples. Checking for normalcy
of the population we ivill use normal quantile plots.
Normal Quantile Plol for \\'ith \'lanipuletrves
data
Normal Quantile Plot for Without Manipulatives Data
tr
tr
s
d
Since the plots do not shou anv severe skew or outliers, we continue under the
assuilption that both popr-tlatior-is are approximately normal.
i
+Fl
|
tll
a6)
15
-t--
ll
with nanipulatives without manipulatives
The boxplots indicates there is no evidence of outliers, an indication that the condition
of normality 'nvould be satisfied.
.
2.
I
Does manipulative increase test scores?
LL,'. /Jt=
/)z
Il,.'. lJL> Pz
3.
Where pr is the mean test score of those using manipulatives and
test score of those not using manipulatives.
a = 0.05
Unauthorized copying or reusing
any part of this page is illegal
pz
is the mean
Section
II
Sample Examination Three-Answers
A ,_(x,-rr)-(ttta.r-T
,'l ru- n
ltz)
6.
Approximately / distribution with d.f. equal to the smaller of
Reiect H" if r> 1.833
7.
test statistic' r =
5.
4l
-s3.37', , 5.4',
/ -To-- - 10
= 1.99
rir- | or nz-
1.
p-value:0.33
1
TVro sample
T for With Manipulatives vs Without Manrpuiatives
N
Mean
StDev
5.40
SE Mean
with manipulatives
1C
84.00
r.1
wi f horrf man'i nrrlatives
10
80.00
3 .37
1.1
T-Test mu with m=mu without (vs >) : T= L.99 P= 0.033 DF= 15
8.
Conclusion: At the 0.05 level of significance there is sufficient evidertce to
conclude that test scores of students using manipulatives are higher than those of
students not using manipulatil'es.
Note: a student using the TI-83 calculator's 2-sample t-test function u'ill obtain
p-value of 0.031 with 18 degress of freedom using the pooled option, or a
p-value of 0.033 rvith 15.085 degrees of freedom using the unpooled option.
a
Because there was no random assignment of students to the two groups, this is an
observational study and caution must be applied in reaching a conclusion. There seems to
be an association between higher scores and use of manipulatives but higher scores may be
due to other underiying conditions that have not been controlled.
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42
SampleExaminationThree-Answers
Section
II
Question Four
The test required for this question is the chi-square test for Independence. We are
looking for statistical evidence that proportion of subjects with low, average and elevated
cholesteroi levels differs for walker and non-walkers.
The assumptions for this test are that all expected counts are at least 5 and that the data
are from a simple random sample. The first condition will be shown to be satisfied when the
calculations are complete. The second condition is not clearly satisfied in the description of
the experiment. W'ere subjects randomly selected to be walkers and not walkers or was the
experiment more likely an observational study? We will proceed as if this condition was
satisfied.
1.
2.
Is there a relationship between aerobic walking and cholesterol levels?
Hu : There is no relationship between walking and cholesterol level
H" : There is a relationship between walking and cholesterol level
3.
4.
5.
0
6.
7.
= 0.05
Test Statist
ic: X2=
t
X2distribution with (rows - 1)(columns - 1) degrees of freedom.
(3- 1X2-I)=2 df.
Criterion for Rejection: Reject Hu for x2> 5.99 for 2 degrees of freedom
Xz = 9,609 Note all expected cell counts above five.
Cholesterol Level
Walkers
Lorl
Average
Elevated
51
86
31
lse2)
(e6)
f+ss)
L-)
94
30
t"j
1.",
Row Total
168
Non-Walkers
518
*t)
427\
15 l
t47
(84)
Column Total
'/4
180
61
315
Expected Counts: in parenthesis
Expected counts can be calculated by rake,n.
,
-^^ ,2
, (r'-#)
x- = ---sgT: *
i5
t
n '',, ,2
,.^
aLt
(so -s6\2
--96-
+
ffi
(r'-ff)' (rr-#)' * 04-84)2
---ZE8 * ---ilE15
15
--T-4-
I
\-" 15 /
+-zn:
l)
p-value
8.
:
= 9'609
0.008 or using table for 2 d.f. 0.005 <p < 0.01
There is evidence to reject the null hypothesis and conclude that level of
cholesterol is related to aerobic walking.
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re
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Section
II
Sample Examination Three-Answers
43
Quesion Five
(a)
Because we have two independent random samples from two distinct populations
(men and rvomen) and rve are looking to see the difference between the
proportion of men and women rvho favor legislation which restrict cell phone use
while driving. the choice of interval u,ould be large sample confidence interval
ri(r_it;-frr!_fr,
forcomparingtwoproportions:formula:(i,,-i,')t,'l#+-'
Assumplions inclttde: Trvo independent sirnple random samples fiom tu'o large
populations , nr ftr , nz fu2, ir r (1 - fut). nz(l - irt) more than 10.
