Download Solution for Homework 4

Homework #4
Statistics 13 5.32. a.
PC-to < I < (0) = .95 where df= 16 Because of symmetry, the statement can be written P(O < I < to);:: .475 where df= 16 ~
P(/?! 10) = .5 - .475 "" .025
10=2.120
b. P(I ~ -10 or I?! 10) = .05 where df = 16 ~
~
2P(1 ?! 10) == .05 P(t?!
(0)
= .025 where df= 16 10=2.120
c. P(I
~ 10 )
= .05 where df = 16 Because of symmetry, the statement can be written P{/?! -10) = .05 where df= 16 10 = -J .746 d. P(t S -/0 or I'?! 10) =:..10 where df = 12 ~
2P{1 ?! 10) = . J0 ~
P(/?! 10) = .05 where df::::: 12 10 = 1.782
e. P(I ~ -to or t?! 10) = .01 where df= 8 ~
2P(t?! to) = .01 ~
P(I ~ 10) = .005 where df =: 8 10 = 3.355
5.40,
a. The point estimate for the average annual rainfall amount at ant sites in the Dry Steppe
region of Central Asia is
=183.4 milliliters.
b. For confidence coefficient .90, a=.10 and al2 = .10/2
A, with df= n - I = 5 - 1 =4,1.0' = 2.132.
c. The 90% confidence interval is:
x
x±tos
J;; ~
183.4±2.l32 20~70
= .05.
From Table IV, Appendix
~ 183.4±19.686~ (163.714,
203.086)
d. We are 90% confident that the average annual rainfall amount at ant sites in the Dry Steppe region of Central Asia is between 1,63.714 and 203.086 milliliters. e. Using MIN ITAB, the 90% confidence interval is: One--Sample T: DS Rain Variable
OS Rain
N
5
Mean
183.400
StOev
20.647
SE Mean
9.234
90% CI 1163.715, 203.085) The 90% confidence interval is (163.715,203.085). This is very similar to the confidence interval calculated in part c.
f. The point estimate for the average annual rainfall amount at ant sites in the Gobi Desert region of Central Asia is x =110.0 milliliters. For confidence coefficient .90, a=.10 and a/2 '" .1012
A, with df= n - I ;: 6 - 1 ;: 5, t.n~;: 2.015.
= .05. From Table IV, Appendix
The 90% confidence interval is:
-
s
X±/o'Tn~11
00
. ±2.015 15.975
J6 ~JtO.O± I 3. 141 ~(96.859, 123.141)
We are 90% confident that the average annual rainfall amount at ant sites in the Gobi
Desert region of Central Asia is between 96.859 and ) 23.141 milliliters.
Using MlNITAB, the 90% confidence interval is:
One-Sample T: GD Rain
Variable
GD Rain
N
Mean
6 110.000
StDev
15.975
SE Mean
6.522
90% CI
\96.858, 123.142) The 90% confidence interval is (96.858, 123.142). This is very similar to the confidence interval calculated above. 5.46
a.
Some preliminary calculations are:
i:.L..:.=
196 = 17.82
n
JJ
l:X2 _~
Sl
=
n-I
n
1
3,734- 196
=
11
11-1
=24.1636 s:::: v'26.1636 = 4.92
For confidence coefficient .95, a= .05 and ai2:::: .05/2 = .025. From Table IV, with df
t.D2S = 2.228. The 95% confidence interval is:
= n - 1 = I ) - 1 '" 10,
i
±taI2~ ~ 17.82±2.228~ ~ 17.82 ±3.3) ~ (14.51,
..;n ..;1)
21.13)
We are 95% confident that the mean FNE score ofthe population of bulimic female
students is between 14.51 and 21.13.,
Some preliminary calculations are:
b. i ;;;
s
1>2 _(LX)
n-I
n
3,164_
=
1982
14-1
14 = 27.9780
=.J27.9780 ;;; 5.29
For confidence coefficient .95, a= .05 and aI2 =: .0512 = .025. From Table IV, with df
I == 13, 1.02' =2.160. The 95% confidence interval is:
= n - 1 = ]4 -
i ±la1z - : - => 14.14 ± 2.160~ => 14.14± 3.05 => (11.09, 17.19) -..In
-..114 We are 95% confident that the mean FNE score of the population of nonnal female
students is between 11.09 and 17.19.
c. We must assume that the populations ofFNE scores for both the bulimic and nonnal
female students are nonnaJly distributed.
Stem-and-Ieaf displays for the two groups are below:
Stem-and-leaf of Bulimia
Leaf Unit .. 1.0 1
3
4
5
(1 )
5
2
2
1
1
1
1
1
:2
:2
:2
45
0 67 0
5
1
1
1
1
1
2
1
-
9
011
3
5
N
4
6
2
6
.. 11 0
33
Stem-and-leaf of Normal
Leaf Unit - 1. 0 7
7
N
14 a
01 33 5
6
899 2 0
2 3
From both of these plots, the assumption of nonnality is questionable for both groups.
Neither of the plots look mound~shaped.
- - _..
