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Submitted by melithdairy on Sun, 2012-03-18 23:27
due on Thu, 2012-03-22 23:24
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Submitted by shahimermaid on Fri, 2012-04-06 00:03
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ch-19, online questions

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Submitted by shahimermaid on Sun, 2012-04-01 04:55
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Ch 7 & 8

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xxxxxxxx xxxxxxxxxxxx were xxxx of Hemoglobin4 xxxxxx and packed cell xxxxxx xxx 21 patients aged xx to xx xxxxx xxxx xxx xx CE6. x

xx Make a scatterplot xx xxx xxxxx xxxxxxxxxxx xxx as the xxxxxxxxxxx xxxxxxxx and Hb xx xxx xxxxxxxxx xxxxxxxxx

Scatter plot is xxx best way xx start xxxxxxxxx then xxxxxxxxxxxx xxxxxxx 2 quantitative xxxxxxxxxx Direction xxxxx xxxx an xxxxxx plot means xxxx as PCV increases, Hb xxxx xxxxxxxxxx This xx a sign xx xxxxxxxx xxxxxxxx xxxxxxx xxx and xxx xxx form of relationship looks like a linear xxxxxxxxx xxx xxxx xxxxxx is xxx strong the xxxxxxxxxxxx xxxxxx From xxxx diagram , xxx xxxxxxxxxxxx xxxxx xxx too xxxxxx but xxxxxxxxxx xx is xxx weak.

b. Calculate xxx xxxxxxxxxxx xxxxxxxxxxx between PCV xxx Hb.

Simple linear regression xxxxxxxx Dependent Variable: xxxxxxxxxxxxxx Variable: PCV

xx = xxxxxxxxx x xxxxxxxxxx PCV

xxxxxx size: xx

r (correlation xxxxxxxxxxxx x xxxxxx

xxx value of x ranges xxxx 0 xxxx xxxx shows a strong xxxxxxxx

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Submitted by shahimermaid on Thu, 2012-03-29 13:14
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ch -19, 3 questions

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In 1998, a san xxxxx reproductive xxxxxx xxxxxxxx 49 xxxx birth to xxx xxxxx xxxxx the age of 40 who xxx xxxxxxxxxx xxxx unable xx conceive.

Find 90% xxxxxxxxxx interval xxx xxx success xxxx at this clinic

Interpret xxxx interval in xxxx context

xxxxxxx what 90% xxxxxxxxxx xxxxx

xx these xxxx refute xxx xxxxxx’x claim of x 25% xxxxxxx rate? Explain.

x

xx have 

        p x 49/207 = xxxx        zc = 1.645        xxx     n x 207 

We xxxxxxxxxx

xxxx +/- xxxxx sq xxx 0.24 x 0.76)/207 =

        xxxx +/-1.645 x xxxxx xxxx x 0.049 xxx xxxx -0.049

  xxxxx and xxxxx

b) We are 90% xxxxxxxxxx based on xxxx xxxxxxx xxxx the proportion xx xxx women xxx xxxx xxxxxxxxxx unable to xxxxxxxx will xxxxxxx a xxxxxxx xxxx at xxxx xxxxxx between 19.1% xxx xxxxxx

c) xx we were to select repeated xxxxxx like this xx would xxxxxx xxxxx xxx of the confidence xxxxxxxxx we created xx xxxxxxx xxx true xxxxxxxxxxx

d) No. xxxxx on x 90% xxxxxxxxxx interval, xxx falls within xxx interval.

xxxxxx routine

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Submitted by shahimermaid on Tue, 2012-03-27 21:32
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ch-18, 2 questions

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4

xx xxxx

xxxxx

xxxx x 0.04

xx = sq xx (pq/n)=0.007

1) xxxxxx is xxxxxx

xx 10% xxxxxxxxx x number of xxxxxxxxx x xxx x xxxx xxx and xxx of failures =0.96 x xxx = 702

xx model xxx p is xxxxxxx xxxxxxx

xxxxxxxxxx that in 732 xxxxxxx 20 are xxxxxx the gene

Probability = xxxxxx = xxxx

xxx xxx xxxxxxx 18 and xx years, xxxxx cholesterol levels xxx xxxxxxxxx have x xxxx of xxxxx xxx a standard xxxxxxxxx xx 40.7 xxxxxx xx data xxxx xxx National Health xxxxxxxx

Mean = xxxxx

SD= xxxx

Find the percentage of xxxxx aged 23, who xxxx a serum xxxxxxxxxxx xxxxx xx xxxxxxxxxxx xxx above). Is this level unusually high?

x xx 210-178.1)/40.7 x 0.784

xxx = 0.784) =0.7823

x >x xxxxx will xx x – 0.7823 x xxxxxx

The xxxxxxxxxx xx males above 210mg/100ml xxxx be 21.77%

In a xxxxx of 25 xxxxx xxxx 23, xxxx is the probability xxxx the mean xxxxx cholesterol xxxxx xxxxxxxxxxxxx xxx above). xx this xxxxx unusually high?

xxxxxxxxxxx that mean serum xxxxxxxxxxx level xxxxxxxxxxxxx or xxxxx = 21.65%

Discuss the xxxxxxxxxxx xxx xxxxxxxxxx xxxx support xxxx xxxxxx and draw a xxxxxxxx xxxxxxx diagram.

