Download Page 1 Statistics 13 Solution for Homework #4 5.30 a. P(

Statistics 13
5.30
a.
Solution for Homework #4
P(-/U < t < to) = .95 where df = 16
Because of symmetry, the statement can be written
P(0 < r <to) = -475 where d f = 16
2
P(t 2 to) = .5 - .475 = .025
to = 2.120
b. P(/ r -to or t 2 to) = -05 where df = 16
2 2P(t 2
to) = .05
P(t 2 to) = .025 where df = 16
2
to = 2.120
c.
P(t 2 to) = .05 where df = 16
Because o f symmetry, the statement can be written
P(/ 2 -to)
= .05
where df = 16
to = -1.746
d.
P(t5-/,,ort2/0)=.IO wheredf=12
2 2 P ( t 2 t o ) = .I0
2 P(t
2 to) = .05 where df = 12
tu = 1.782
e.
P(t 5 -to or t 2 to) = .01 where df = 8
2P(t2to)= .O1
2 P(t
2 to) = .005 where d f = 8
to = 3.355
7-34
Some preliminary calculations:
x x 2
s2 =
icxP
--
-
n-l
"
6.442
7.1 804 - -=.0536
=
6--I
a.
For confidence coefficient .95, a=.05 and d 2 ' = .05/2 = ,025. From Table VI,
Appendix A, with df = n - 1 = 6 - 1 = 5, t.02~
= 2.571. The confidence interval is:
We are 95% confident that the true average decaj rate of tine particles produced fro171
oven cooking or toasting is between 3 3 0 and 1.316.
540
b.
The phrase "95% confident" means that in repeated sampling, 95% of all confidence
intervals constructed will contain the true mean.
c.
In order for the inference above to be valid, the'distribution of decay rates must be
normally distributed.
a.
Sonie preliminary calculations are:
For confidence coefficient .95, a=.05 and d 2 = .05/2 = .025. From Table V1, with df
= n - 1 = 1 1 - 1 = 10, t.025 = 2.228. The 95% confidence interval is:
We are 95% confident that the rnean FNE score of the population of bulimic female
students is between 14.5 1 and 21.13.
b.
Some preliminary calculations are:
For confidence coefficient .95, a=.05 and d 2 = .05/2 = .025. From Table VI, with df
n - 1 = 14 - 1 = 13, t.025
= 2.160. The 95% confidence intewal is:
=
We are 95% confident that the mean FNE score of the population of normal female
students is between 1 1.09 and 17.19.
c.
We must assume that the populations of FNE scores for both the bulimic and normal
female students are normally distributed.
Stem-and-leaf displays for the two groups are below:
Stem-and-leaf of Bulimia
Leaf Unit = 1.0
Stem-and-leaf of Normal
Leaf Unit = 1.0
N
N
=
=
11
14
From both of these plots, the assumptioh of normality is questionable for both groups.
Neither of the plots look mound-shaped.
5.46
a.
The sample size is large enough if the intewal j
+ 30;
does not include 0 or 1.
Since the inter\'al lies within the interval (0, I), the normal appl-oxi~l~ation
will be
adequate.
b.
For confidence coefficient .90, a= .I0 and d2 = .05. From Table IV, Appendix A,
1.645. The 90% conf dence interval is:
z.0~
=
5.52
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
reasollable assumption.
a.
The population of interest is all shoppers in Muncie, Indiana.
b.
The characteristic of interest is the proportion of shoppers who'think "Made in the
USA" means 100% of US labor and materials.
c.
The point estimate for the proportion of slioppers who think "Made in the USA" means
x
64
100% of US labor and materials is f, = - = -= .604 .
n 106
'To see if the sample size is sufficiently large:
Since this interval is wholly contained in the interval (0, I), we may conclude that the
normal approximation is reasonable.
For confidelice coefficient .90, a=.10 and d 2 = ,1012 = .05. From Table IV, Appendix
A, zo5= 1.645. The 90% confidence interval is:
z58
c.
We are 90% confident that the proportion of shoppers who think "Made in the USA"
means 100% of US labor and materials is between .526 and .682.
d.
"90% co~~fidence"
means that in repeated samples of size 106,90% of all confidence
intervals formed for the proportion of shoppers who think "Made in the USA" means
100% of US labor and materials will contain the true population proportion.
a.
Some preliminary calculations
@ += x l n = 118511669= .7lO
We must check to see if the sample size is sufficiently large:
Since the interval is wholly contained in the intel-val (0, I), we may assunle that the
normal appro xi ma ti or^ is reasoilable.
For confidence coefficient .90, a=. I0 and ui2
A, zos = 1.645. The confidence interval is:
= .I012 = . 0 5
From Table IV, Appendix
!
We are 90% confident that the true proportion of retail stores and supermarkets with
electronic scanners that pass the FTC price-check test is between ,692 and ,728.
770
b.
Since the 90% confidence interval does not contain .45, we are 90% confident thal the
proportion of stores that pass inspection now exceeds .45.
a.
The confidence level desired by the researchers is .95.
b.
The sampling error desired by the researchers is SE = .001.
c.
For confidence coefficiellt .95, a='05 and ui2 = .05/2 = .025. Froin Table IV,
Appendix A, z 025 = 1.96.
The sample size is
576
11
=
a.
The confidence interval might lead lo an erroneous inference because it is so wide. The
width of the interval is ,459, which is very wide for making inferences.
b.
For confidence coefficient .95, a=0 5 and ui2 = .05/2 = .025. From Table 1V,
Appendix A, 2 0 2 5 = 1.96. From the previous study, we will use j = 10118 = .5555 to
estimatep.
n =z A , -- 1.96*(.5555)(.4445) = 592.85
(SE)'
.04'
8.16
-
(zU2,)'a2- 1 .962(.005)2
= 96.04 97 .
(SE)'
.0012
a.
-
593
A Type I 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, helshe is innocent.
A Type 11 error is accepting the null hypothesis when it is false. In this case, we would
be concluding that the accused is innocent when, in fact, helshe is guiIty.
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 n~urderagain.
c.
In a jury trial, a is assunled to be smaller than p. 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 snlall.
d.
In order to get a unanimous vote to convict, there has to be overwhelmi~lgevidence of
guilt. The probability of getting a unanimous vole of guilt if the person is really
innocent will be very small.
e.
If a jury is predjuced agairlst 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 P will increase. The
probability of declaring a guilty person innocent will be larger if the jury is prejudiced
against a guilty verdict.
6.26
a.
b.
The rejection regipn requires a- .O1 in the lower tail of the z distribution. From Table
IV, Appendix A, z.0, = 2.33. The rejection region is z < -2.33.
x-P"
The test statistic is z = -- D
c.
19.3- 20
= -.40
11.9
Since the observed value of the test statistic does not fall in the rejection region
(z = -.40 4: -2.33), Hn is not rejected. There is insufficient evidence to indicate the
mean number of latex gloves used per week by hospital e~nployeesdiagnosed with a
latex allergy from exposure to the powder on latex gloves is less than 20 at a= .01.