These are my work from taking 'Statistics' class at Hudson County Community College. (Mar 2015-May 2015)
- Book : Understanding Basic Statistics
- ISBN-13:9780495831488, ISBN-10:0495831484
- For students, please use these as resources, do not plagiarize.
- If you have a question, please leave a comment.
- Graphs and Formulas may not shown.
Solutions part 1.
http://jieunkimresume.blogspot.com/2015/05/understanding-basic-statistics.html
Solutions part 2.
http://jieunkimresume.blogspot.com/2015/05/understanding-basic-statistics_21.html
(Section9.1) #5 (page
364)
Jieun Kim 6-2
(a) 
:µ = 60kg
:µ = 60kg
(b)
:µ < 60kg
:µ < 60kg
(c):µ > 60kg
:µ > 60kg
(d):µ ≠60kg
:µ ≠60kg
(e)
In part (b), 
:µ < 60kg correspond to
the P-value area is on the left. Because the probability of getting a test
statistic lower than 
.
:µ < 60kg correspond to
the P-value area is on the left. Because the probability of getting a test
statistic lower than .
In part (c), :µ > 60kg correspond to
the P-value area is on the right. Because the probability of getting a test
statistic higher than .
:µ > 60kg correspond to
the P-value area is on the right. Because the probability of getting a test
statistic higher than .
In part (d), :µ ≠60 correspond to the
P-value area is on both sides of the mean. Because the probability of getting a
test statistic either lower than -
or higher than 
:µ ≠60 correspond to the
P-value area is on both sides of the mean. Because the probability of getting a
test statistic either lower than - or higher than
#7 (page 364)
(a) :µ = 16.4 feet
:µ = 16.4 feet
(b):µ >16.4 feet
:µ >16.4 feet
(c):µ <16.4 feet
:µ <16.4 feet
(d):µ ≠16.4 feet
:µ ≠16.4 feet
(e)
In part (b), :µ >16.4 correspond to
the P-value area is on the right. Because the probability of getting a test
statistic higher than .
:µ >16.4 correspond to
the P-value area is on the right. Because the probability of getting a test
statistic higher than .
In part (c),
:µ <16.4 
correspond to the P-value
area is on the left. Because the probability of getting a test statistic lower
than .
:µ <16.4 correspond to the P-value
area is on the left. Because the probability of getting a test statistic lower
than .
In part (d), :µ ≠16.4 correspond to the
P-value area is on both sides of the mean. Because the probability of getting a
test statistic either lower than - or higher than
:µ ≠16.4 correspond to the
P-value area is on both sides of the mean. Because the probability of getting a
test statistic either lower than - or higher than
#9 (page 365)
(a)
The level of significance 
=0.01
=0.01
The null hypothesis :µ = 4.7
:µ = 4.7
The alternate hypothesis :µ >4.7
:µ >4.7
I will use the right-tailed because :µ >4.7 states that the
parameter is greater than the value claimed in .
:µ >4.7 states that the
parameter is greater than the value claimed in .
(b)
Normal distribution.
Since the x distribution is normal and σ is known, I will
use the normal distribution with z.
=5.38
z=
= 
=
=
=0.89≈0.9
= ===0.89≈0.9
The value of the sample test statistic is 0.9.
(c)
P(z>0.9)=1-0.8159=0.1841
(d) P-value of 0.1841 > 0.01 for
It fails to reject 
(e) The sample data are not significant at the 
level. At this point, there
is not enough evidence to reject claim that average yield for Australian bank
stocks equals average yield for all stocks.
level. At this point, there
is not enough evidence to reject claim that average yield for Australian bank
stocks equals average yield for all stocks.
(Section 9.2) #7 (page 378)
(a)
The level of significance =0.01
=0.01
The null hypothesis :µ = 16.4feet
:µ = 16.4feet
The alternate hypothesis :µ >16.4feet
:µ >16.4feet
(b)
Standard normal distribution. Because µ and σ are given.
z==
=
=
≈1.54
===≈1.54
(c) P(z>1.54)=1-0.9382=0.0618
(d) P-value 0.0618 > 0.01 (
It fails to reject
(e) The sample data are not significant at the level. At this point, there
is not enough evidence to say that the average storm level is increasing.
level. At this point, there
is not enough evidence to say that the average storm level is increasing.
