Download 12) The mean top of knee height of a sitting male is 20

12) The mean top of knee height of a sitting male is 20.7in.
H0:  =20.7, H1:   20.7
15) Plain M&M candies have a mean weight that is a least 0.8535g.
H0:  <20.7, H1:   20.7
16) The percentage of workers who got a job through their college is no more than 2%
H0: p>0.02, H1: p  0.02
19) Right-tailed test;a=0.01
Critical value=2.33
20) Left-tailed test;a=0.05
Critical value=-1.645
12) Travel Through the Internal.Among 734 randomly selected internel users,it was
found that 360 of them use the internet for making travel plans(based on data from a
Gallup poll).Use a 0.01 significance level to test the claim that among internet users,less
than 50% use it for making travel plans.Are the results important for travel agents?
H0: p  0.5 H1: p  0.5
z-statistics=(360/734-0.5)/sqrt((360/734)*(374/734)/734)=-0.52
critical value=-z(0.01)=-2.33
since z- statistics>-2.33, we can not reject H0.
Conclusion: among internet users, at least 50% use it for making travel plans.
It is very important to travel agent, with many people using internet to make travel plan,
the travel agent should improve their website to attract more customer.
13) Percentage of E-mail users.Technolongy is dramatically changing the way we
communicate.In 1997,a survey of 880 U.S. households showed that 149 of them use email(based on data from The World Almanac and Book of Facts).Use those sample
results to test the claim that 15% of U.S. households use e-mail. Use a 0.05 significance
level.Is the conclusion vaild today?Why or why not
H0: p  0.15 H1: p  0.15
z-statistics=(149/880-0.15)/sqrt((149/880)*( 731/880)/880)=1.53
critical value=z(0.025)=1.96
since z- statistics<1.96, we can not reject H0.
Conclusion: there is 15% of U.S. households use e-mail in 1997.
I do not think the conclusion is valid today since people who use emails increase by a lot
in the past few years.
13) Perception of time.Randomly selected statistics students of the author participated in
an experiment to test their ability to determine when 1 min(or 60sec) has passed.Forty
students yielded a sample mean of 58.3 sec.Assuming that a=9.5 sec,use a 0.05
significance level to test the claim that the population mean is equal to 60 sec.Based on
the result ,does there appear to be an overall perception of 1 min that is reasonably
accurate?
H0:   60 H1:   60
t-statistics=(58.3-60)/9.5*sqrt(40)=-1.13
critical value=t(0.025,39)=2.02
since -2.02<t- statistics<2.02, we can not reject H0.
Conclusion population mean is equal to 60 sec.
17)Blood Pressure Levels.When 14 different second-year medical students at bellevue
hospital measured the systolic blood pressure of the same person,they obtained the results
listed below(in mmHg).Assuming that the population standard deviation is known to be
10 mmHg,use a significance level to test the claim that the mean blood perssure level that
is too high because it is 140 mmHg or greater.Based on the hypothesis test results,can it
be safely concluded that the person does not have hypertension?
138,130,135,140,120,125,120,130,130,144,143,140,130,150
Sample mean=133.93
H0:   140 H1:   140
z-statistics=(133.93-140)/10*sqrt(14)=-2.27
critical value=z(0.05)=1.645
since z- statistics<1.645, we can not reject H0.
Conclusion the mean blood perssure level that is not too high at 0.05 level. We cannot
safely conclude that the person does not have hypertension.
19)Appendix B Data set: Weights of Quarters. Use the weights of the post-1964 quarters
listed in data set 14 from appendix B.Assuming that quarters are minted to produce with
0.01 significance level to test the claim that the quarters are from a population with mean
of 5.670g.Do the quarters appear to be manufactured according to the U.S. mint
specification that mean is equal to 5.670g.?
No data!
11) Hypothesis Test for Magnet Treatment of Pain.Researchers conducted a study to
determine whether magnets are effective in treating back pain,with results given
below( based on data from Biopolar premanent magnets for the treatment of Chronic
Lower Back Pain:A pilot study" by Collacott,Zimmerman,White and Rindone.Journal of
the American Medical Assocation,Vol.283.No.10). The values repersent measurements
of pain using the visual analog scale.Use a 0.05 significance level to test the claim that
those given a sham treatment(similar to a placebo)have pain reductions that vary more
than the pain reductions for those treated with magnets.