Note: In probiems that deal rvith proportions, some texts use
ntf)t,ttz
pr,n'(l -ir'),n,(l -p,)are all greater than 10 when checking
assumptions, r.r'hile other texts use the value of 5. Please be advised that as long
as the student shou's that the assumptions har e been checked, either value is
acceptable.
bt=
l?n
Iuj
2i#
)uu= 0.-16 tit.- = li]
500
= rl. jSU
tt
- 500
rz:=
5C)0
Clearly a/l
o.f the conclition.y ctre .satis/ietl. \\'e also asslllre that there are at least
5000 rnen and 5000 \\omen in the countn
For a 95%o confidence interval. the value of z = 1.96
The interval.for p,,,,- pnon*i.r (0.0109, 0.1331).
(b) We are 95% confident that the difference in population proporlions is between
0.0109 and 0.1331. Because the interval does not contain 0, $'e are confident that
men have a stronger preference for legislation that would restrict the use of cell
phones while driving.
(c)
A99% confidence interval would change the value of z in the formula above
from 1.96 to 2.576 thus making the interval wider. The new interval would
become: (-0.0083, 0.15229). This wider interval would contain 0 and the
conclusion reached concerning whether there u,as a difference in the proporlion
of men and women who favored legislation would be different than in part (a),
i.e. there r.vould appear to be no statistically significant difference between the
two groups.
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44
Sample Examination Three-Answers
Section
II
Question Six
(a)
0'60' using the
+
The probability of B or o blood is P(B) + P(O) = 0.15 0'45 =
B or O blood
probability of b.OO, \ve can assign six of the ten integers to be a
is to
solution
possibie
iype and assign four to be some other blood type. One
the random
urrlgn 0 throigh 5 as blood type B or O. We then need to use
values
three
at
number table and count how many digits it takes to arrive
are needed to get three
betu,een 0 and 5. This would give us how many donors
donors with blood tYPes B or O'
(b)Startingintheupperlefthandcornerw.euses,4,l,T'7and0.Westopat0
0 and 5
have found three values 4, 1, and 0 that are values between
'ne
type. It took 6 donors to
and represent three donors with type B or O blood
again where we left off-"uve
arrive at three uith the desired blood type" Beginning
at I because we have founc
use the number 6.7.5. 7.1.1,6, and 1. we stop
donors before finding
three integers bet$'een 0 and 5. l'his time it took eight
we again start where we
three with the desired blood type. For the last simulation
these numbers are
left ofr. The numbers begin rvith 3, 1, and 5. All three of
donors with type B or O
integers betu,een 0 and 5 and therefore u'e got three
that it took
first tl.rree numbers. Three trials of our simulation found
blood in
'ur
the blood type B or O'
six, eight and three clonors io get three donors r'vith
because
(c)
protrability of getting a
Distribution B is the correct distributionBecause the
probability of finding a donor
donor *,ith tlpe B or o blood is greater tharr the
donors with
nith t.vpe A or AB blood. it $ou1d take fe$'er donors to find three
mean and a lower maximum
type ts or o blood t1,pe. Distribution B has a lorl'er
as well.
(d)UsingdistributionB,theexpectetlnumberisfoundusingtheformula:
(.x) can be found by raking the frequencies for
E (ri= x , /(r). The vah_ies of l
of the frequencies which
each:rvalue of the histogram and dividing by the sum
isl00.Thevaluesofxaretlrevaluesonthehorizontalaxisofthehistogram.
Thisgivesus:0.22(3)+0.27(4)+0.21(5)+0.15(6)+0.07(7)+0.05(8)
+0.02(9) + 0 0i (i1) = 4'81
(e)
of type o or B donors
Blood bank A would be more likely to have a percentage
donors each day as a
to be more than seventy-five percent. lf r.r'e treat the
o anci B donors for blood bank
sample. the distribution of the percentage of type
Awouldbeapproximatelynormallydistributedwithameanof0.60anda
standard deviation
"t m
= 0.069. The distribution of the percentage
typeoandBdonorsforbloodbankBhasameanof0.60andastandard
deviation"r/W=oo35'
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of
Section
II
Sample Examination
Three-Answers
45
We can find the appropriate probabilites by using the z-score table and the
) l7
fbrmula for z-scores. The z-score for blood bank A would be 0'7i =-p='60 =
0.069 -L'tt'
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The z-score for blook bank B
'.
curve to the right of 2.17 is 0.0150.
= 4.29.The area under the normal
T'he area under the curve to the right of 4.29
is less than 0.0002.
OR
Using the TI-83 calculator for blood bank A:
Press: 2nd distribution
Use 2: normalcdf as shorvn below:
norma L sdf ( E. 75 ' I
E16' 9.6' B.869)
.8148557881
I
Using the TI-83 calculator for blook bank B:
Fress: 2nd distribuiion
Use 2: normalcdf as shown belou,:
rdf { B. 75 I
8.6' E.835) '-5
nBFr'ia l
E16'
T
9. 11484S649e
The probability of having at ieast T5ohtype B or O donors is greater for blood
bank A than for biood bank B.
This also stands to reason since there are f-ewer donors at Blood Bank A and the
Law of Large Numbers implies that the larger sample (Blood Bank B) is more
likely to have a sample proportion that is closer to the true population proportion
of 6A9'o. The higher variability in Blood Bank A's distribution is caused by the
smaller sample size. With the higher variability', the chances of obtaining a
sample percentage of 75o/, or more are higher for Blood Bank A.
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