1) sample is random

2) 10% xxxxxxxxx x number of successes and xxx of failures xxx more than 10


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Submitted by shahimermaid on Fri, 2012-03-23 12:35
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ch 5 & 6

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xx xxx book, there xx xxx with xxxxxxxxx multimedia xxxxxxxxxx xxx x = xxxxx xxx find the xxxx

x have mentioned this line in xx xx xxx can xxxx xxxx xx xxxx xx xxxx book and xxxx use it. x have xxxx the table and find xxx xxxxx


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Submitted by shahimermaid on Wed, 2012-03-21 19:17
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ch-17 2 questions

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xx xxxx x Probability xxxx the citizen xxxxx xxxxx dentist twice x year=p= xxxxxx

Q= xxxxxxx xxxxx n= 10

xxxxxxx x successes in x trial xx 10 xxxxxx x successes and x xxxxxxx

xxxx x xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx . xxxxxxxxxxx

Probability xxxx exactly six see their dentist xxxxx x year x xxx x (0.60)6 x (0.40)4

= 210 x xxxxxxxx x 0.0256 x 0.25 x xxx

At xxxxx xxx means P(6)+P(7)+P(8)+P(9)+P(10)

xxxxx statcrunch, P=>6, 0.633

xxx At xxxx 6, xxxxx xxxxxxxxxx P<=6, 0.62

xxxxx np = 10 x 0.6= 6

SD x sq rt xxxxx x xx rt xxx x .6 x xxxx 1.55

xx HE x

N= xxxxx P=(0.6) , xxxxxxx

Mean = xxx 1000 x xxxx 600

xx x xx rt(npq)= sq xx (1000 x xxx x xxxxx xxxxx

x x xxxxxx xxxxx is a close xxxxxx xxxxxxxxxxxxx xxxx xxx x xxxxx number of trial

x xxxxxxxx model is approximately normal, xx we expect xx xxxxx xx successes xxx xx xxxxxxxx

np≥10 and xx ≥xx

xx this case np= 1000x xxxx 600 which xx ≥10 and xxx xxx ≥10, xx xx xxx say xxxxxx model approximates xxxxx

68 – xx – 99.7 rule

The

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Submitted by shahimermaid on Wed, 2012-03-21 12:53
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ch-4 2 questions

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xxxxxxxxx xxxxx x snapshot of the xxxxxxxxx xx xxx variable xxxxxxxxx at various value but xxx xxxx and xxxx xxxxxxx is xxxxxx xxxxxx as xx depicts xxx xxxxxxxxxx xxxxxxx

Variable: xxxx xxxxxxx xxxxx xx x digit(s) to the xxxxx of xxx xxxxxx 2 : xxx x x x 568 4 : x : 558889 5 : 123 5 : xxxx : xxxx : 6 7 x xx

Stem xxx xxxx display organizes data into groups called xxxxx and the xxxxxx xxxxxx xxxx xxxxx xxxxxxxx out xx thr right on each xxxx xxxx xxxxx xxx the data xx distributed and where concentration of data occurs. In xxxx xxxxx it xx xx the xxxx of 40’xx

xx xxxxx xxxxx

median= xx

c) xxxxxxxx

xxxxxxxxx 138.58

SD= xxxxx

Summary xxxxxxxxxxx

xx

Column

Mean

xxxxxxxx

xxxx Dev.

Median

Range

var1

xxxxx

138.57631

11.771844

50

45

xxx xxxxxxx xxxxx is xxxx xxxxxxx xxxxxx xxx xx the line xx 40 xxx mean, median are xxxxx xxx xxxxxx xxxxxxx

Q4B

b)The space between histogram are xxxxxx gap xx the xxxxx indicating x region xxxxx xxxxx are xx xxxxxxx

c) xxxx number

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Submitted by shahimermaid on Tue, 2012-03-20 16:26
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There are 2 question of Ch-2 and 2 of Ch-3

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Ch-2 x 4, xxxxxx xx a quantitative variable

xx xx 5W’s – xxxx xxxxx When, xxxx xxx Why

Who- xxxxxxx researchers at a large xxxxxxxxx

What- xxxxxxxx of xxxxxxxx to xxx xxxxxxxxx

xxxx – from xxxx xx 2005

xxxxx xxxxxx patients undergoing xxxxxxxxxxxx

Why- - Impact xx new type xx xxxxxxxxxxxx on four xxxxx of blood xxxxxx

xxx data xx categorical, xx categories on xxx basis xx age, sex, xxxx xx cancer, stage xxx xxxxxxx person responded to xxx treatment.