Reduction in pain level after sham treatment:n=20,x=0.44,s=1.4
Reduction in pain level after magnet treatment: n=20,x=0.49,s=0.96
H0:
1

 1 H1: 1  1
2
2
F-statistics=1.4^2/0.96^2=2.13
critical value=F(0.05,19,19)=2.17
since F- statistics<2.17, we can not reject H0.
Conclusion those given a sham treatment do not have pain reductions that vary more than
the pain reductions for those treated with magnets.
15) Weights of Quarters.Weights of quarters are used by vending machines as one way to
detect counterfeit coins.Data set 14 in Appendix B includes weights of pre-1964 silver
quarters and post-1964 quarters.Here are the summary statistics:pre1964:n=40,x=6.19267g,s=0.08700g:post1964:n=40,x=5.63930g,s=0.06194.Use a 0.05
significance level to test the claim that the two population of quarters have the same
standard deviation.If the amounts of variation are different,vending machines might need
more complicated adjustments.Does it appear that such adjustments are necessary?
H0:
1

 1 H1: 1  1
2
2
F-statistics=0.087^2/0.06194^2=1.97
critical value=F(0.05,39,39)=1.71
since F- statistics>1.71, we can reject H0.
Conclusion: standard deviation are different. Machines need more adjustments
Sect 10-2.Testing for linear Correlation. In exercises 17,21,23.construction a
scatterpolt,find the vaule of the linear correlation coefficient r,find the critical value of r
from table a-6 by using a=0.05,and determine whether there is a linear correlation
between the two variables.Save your work because the same data sets will be used in
section 10-3 exercises.
17) Bear Chest Size and Weight.Listed below are the chest sizes(in inches) and
weights(in pounds) of randomly selected bears that were anesthetized and
measured(based on data from Gary Alt and Minitab,Inc).Because it is much more
difficult to weigt a bear than measure its chest size,the presence of a correlation could
lead to a method for estimating weight based on chest size.Is there a linear correlation
between chest size and weight?
Chest Size | 26 45 54 49 35 41 41 49 38 31
Weight | 80 344 416 348 166 220 262 360 204 144
Weight
500
400
300
200
100
0
0
20
40
60
r=0.98
critical value=0.632
since 0.98>0.632, we conclude that there is a linear correlation.
21) Buying a TV Audience.The new york post published the annual salaries(in millions)
and the number of viewers(in millions),with results given below for Oprah
Winfrey,David Letterman,Jay Leno,Kelsey Grammer,Barbara Walters,Dan Rather,James
Gandolfini,and Susan Lucci,repsectively.Is there a correlation between salary and number
of viewers? Which of the listed stars has the lowest cost per viewer?Highest cost per
viewer?
Salary | 100 14 14 35.2 12 7 5 1
Viewers | 7 4.4 5.9 1.6 10.4 9.6 8.9 4.2
Viewers
12
10
8
6
4
2
0
0
50
100
150
r=0.12
critical value=0.707
since 0.12<0.707, we can not reject H0.
Conclusion: there is no linear correlation between salary and number of viewers.
Susan Lucci has the lowest cost per viewer.
Kelsey Grammer has the highest cost per viewer.
23) Temperatures and Marathons.In"The Effects of Temperature on marathon runner's
Performance"By david Martin and John Buoncristiani(Chance,Vol 12,No 4),high
temperatures and times (in minutes)were given for women who won the New York City
marathon in recent years.Results are listed below.Is there a correlation between
temperature and winning time?Does it appear that winning times are affected by
temperatures?
x(temperatures) | 55 61 49 62 70 73 51 57
y(time) | 145.283, 148.717, 148.300, 148.100, 147.617, 146.400 , 144.667, 147.533
y(time)
149
148
147
146
145
144
0
20
40
60
80
r=0.18
critical value=0.707
since 0.18<0.707, we can not reject H0.
