Practice
I2.lB
AP Statistics
Name:
LeRoy, a starting player for a major college basketball team, made only 40o/o of his free throws
last season. During the summer, he worked on developing a softer shot in hopes of improving
his free throw accuracy. In the first eight games of this season, LeRoy made 25 free throws in 40
attempts.
1.
You want to investigate whether LeRoy's work over the summer will result in a higher
proportion of free-throw successes this season. What conclusion would you draw atthe
0.05 level about LeRoy's free throw shooting? Justify your answer.
Hoip = 0.4
Ho:p > 4.4
p:true proportion of'free
a:
thror.vs LeRoy makes
sample one proportion z-test d or0.05
sample P(p > .625) - P(z > 2.9) = 0'0018
Assumptions: *random
-pop. ls at least lOx
-25 X0 -15 >0
Since p<oq it is statistically significant,
therefbre reject Hr. There's enough evidence to
suggest that LeRoy's work over the summer
helped his free throws
2.
Construct and interpret a90%o confidence interval for the proportion of free throws LeRoy
will make this season.
One proportion z-interval
Assumptions: Same as above
!
A.625 +l- 0. 1 26:(0.499,0.75 I )
We are 99a/a confident the true proportion of ffee throws made by LeRoy is b/rv
50o/o and 75o/o
Hypothesis Testing with Proportions
AP
Stotistics
Multiple Choice
you ore testing the hypotheses stotements He: p = .5 d Ho: p < .5, then you should perform o
b) right-toil
c) two-toil
d) none of the obove.
If
1)
test
o) EfJEI€E
Since H"
test
.' p < "5 we look at the left tail
2) A seaside resort cloims thot 75% of the doys during its peok seoson are clear. A consumer group f eels this cloim is too
high so they exomine o rondom somple of 150 doys over the lost two peok seosons, f inding 105 cleor doys. At the 0.05
signif iconce level. whot conclusion con be mode?
o) Reject He since the p-volue is greater thon cr
b) Re
since the n-volue is less thon a
c)
d)
e)
t
the p-volue is grealer thon cr
Foil to reject Ho since the p-volue is greoter thon a
Accept He since the p-volue is greoter thon a
Foil to
^ 1n5
p = *Jl =./Q
' 150
Hs since
.70
Z=*::
i
* .75
n-t
j.rct.z3J
rbo
--t
=-1
\l
Since the p-value ls greater than
.41
p - value
:
N"o,i*100, *1.41) = .A7
tt we fail to relectH"
3) A deporfment store ot o lorge moll cloims thof over 60% of the moll's visitors shop ot thot store. A consumer group
doubts this cloim. A hypothesis IesJ of doto from o rondom somple of moll visitors reports o p-volue of 0.11. Which of the
following is o correct interpretotion for this p-volue?
o) The probobility thot the null hypothesis is true is .11.
b) The probobility thot the olternotive hypothesis is true is
ion is .11.
c) The probobility of qettinq the
d)
e)
The probobility
of getting
.11.
o somple proportion os extreme or more thon octuolly observed is .11, ossuming p = .6.
The probobility of getting o populotion proportion os extreme or more thon octuolly observed is
.11.
We never say that the nu// hypothesls is true.
We sametimes say that there ls sufficrent evidence to "stlggest"
that th* alternative hypathesis is true.
c ls iust plarn wrong"
e would probably be ok if we didn't also have d as a choice.
stotements is true?
o) The choice of the ohernqtive hypothesis depends on the objectives of the study.
b) Hypotheses ore written obout the somple doto.
c) The null hypothesis should be whot you ore trying to prove.
d) The olternotive hypothesis moy be written using < or >.
e) Tf the null hypothesis is rejected,fhere is strong stqtisticol evidence thot the Hs is true.
5)
Ag Trotfernewsletter reported thot, notionolly,90% of odults drink milk. A regional formers'orgonizotion plonning
new morketing compoign ocross its multi-county oreo wonts to know if their region is different thon the notionol proportion.
fn o rondom somple of 750 oduhs living in the region,657 odults soid thot they drink milk. Whot is the test stotistic for
the oppropriote hypothesis test?