5.52
_----------­
The sample size is large enough ifboth np ~ 15 and nq 2: 15.
a.
np=144(.76)=109,44 and nq=144(.24)=34.56. Since both of these numbers are
greater than or equal to 15, the sample size is sufficiently large to conclude the normal
approximalion is reasonable.
b. For confidence coefficient .90, a=< .10 and al2 = .05. From Table III, Appendix A, z.o, = 1.645. The 90% confidence interval is; pq
l76(.24) =>.6±.OS9
7
p±zo,~pq
-lIlIp±t.645 ~ -=>.76±1.645
. n
n
144
A
A
==(.701, .819)
5.58
c. We must assume the sample was randomly selected from the population of interest. We
must also assume our sample size is sufficiently large to ensure the sampling
distribution is approximately normal. From the results of part a, this appears to be a
reasonable assumption.
a.
The parameter of interest is the true proportion of fillets that are really red snapper.
b. The sample size is large enough if both np 2: IS and mi 2: IS .
np = 22(.23):; S.06 and nq = 22(.77):: 16.94. Since the first number is not greater
than or equal to 15, the sample size is not sufficiently large to conclude the normal
approximation is reasonable.
c. The Wilson adjusted sample proportion is
\
- == x+ 2
p n+ 4
=~=.2.= .269
22+ 4
26
For confidence coefficient .9S, a= .05 and aJ2 = .OS/2 = .02S. From Table Ill,
Appendix A, Z.02S = 1.96. The Wilson adjusted 9S% confidence interval is:
p ± ZIl25 ~p{l- p) => .269 ± 1.96
n+4 d. - 62
).
.269(.731) => .269 ± .170 => (.099, .439)
22+4
We are 95% confident that the true proportion of fish fillets purchased from vendors
across the U.S. that are really red snapper is between .099 and .439.
F'Irst,
we compute p:
, P == -x = - 88 == .175
11
504
A
The sample size is large enough if both
lip ~ 15
and
nq ~ 15 .
and nq =504(.825) =415.8. Since both of the numbers are greater
than or equal to IS, the sample size is sufficiently large to conclude the normal approximation
is reasonable.
lip =504(.175) =88.2
For confidence coefficie"nt .90, a == .10 and aJ2
= 1.645. The 90% confidence interval is:
= .10/2 =
.05. From Table III, Appendix A.
=.05
.
F¥Jq .
6
/pc] .175(.825)
p±=05 -=>p±l. 4\,-=>.175±1.645
II
V 11 504
'?
=>.175±.0~8=>(.14
7'" 3
•. _0)
We are 90% confident that the true proportion of all ice melt ponds in the Canadian Arctic
that have first-year ice is between .147 and .203.
5.74 a.
For confidence coefficient .95, a= .05 and a/2 = .025. From Table m, Appendix A.
%.023 == ) .96.
..
The sample SIze
IS
n=
(za/z)2 pq
(SE)2
(J.96)2(.3X·7) - 224)
.06 2
.
=
You would need to take n = 225 samples.
b.
To compute the needed sample size, use:
n:: (za/zt pq
(SEi
=(1,96)2 (.5X.S) =266.8::;: 267
.06
2
You would need to take n =267 samples.
225
""
5.80
a. The confidence level desired by the researchers is .95.
b. The sampling error desired by the researchers is SE =: .001.
c. For confidence coefficient .95, a:;: .05 and aI2 = .05/2 = .025. From Table III,
Appendix A, Z.02~ == 1.96.
..
(z )40'2 1.964(.005)4
The sample Size IS n = .O~ 1 ==
1
== 96.04 == 97 .
(SE) .001
6.18 a.
AType 1 error is rejecting the null hypothesis when it is true.
In a murder trial, we
would be concluding that the accused is guilty when, in fact, he/she is innocent
A Type n error is accepting the nuJ] hypothesis when it is false. In this case, we would
be concluding that the accused is innocent when, in fact, helshe is guilty.
b. Both errors are bad. However, if an innocent person is found guilty of murder and is
put to death, there is no way to correct the error. On the other hand, if a guilty person is
set free, helshe could murder again.
c. In a jury trial, 0. is assumed to be smaller than fl The only way to convict the accused is
for a unanimous decision of guilt. Thus, the probability of convicting an innocent
person is set to be small.
d. In order to get a unanimous vote to convict, there has to be overwhelming evidence of
guilt The probability of getting a unanimous vote of guilt if the person is really
innocent will be very small.
e. If a jury is predjuced against a guilty verdict, the value of a will decrease. The
probability of convicting an innocent person will be even smaller if the jury if
predjudiced against a guilty verdict.
f. If a jury is predjudiced against a guilty verdict, the value of ,BwiJ) increase. The
probability ofdeclaring a guilty person innocent will be larger if the jury is prejudiced
against a gUilty verdict.
6.28 a.
The rejection region requires a = .01 in the lower taB of the z distribution. From Table
III, Appendix A, Z,OI =2.33. The rejection region is z < -2.33.
' · ·IS Z== x-Po
b . The test StatlstlC
- : : : 19.3-20 =-.40
0'-x
11.9
_
_
J46
c. Since the observed value of the test statistic does not fall in the rejection region
(z =-.40 /:. -2.33), Ho is not rejected. There is insufficient evidence to indicate the
mean number of latex gloves used per week by hospital employees diagnosed with a
latex allergy from exposure to the powder on latex gloves is less than 20 at a = .01.
,