Conclusion: there is no linear correlation between temperatures and time.
it appear that winning times are not affected by temperatures
Sect 10-3.Finding the Equation of the Regreesion Line and Making Predictions.Exercises
17,21,23. use the same data sets as exercises 17,21,23 in section 10-2.In each case,find
the regression equation,letting the first variable be the independent (x) variable.Find the
indicated predicted values.Caution: When finding predicted values,be sure to follow the
prediction procedure described in the section.
17) Bear Chest Size and Weight.Find the best predicted weight ( in pounds) of a bear
with a chest size of 50 in.
Chest size| 26 45 54 49 35 41 41 49 38 31
Weight | 80 344 416 348 166 220 262 360 204 144
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.98
R Square
0.97
Adjusted R Square
0.96
Standard Error
21.18
Observations
10.00
ANOVA
df
Regression
Residual
Total
Intercept
Chest size
1.00
8.00
9.00
SS
MS
104666.6853 104666.7
3587.714746 448.4643
108254.4
Coefficients Standard Error
t Stat
-251.95
33.81405547 -7.45098
12.38
0.810373821 15.27708
F
Significance F
233.3890906
3.34339E-07
P-value
7.25773E-05
3.34339E-07
Lower 95%
Upper 95% Lower 95.0% Upper 95.0%
-329.923221 -173.9725177 -329.923221 -173.9725177
10.51141812 14.24886889 10.51141812 14.24886889
Y=12.38x-251.95
When x=50, y=367.05
21) Buying a Tv audience.Find the best predicted number of viewers for a television star
with a salary of $2 million.(In the table below ,the salaries are in millions of dollars and
the numbers of viewers are in millions.)
Salary | 100 14 14 35.2 12 7 5 1
Viewers | 7 4.4 5.9 1.6 10.4 9.6 8.9 4.2
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.118259475
R Square
0.013985303
Adjusted R Square
-0.150350479
Standard Error
3.265791221
Observations
8
ANOVA
df
Regression
Residual
Total
Intercept
Salary
1
6
7
SS
0.907646193
63.99235381
64.9
MS
F
Significance F
0.907646193 0.085101998
0.780322179
10.6653923
Coefficients Standard Error
t Stat
P-value
6.760141042
1.458895818 4.633738035 0.003563735
-0.011058068
0.037906123 -0.291722467 0.780322179
Lower 95%
Upper 95% Lower 95.0% Upper 95.0%
3.190351582 10.3299305 3.190351582
10.3299305
-0.103811008 0.081694873 -0.103811008 0.081694873
Y=-0.011x+6.76
When x=2, y=6.74
23) Temeperatures and Marathons.Find the best predicted winning time for the 1990
marathons when the temeperatures was 73 drgrees.How does that predicted winning time
compare to the actual winnning time of 150.750min?
x(temeperatures)| 55 61 49 62 70 73 51 57
y(time) |145.283,148.717,148.300,148.100,147.617,146.400,144.667,147.533
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.183127465
R Square
0.033535669
Adjusted R Square
-0.12754172
Standard Error
1.565565471
Observations
8
ANOVA
df
Regression
Residual
Total
Intercept
x(temperatures)
1
6
7
SS
MS
0.510287415 0.510287415
14.70597146 2.450995243
15.21625888
Coefficients Standard Error
t Stat
145.1862046
4.180966256 34.72551456
0.031647203
0.069358412 0.456285003
F
Significance F
0.208196004
0.664235397
P-value
3.79973E-08
0.664235397
Lower 95%
Upper 95% Lower 95.0% Upper 95.0%
134.9557488 155.4166605 134.9557488 155.4166605
-0.138066716 0.201361123 -0.138066716 0.201361123
Y=0.032x+145.19
When x=73, y=147.53
When y=150.75, x=173
11) Deaths from Car Crashes.Randomly selected deaths from car crashes were obtained
and the results are included in the table below(based on data from the insurance institute
for highway safety).Use a 0.05 significance level to test the claim that car crash fatalities
occur with equal frequency on the different days of the week.How might the results be
explained?Why does there appear to be an exeeptionally large number of car crash
fatalities on saturday?
