1. No category

Sample Examination Four-Answers
Section
I
I
. (D) The maximum and minimum of Set X are 100 and 50, respectively, for a range of
50. The maximum and minimum of Set Y are 100 and 57, respectively. fbr a range of
43. Therefore. the range in Set X > the range in Set Y.
For Set X. using the 1.5(lQR) rule, 50 is not an outiier. The interquartile range (lQR)
is defined to be the 75th percentile minus the 25th percentile. In this case, the 75th
percentile is approximately 90, and the 25th percentile is approximately 67, for an
interquartile range of 90 -67 =23. Multiplying 1.5(23)=34.5. If this value is
subtracted from the 25th percentile (67);the lower bound of 32.5 is obtained. Since 50
is higher than this boundary value, 50 is not an outlier.
The amount of data in the two sets cannot be determined from a boxplot.
The interquartile range of Set Y is visually larger than than the interquartile range of
Set X.
The 5-number summarv in a boxplot includes the minimum, quartile 1. median, quartile
3, and maximum.
2.(B)
Since x= 2 and
)'=
1'1.
g'e hare 11=2.2t+
I
l0 = 2nt
5--nt
Therefore, the equation of the regression line is
)=
5x+ 4 and y= 5x+ 4
3. (C) Since the mean(7o/o) is much lower than the median (12%), this implies that the
distribution is sker.ved left. The mean is affected by extreme values.
4.(B) Using the formula
,=+
left-hand area of 100% -
and the critical value 0.675 (which corresponds to a
25o/o = 15%)
6-t5=r+L
41.25=475-p
42j.15= 1t = 4Zg
,,
t.
o=70
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Sample Examination Four-Answers
48
Section
12. (B) This scenario has a binomial distribution.
success, then P(success) =9.15
The formula for the binomial is:
P(X= k) = (,C)p*(l - p)'
where
-
I
If we call an underweight package a
o
,C*=;/L
" ft!(n - ft)l
So, P (3) = ( sCr) 0. I 5r (0.8s
)'
= i t o ) (0.003375)
(o.t zzs) = 0.02438437 5
P(4) = (,c,)0.1s' (0.85)'= 1s;10.0005062s)(0.8s) = 0.0021 srs62s
"
P ( s) = (,c,) o. 1 5' (0.85) = ( t) (o.oo 007 s%7 s)( I ) = o.oooo7 se37 s
P (x' 2) =P (:) + e (+) + e (s) = o.oz00
Using the TI-83: I
13.
(c), =- - !
sl
-
[Binomcdf (s,0.15.2 )] =0'0266
3:- = 3.?j
--l1.53r
Jn
,'25
=
7:7
ftr:8
sr: L53
5
Using the /-table with 25
- l = 24 degrees of freedom, we see that 0.001 < p < 0'0025
t4. (E) If the subjects do not know which pill they are taking, the experiment is singleblinded. If neither the subject nor the experimenter knows which pill is being taken, the
experiment is double-blinded.
15.
(E) A is a normal or rounded distribution.
B is a rectangular distribution.
C is a left-skewed distribution (skewness follows the tail)'
D is a histogram with no set pattern.
16. (B) With stratified sampling the population is subdivided into at least two different subpopulations. A random sample is then drawn from each sub-population.
17.
(c)
Expected value is:
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50
Sample Examination Four_Answers
Section
22. (A) 0.0038
Find the z-score by substituting into
24)
the equatio n r=Q--lt)
, -(BLook up on the normal distribution rabre
the u..u .o..ffi,nrXo=
The area =0.003g or 0.3ga/o
OR
on the TI-83 calculator: 2nd vars 2: Normalcdf
hit enter type and in (lower bound,
upper bound, mean, standard deviation)
g.24,6)=
(_ 1000,
0.003g30 or 0.3g%
$*=-'uuu
23. (D) 9.55
The standard deviation is the square
root of the variance.
Variance=16,o=fi6=4
The 35th percentile corresponds to 0.35
which is the area under the normal
cun,e.
Looking up the area of 0.35 'nve find
p)
z=-0.39. Substitute in the z_formura
' \vL\"ut. {,_(r- -z(s
therefore -0.39 -p)
Multiplying both sides by,. 4, rve get + (- 0.30)
= S _ 1t thensolve for p.
F = 9.56
OR
Use the TI-83 calculator to find the.z-score
2nd vars 3: Invnorm hit enter and type
in
(left-hand area. mean. srandard deviation)
(0.:S.0, l)=_ 0.3g53 which is the
z_score.