#9 (page 378)
(a)
The level of significance =0.05
=0.05
The null hypothesis :µ = 41.7
:µ = 41.7
The alternate hypothesis :µ ≠ 41.7
:µ ≠ 41.7
(b)
I will use standard normal distribution.
z==
=
=
≈-1.99
===≈-1.99
(c)
P-value=2P(z<-1.99)=2(0.0233)=0.0466
(d) (P-value=) 0.0466
≤ 0.05 (=)
)
It rejects
(e) At the 5% significant level, there is sufficient
evidence to say that the average number of e-mails is different with the new
priority system. So, the new priority mail system has an effect on the mails.
Jieun Kim 7-1
#4 (page 130)
(a) near -1
Two variables are positively correlated.
(b) near 0
There is no linear relation between two variables.
(c) near 1
Two variables are negatively correlated.
#7 (page 131)
(a) I don't think increasing consumption of diet soda drinks
causes traffic accidents. If the cause is a consumption of beer, it could
causes traffic accidents. But, the consumption of diet soda drinks normally
doesn't affect traffic accidents.
(b) Increasing population of young generation.
Because young people consider their diet, so, it will
increase consumption of diet soda drinks. It also increase traffic accidents,
because young people who just got a driver license don't have enough driving
experience.
#9 (page 131)
(a) No. Because there are no negative correlation between
annual income and time record to run 1 mile.
(b) Inflation is a lurking variables for increasing annual
income.
Getting a training for running is a lurking variables for
time record to run 1 mile.
# 11 (page 131)
(a)
(b) Strong Because 0.7<r<0.99, Positive Because
0<r<1.
(c) r= ====≈0.972
====≈0.972
As x increases, y should increases. Because r≈0.972. If r is
between 0 and 1(0<r<1). The x and y values have a positive correlation.
#14 (page 132)
(a)
(b) This correlation is moderate. Because r=0.51 is 0.4<r<0.6.
This correlation is positive. Because r is 0<r<1.
(c)
r===
==
==
=
increase
Because this correlation is positive correlation.
Jieun Kim 7-2
Read all of the highlighted information on pp. 135-143 as
well as complete the suggested Study Guide exercises #3 (page 141-142), and #4
(page 143).
#7 (page 145)
(a)
(b) =16+33+50+28+50+25=202
=16+33+50+28+50+25=202=2+3+6+5+9+3=28=256+1089+2500+784+2500+625=7754=4+9+36+25+81+9=164=32+99+300+140+450+75=1096
r====0.86
===0.86
(c)
=202/6=33.67
a=4.67-0.161x33.67=4.67-5.421=-0.751
b=0.161
0.161=-0.751+0.161x
(d)
(e)
74% of variation can be explained and 26% of variation
cannot be explained.
(f)
=-0.751+0.161x=-0.751+0.161x40=-0.751+6.44=5.689≈5.69(in
hundreds)
#9 (page 145)
(b)
=27+44+32+47+23+40+34+52=299=30+19+24+13+29+17+21+14=167=729+1936+1024+2209+529+1600+1156+2704=11887=900+361+576+169+841+289+441+196=3773=810+836+768+611+667+680+714+728=5814
r=
(c)
=299/8=37.38167/8=20.88
a=20.88-(-0.60)x37.38=20.88-(-22.43)=43.3
b=
=43.4-0.60x
(d)
(e)
≈0.895
89.5% of variation in y can be explained by corresponding
variation in x and the least-squares line.
10.5% cannot be explained.
(f) =43.4-0.60x=43.4-0.60x38=43.4-22.8=20.6mpg
=43.4-0.60x=43.4-0.60x38=43.4-22.8=20.6mpg
#12 (page 146)
(a)
(b)
=37+47+57+67+77+87=372=5+8+10+16+30+43=112=1369+2209+3249+4489+5929+7569=24814=25+64+100+256+900+1849=3194=185+376+570+1072+2310+3741=8254
r=
(c)
=372/6=62112/6=18.67
a=18.67-0.75x62=18.67-46.5=-27.83
b=
=-27.83+0.75x
(d)
(e)
≈0.889
88.9% of variation in y can be explained by corresponding
variation in x and the least-squares line.
11.1% cannot be explained.
(f) =-27.83+0.75x=-27.83+0.75x70=-27.83+52.5=24.67
=-27.83+0.75x=-27.83+0.75x70=-27.83+52.5=24.67
24.67%
Solutions part 1.
http://jieunkimresume.blogspot.com/2015/05/understanding-basic-statistics.html
Solutions part 2.
http://jieunkimresume.blogspot.com/2015/05/understanding-basic-statistics_21.html
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