The
o) P = 0.0285
The approprrate hypothesis test
.
p'
f.F.7
II'
750
=.876
is
c) 0 =0.870
H. .'F = .90, H- .'p '. .90
The lest stattstic = Z =
p*p
rff:p)"
It
*.90
=
.876
****ry-
le{
d) z = 2.191
^ 4^4
= *Z.lV
I
1)
\ zso
o
6) Which of the colculoted volues of o test stctistic would have the lorgest p-volue?
o\ z = 2.O5 from o riqht-toiled tesf
b) z= 2.O5 from o left-tsiled fest
c) z= 2.O5 from o two-toiled test
d) none, oll the p-volues ore eguol
This ls a trlck questiCIn as stated.
A left tailed test with a lower bound of -/00, and an upper bourld of 2.05 has
a very hlgh p-value, but rs a very unlikely test
A better guestian { and easier) would haveZ = -2.45 for answer choice b,
then the correct answer would be c because you would have a p-value twice
as largs for a two tailed test
7) Public heolth officiols believe thot 90% of children havebeen voccinated ogoinst meosles. A rondom survey of medicol
records ot schools ocross fhe country found thot only 89.4%hovebeen voccinoted. Whot ore the oppropriote hypotheses to
test if the public heolth off iciols' belief is too high?
o)
He:
d)
p=.9ondHo:p>.9
He:
F =.9ondH":
p+.9
b)He: F=.9ondH":p..9
=.9qndH^:p<.
c) He: p=.9ondHo:p+.9
The a/ternative hypothesis is what we are trying to show might be true.
ln this case, that the number of vaccinated children is less than 90%
8) On overoge,afair die should giveyou o five on one roll out of six, or l/6 of the time. You hove o die thot wos run over by
o cor, ond you wont to see if it's producing the expected number of fives. You roll the die 60 times ond get 7 fives. Whot is
ihe volue of the stondord deviotion of the somple proportion you'd expect if the die were foir?
,,W
o)W
o,F6m
.)Fo(161s6)
lf you suspect that the dlce are not fair then H. t 9 = 1/ 6, H" : p +
When yCIu calculate the standard deviation you use p
1/
6.
,qlqf
\
o:11/olo
160
-
i
9) A left-sided hypothesis test ollows you to reject He ot the 0.05 levelof significonce. Suppose a90% confidence intervol
is constructed with the sample doto. Which of the following stotements is true?
o) A90% confidence intervol will not contoin the true proportion.
b) A 90% confidence intervol will contoin the true proportion.
c) There is no woy 1o fell if the true proportion will be inthe90% confidence intervol sincewe do not know onything
obout the
d)
doto.
There is a90% chonce thot the true probobility will be in the confidence intervol.
from oll its checking occounts ond found thqt 45 were overdrown ot
leosf once in the post two yeors. The bonk manoger reported thot the bonk hos o significontly lower percentoge of
overdrown checking qccounts compored to the notion os o whole. If 20% of occounts notionolly ore overdrawn, which of the
following would be the correct procedure to verify the monoger's cloim?
10) A bonk randomly selected 300 checking occounts
, . =_TT
.15_.2
o)
b) z =
m
o=
'
11)
lL=.ls
300
.2
-
,,
.2 - .t5
d)z=----
)z= :.t5-.2
.15
J5c85)
l.z(.e)
/.15(.85)
3oo
tr
1/ 3oo
z
Which of Ihe following stotements is true?
a) Thelevel of significonce is the omount of evidence needed to occept the null hypothesis.
on of the time thqt vou willfoil to reiect the null
thesis.
b) The level of siqnif icon ce is the
c) Thelevel of significonceisthe probobility thotyou willreject the null hypothesis, ossuming it is true.
d) The level of signif iconce musf be 0.05.
e) The level of signif icon ce is the probobility of obtoining o somple proportion as extreme ot mote exfreme thon whot is
observed.