Substitute z-score in a[rore equatior and
the answer is 9.54r3
24 (D) 149. The degrees of freedom are
/r
-
r rvhere n is the sampre size. (1501)= 149
25 (D) 1'746' use the table of l-distribution
critical values, with degrees of freedom
(df)
n - | = 17 - | = 16 and confidence level
90oh, t = 1.746
26' (D) A normal curve is always symmetrical
and is never skewed. It only has a
mean of
zero and a standard deviation of one rvhen
it is a,stanclardnormal curve.
27.
B)3/l\'/!\'
\o/ \6/
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=
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Sample Examination Four-Answers
52
Section
I
33. (A\ 2
The standard error ir
-F.
v'll
Since the standard deviation equals the square root of the
variance,
o = /tOO or 10.
Substituting into the standard error formula we get:
l9=r
Jzs
34. (C) (0.2177,0.4489)
Substitute into the formuia p +
^
P=
7
15
A<
The z-score for a 90oh interval is 1.645
't<
/l 15 tr3u
/\Jli\+-.\r
r
'^",=
i+tt.645,!
4)
|
+) 0.3333 + 0.1 156 = ( u l 171.0.4459)
OR
Use the TI-83 calculator: STAT tests A: 1-Prop Z-intewal.
Hit enter and type in
-- 1S
n=45
C-level: 0.90 arrow dorvn to calculate. hit enter. rne get (,0.2177,0.4489)
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Sample Examination Four-Answers
54
Section
Section
II
II
Question One
p= pt,
of boys who admit to cheating
pz=proportion of girls rvho admit to cheating
Assumptions: Population is at least l0 times the sample
Ho'"
t1^'.p'>
nfu
p'
pr =proportion
Forp:550(0.434)=2gg
>10
For
n(l
-p)>10
p2
:430(0.434) =
ffi = 0.46
g.4
'pr= ffi
+Jt.r =
e=
1gZ
a=ffiffi=0.434
Forpr:550(0.566)=3tt
For pz:430(0.566) = 2qZ
Note: In problems that deal with proportions, some texts use np and n (l
- n) > to
rvhen checking assumptions, while othertexts use the r,,alue of 5. please be advised
that as long as the student shorvs that the assumptions have been checked, either
value is acceptable.
0.46
-
0.4
0431(osoo)(#-r*)
=
1.88
TI-83: 2-proportion ;-test, r= 1.88. p = Lr.03
All assumption check. Also assume SRS.
z:
1.88
Conclusion: A z-score of 1.88 results in ap-value of 0.03. Due to the smallp-value,
to be enough er.,idence ar d, = 0. 0 5 to rej ect Ho and state that boys are more
Iikely to admit to cheating than girls.
the re appears
{)uestion Two
(a)
Assumptions: The data are an SRS from the population of interest. The
popuiation is at least I 0 times larger than the sample (Rule of Thumb I i/ is so
).
large that both the count of successes np and the count of faiiures ,(l-it)are 10
or more (Rule of Thumb 2).
(b)
The first assumption appears to have been violated, since using a random sample
of high school male students omits the female students and those students under
high school age.
The second assumption appears valid as we assume thatthere are at least l0 X
750 = 7500 students in the public schools of the ma,ior US city.
The 3rd assumption appears valid as 82 successes and 668 failures are both l0 or
more.
Note: students may choose to use 5 in place of 10.
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56
Sample Examination Four-Answers
Section
II
Question Five
This is a one-proportion z-test.
Assume 40 kicks constitute an SRS of all kicks from that ranse.
Assumptions: npo>10 and n (t -p,) >10
40(0.+s)=
13
40(0.ss\=22
So conditions are OK.
problems that deal with proportions, some texts use nro and
l0
"(I -f,)>
rvherr checking assumptions, while other texts use the value of 5. Please be advised that
as long as the student shorvs that the assumptions have been checked, either value is
Note:
ii-r
acceptable.
i:io: P = 0.2i5
LJa: p > 0.45
1t
D=?-;=U.tt
-+u
where p : tire population
p-vaiue = 0.028
0.028 <ri.05
o=
proportion of successfui field goals from that range
0.0-5 level
Conclusion: The lor', p-raiue slrosests that rre can reiect F{o. The data suggests that
Stiiafi's summer n'ork made an improrenrenr in his field soal kicking sllccess.
oR.
Use {ire
Ti-83: 2nd vars 5:
I-{it enter and type in
po= 0.45
x=24
n'= 4A
Pro;;>P
o
= .9t)l
p-v:riue = 0.02E
z
I
1
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