Unlock document.
This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
29. a. 75 seconds is 1.25 minutes, so
22
2
(1.96) (4) 39.3380
(1.25)
n==
Use n = 40
22
2
(1.96) (4) 61.4656
1
30. Planning value from previous study:
2000
=
22 22
.025
22
(1.96) (2000) 1536.64
(100)
z
nE
= = =
Use n = 1537 to guarantee the margin of error will not exceed 100.
31. a.
p
= 100/400 = .25
b.
(1 ) .25(.75) .0217
400
pp
n
−==
( )
1
33.
22
.025
22
*(1 *) (1.96) (.35)(.65) 349.59 Use 350
(.05)
z p p
nn
E
−
= = = =
34. Use planning value p = .50
2
2
(1.96) (.50)(.50) 1067.11 Use 1068
(.03)
nn= = =
.05
2000
.88 .0120
or .868 to .892
d. Margin of Error
.025
(1 ) .88(1 .88)
1.96 .0142
2000
pp
zn
−−
==
95% Confidence Interval
b. Margin of error:
.025
(1 ) .24(1 .24)
1.96 .0265
1000
pp
zn
−−
==
c.
.0265p
Confidence Interval: .24 .0265 or .2135 to .2665
d.
530/1000 .53p==
.025
(1 ) .53(1 .53)
1.96 1000
pp
zn
−−
=
=.0309
b.
104/162p=
= .6420
Margin of error =
(1 ) (.642)(.358)
1.96 1.96 .0738
162
pp
n
−==
2
2
1.96 (.50)(1 .50) 1067.11
.03
November
2
2
1.96 (.50)(1 .50) 2401
.02
n−
==
Pre-Election
2
2
1.96 (.50)(1 .50) 9604
.01
n−
==
43. a. Margin of Error =
/2
(1 ) (.90)(.10)
1.96 .0201
857
pp
zn
−==
95% Confidence Interval: .90 .0201 or .8799 to .9201
b. Margin of Error = 1.96
(.67)(.33)
857
= .0315
95% Confidence Interval: .67 .0315 or .6385 to .7015
c. Margin of Error = 1.96
(.56)(.44)
857
= .0332
33.77 4.00 or $29.77 to $37.77
45. a.
x
t.025
( / )s n
df = 63 t.025 = 1.998
252.45 1.998
( . / )7450 64
252.45 18.61 or $233.84 to $271.06
b. Yes. the lower limit for the population mean at Niagara Falls is $233.84 which is greater than
$215.60.
b.
x
± margin of error
Estimate of total expenditures = 92(1873) = 172,316
In dollars, we estimate that $172,316 million dollars are spent annually by Americans of age 50 and
over on restaurants and carryout food.
d. We would expect the median to be less than the mean. The few individuals that spend much more
than the average cause the mean to be larger than the median. This is typical for data of this type.
Margin of error =
.025
(1 ) .40(1 .40)
1.96 .0554
300
pp
zn
−−
==
95% Confidence Interval: .40 .0554 or .3446 to .4554
b. From the sample of 300 Russians, use Excel to find that the sample mean annual consumption
Margin of error =
.025
3929
1.96 445
300
s
tn==
95% Confidence Interval: 2786 445 or 2341 to 3231
c. From the sample we see that 120 Russians smoke. The annual consumption per person for those
who smoke is given by
835800/120 6965x==
. With 365 days in a year, an estimate of the
number of cigarettes smoked per day by those Russians who smoke is 6965/365 = 19. A cigarette
pack holds 20 cigarettes. So, the typical Russian smoker smokes almost a pack a day.
b. Using Excel shows that the sample mean and sample standard deviation are
187x=
and
178.6207s=
178.6207
s
90% Confidence Interval: 187 14.69 or $172.31 to $201.69
c. There were 136 employees who had no prescription medication cost for the year.
136/ 400 .34p==
49. a. Using Excel:
66.93x=
The amount of playing time is approximately 11 minutes. So the time standing around is over 6
times as much. You may find this difference surprising; the authors did.
b. Using Excel: s = 4.4943 minutes
c.
.025 /x t s n
df = 59
.025
t
= 2.001
66.93 2.001 (4.4943/
60
)
210617 107
2 Use
52.
n n= = =
( . ) ( ) .
196 675
100 17503 176
2 2
2 Use
53. a.
pp p
n
−
196 1
.( )
(.47)(.53)
Margin of error =
.025
(1 ) (.93)(.07)
1.96 .0224
500
pp
zn
−==
95% Confidence Interval: .93 .0224 or .9076 to .9524
b.
p
= .21
(1 ) (.21)(.79)
500
pp
−==
95% Confidence Interval: .21 .0469 or .1631 to .2569
c. The margin of error is larger in part (b) for two reasons. With
p
= .21, the estimate of the standard
error is larger than with
.930.p=
And
.005
z
= 2.576 for 99% confidence is larger than
.025
z
= 1.96 for
95% confidence.
p
= 315/750 = .42
Margin of error
(1 ) .42(1 .42)
1.96 1.96 .0353
750
pp
n
−−
= = =
95% Confidence interval: .42 .0353 or .3847 to .4553
57. a.
2
2
(1.96) (.3)(.7) 2016.84 Use 2017
(.02)
nn= = =
b.
p
= 520/2017 = .2578
b.
pp p
−
196 1
1993
.( )
(.3101)(.6899)
1993
.3101 .0203 or .2898 to .3304
c.
2
2
*(1 *)z p p
nE
−
=
2
2
(1.96) (.3101)(.6899) 8218.64 Use 8219
(.01)
zn= = =
No; the sample appears unnecessarily large. The .02 margin of error reported in part (b) should
provide adequate precision.