€ is not correct because the p-value ls the probability of obtaining a
sample prapoftlofi as extreme or more extreme than what is observed
flot g
12) Suppose you wont to esfimote the proportion of polydoctyl cots (cats with extro toes).
You exomine 419 rondomly
selected cots, ond find56 polydoctyls. You hove no prior ossumptions to test - you only wqnt to estimote the true proportion
of polydoctyl cots. Should you corry out o hypothesis test using p =U%rros you null hypothesis?
No, 419 isn't
b) No, since you wont fo estimote the proportion of polydoctyl cots, you should construct q confidence intervol.
c) No, your null hypothesis should be: The proportion of cots who ore polydoctyl is .5, since polydoctyl is o genetic
defect.
d) Yes, your null hypothesis
should be: The proportion of cots who ore polydoctyl is 56/419.
You are nol trying to prove or disprove a
claim.
You are trytng to extimate
a parameter front a sample p{oportlon.
B) Af f er o f rost worning wos issued, the owner of a lorge orange grove wonted to test to see if more than 20% of his trees
suffered consideroble domoge due to the frost. fn o rondom somple of 100 trees from his grove, 26 trees showed
consideroble domoge. fs there sufficient evidence to suggest thot more thon 20% of the lrees suffered consideroble
domage?
o)
Yes,
would
t the
null
Sotc[=
b) Yes, you would re, the null hypothesis qt cr = .1.
c) No, you would occept the null hypothesis ot cr = .05.
d) No, you would occept the null hypothesis ot a = .01.
e) No, you would foil to reject the null hypothesis ot cr : .1.
^26
.26 -.20 , _
p---_.lo
p-va/ue=N,.,
'
100
rl
i't
!
r)
(1 .5,100 ) = .967
100
lf <t- = 1A lhen the p-value is less than u so answer b would be correct
Yrsu never accept the nul/ ltypothesis
14) Tn perf orming o two-sided hypothesis test, you colculoted
value?
o)
p
.0179
thot the test stotistic z = -2.1. Whqt is the oppropriote pc).9821
.0357
d).5
* value = 2 * Ncdf ( *10a,*2.1) = .0lgz
t5) A group of scientists wonted to estimote the proportion of geese returning to the some site for the next breeding
seoson. Suppose they decided to increcse the somple size from 2OO to 500. How will this affect the distribution of the
somple proportion?
The distribution of the
will be more
will be less spreod out.
b) The distribution of fhe
c) The spreod of the distribution of the somple proportion will remoin unoffected.
d) The distribuf ion of the somple proportion will more closely resemble the binomiql distribution.
e) The distribution of the somple proportion will be more strongly skewed.
The standard deviation will be smaller, so the distribution of the sample
proportion wi// be less spread ouL
Practice
l2.lA
AP
Statistics
Name:
We hear that newborn babies are more likely to be boys than girls, presumably to compensate for
higher mortality among boys in early life. Is this true? A random sample found 13,173 boys
among 25,468 firstborn children.
1.
Does this sample give sufficient evidence that boys are more common than girls in the entire
population? Carry out an appropriate test to support your answer.
(Hint: We can't use a 2 prop. because they are nat2 independent samples)
Ho:P = 0.5
Ho:p > 0.5
p:tme proportion of newborn children rvJro are boys
Assumptions: -random sarn-ple one proportion z-test w/ e0.005
P(p > .51.7) * P(z > 5.502) = 0.000000018
-pop. ls at least 10x sample
-13173X0 -12295>0
Since p<oq it is statistically
significant,therefore reject Ho. There's
enough evidence to suggest that newborr
boys are more common than girls.
2,
Construct and interpret a99o/o confidence interval for the population proportionp.
One proportion z-interual
Assumptions: Same as above
0.5 1 7
!
+/- 0.008:{0.509,0.525)
We are 99o/o conttdent the true proportion of-newborn boys is b/w 5A.9o/o and
52.5%