xxxxxxx 3: Displaying xxx Describing xxxxxxxxxxx Data

xxxx

xxxxxx

xxxxx xxxxxxx

27.9

xxxxxx

22.8

Circulatory diseases xxx Stroke

xxx

Respiratory Diseases

5.2

xxxxxxxxx

xxx

Other xxxxxx

33.2

xxx chart

Q3 xxx

Age

xxxxxxxxxxxxxxxxx

Blood Pressure

Under xx

xxxxx

0ver xx

xxxxx

Low

xx

37

31

95

xxxxxx

48

xx

93

xxx

xxxx

23

51

73

147

total

xx

xxx

197

xxx

Percentage of xxxxxxxx over 50 x 197/474*100 x xxxxx

Percentage of xxxxxxxx having xxxx xxxxx xxxxxxxx x 147/474*100 =31.0%

xxxxxxxxxx xx xxxxxxxx xxxx xx xxx with xxxx blood pressure = xxxxxxxxxxxxxxxx

Percentage xx patients xxxx xxxx xxxxx xxxxxxxx xxxx over 50 x 73/147*100=49.7%

Percentage xx xxxxxxxx over 50 xxx High xxxxx xxxxxxxx = xxxxxxxxxxxxxxxx

Yes, xxxxx xx xxxxxxxxxxxx xxxxxxx age xxx xxxxx xxxxxxxx xxxxxxx xxxxx patients are under xxx 37.8% xxxxxxxx are between 30-49 and xxxxx xxxxxxxx are above xxx xx xxx age xxxxxxxxxx blood xxxxxxxx patients also increase.


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Submitted by shahimermaid on Tue, 2012-03-20 11:37
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ch 16, 2 question in attached file

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16.HE.A:

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

X

xxxx

xx xxxxxxxxx

Deviation xxxxxxx

xxxxxxx xxxxxxxxx

variance

xx

x

xxxx

0

xxxxxx

x

xxx

0.45

1

xxxx

xxxx

1-2=-1

1

xxxx

0.5

x

xxxx

0.7

2-2=0

x

x

0

x

0.15

xxxx

3-2=1

x

0.15

0.39

4

xxxx

xxx

4-2=2

4

xxx

xxxx

x

xxxx

xxxx

xxx =3

x

0.45

0.67

Total x xxxxxxxxxxxxxxxxxx

xxxxxxxx value xx a xxxxxx variable xx xxxxx by xxxxxxxxxxx xxx value xx x with the probability xxxx xxx adding them.

Expected number of patients can xx xxxxx xx xxx table in 3rd xxxxxxx xx xxxxxx xxxxxx be xx fraction, we round xxxx

xxxxx in xxxxx

The xx xxxxx xxxxxx expect x patients to be admitted on a xxxxx day.

xxxxxxxx x 1.65, xxx xx xx xxxxxxxx = 1.28

x

xxxxxxxxxxxxxxxxxx

X

xxxx

E(x)

deviation

Squared xxxxxxxxx

variance

0

0.0000

0

xxxxxxxxx -3.9598

15.68

0

x

0.0001

xxxxxx

1-3.9598= -2.9598

xxxx

xxxxxxxx

2

xxxxxx

xxxxxx

xxxxxxxxx xxxxxxx

3.84

0.002304

3

0.0387

0.1161

3-3.9598= xxxxxxx

0.92

0.035604

4

xxxxxx

xxxxxx

xxxxxxxxx xxxxxx

0.0016

xxxxxxxx

xxxxx xxxxxxxxxxxxxxxxxxxxxx

Expected value= xxxxxx

xx x 0.2


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Submitted by shahimermaid on Sun, 2012-05-06 07:24
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final submission

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x have xxx done xx C

22.HE.A:

xxxx xxxxxxx x 95% xxxxxxxxxx xxxxxxxx xxx xxxxxxx the xxxxxxxx xx xxxxxxxx

a) xxxxxxx xxxxx design xxx xxxx to collect xxx xxxx

xxxxx

xxx xxxx xx xxxxxxxxxx x 67 xxxxx p1= 0.19, xxx xxxx

xx xxxxxxxx xxxx xxxxx xxxxxxxxxxx 26 died. p2= 0.29, xxx 0.71

xx xxx xxxxx xx xx xxxxxxxxxxx between xxxxxxxxxx and death

xxx people xxxx depression xxx xxxxx

c) xxxxxxxxxxx are

Independence Assumption: xxxxxx each group, xxx data xxxxxx xx based on xxxxxxx for xxxxxxxxxxx xxxxxxxxxxxx

xxxxxxxxxxxxx xxxxxxxxxx The xxxx xx xxxx group should be xxxxx xxxxxxxxxxxxx xxx at random from x homogeneous xxxxxxxxxx or generated by a xxxxxxxxxx xxxxxxxxxxx xxxxxxxxxxx

xx xxx xxxx xxx xxxxxxx xxxxxxx xxxxxxxxxxxx xxx sample should not xxxxxx xxx of xxx population.

xxxx xxxxxx xxxxxx xx xxx enough that xx xxxxx xx xxxxxxx xxx 10 xxxxxxxx xxx there.

d) SD1= √xxxxxxxxx √0.19 x 0.81/361=0.0261

xxxx √p2.q2/n2= √xxxx x 0.71/89=0.0481

xxxxxxxx error in xxx difference xx two sample

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