Day |Sun mon tues wed thurs fri sat
Number of fatalities|132 98 95 98 105 133 158
Chi-square statistics=(132-117)^2/117+(98-117)^2/117+(95-117)^2/117+(98117)^2/117+(105-117)^2/117+(133-117)^2/117+(158-117)^2/117=30
Critical value=chi-square(0.05,6)=12.6
Since Chi-square statistics> Critical value, we reject the conjecture that car crash
fatalities occur with equal frequency on the different days of the week. We see that there
are many crash on Friday, Saturday and Sunday because many people drive on these days.
21) M&M Candies.Mars.Inc,claims that its M&M plain candies are distributed with the
following color percentages: 16% green ,20%orange ,14% yellow , 24% blue , 13% red ,
and 13% brown.Refer to data set 13 in appendix B and use the sample data to test the
claim that the color distribution is as claimed by Mars.Inc.Use a 0.05 significance level.
No Data.
11) Accuracy of Polygraph Tests.The data in the accompanying table summarize results
from tests of the accuracy of polygraphs(based on data from the Office of Technology
Assessment).Use a 0.05 signficance level to test the claim that whether the subject lies is
independent of the polygraph indication.What do the results suggest about the
effectiveness of polygraphs?
Polygraph indicated truth Polygraph indicated lie
Subject actually told the truth | 65 15
Subject actually told a lie | 3 17
Actual
65
3
68
Expected
54.4
13.6
15
17
32
80
20
100
25.6
6.4
Chi-square statistics=(65-54.4)^2/54.4+(15-25.6)^2/25.6+(3-13.6)^2/13.6+(176.4)^2/6.4=32.27
Chi-square critical value=chi-square(0.05,1)=3.84
Since Chi-square statistics>critical value, we reject subject lies is independent. Polygraph
is good to detect subject who actually tell a lie, but not good to detect subject who
actually tell a truth.
17) Occupational Hazards. Use the data in the table to test the claim that occupational is
independent of whether the casue of death was homcide .The table is based on data from
the U.S. department of labor,bureau of labor statistics.Does any particular occupation
appear to be most prone to homicides?If so ,which one?
Police Cashiers Taxi Drivers Guards
Homicide | 82 107 70 59
not homicide | 92 9 29 42
Actual
82
92
174
Expected
112.92
61.08
107
9
116
70
29
99
59
42
101
75.28
40.72
64.25
34.75
65.55
35.45
318
172
490
Chi-square statistics=65.52
Chi-square critical value=chi-square(0.05,3)=7.82
Since Chi-square statistics>critical value, we reject occupational is independent. Police
and Cashiers appear to be most prone to homicides.
Sect 12-2 In exercise 11.use the listed sample data from car crash experiments conducted
by the national transportation safety administration.New cars were purchased and crashed
into a fixed barrier at 35mi/h,and the listed measurements were recorded for the dummy
in the driver's seat.The subcompact cars are the Ford Escord , Honda Civic , Hyundai
Accent , Nissan Sentra,and Saturn SL4.The compact cars are Chevrolet Cavalier , Dodge
Neon , Mazad 626 DX ,Pontiac Sunfire , and Subaru Legacy.The midsize cars are
Chevrolet Camaro , Dodge Intrepid ,Ford Mustang , Honda Accord , and Volvo S70. The
full size cars are Audi A8, Cadillac Deville, Ford Crown Victoria, Oldsmoblie Aurora,
and Pontiac Bonneville.
11) Head injury in a car crash.The head injury data(in hic) are given below.Use a 0.05
significance level to test the null hypothesis that the different weight categories have the
same mean.Do the data suggest that larger cars are safer?
Subcompact: 681 428 917 898 420
Compact: 643 655 442 514 525
Midsize: 469 727 525 454 259
Full size: 384 656 602 687 360
Anova: Single Factor
SUMMARY
Groups
Subcompact:
Compact:
Midsize:
Full Size
Count
5
5
5
5
ANOVA
Source of Variation
Between Groups
Within Groups
SS
88425
475323.2
Total
563748.2
Sum Average
3344
668.8
2779
555.8
2434
486.8
2689
537.8
Variance
58542.7
8272.7
28110.2
23905.2
df
MS
F
P-value
F crit
3
29475 0.992167014 0.421569916 3.238871522
16 29707.7
19
Sicne F-statistics<critical value, we cannot reject the conjecture that the different weight
categories have the same mean. The data do not suggest that larger cars are safer.