exam interpretation

8/4/2019 Exam Interpretation

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Name: N.V. Shashidhar Parimi Stat Exam Date:

04/04/11

Problem 3.1.1 Is the graphical representation calculation about Explaining Variability.

B/w BG Girls and BG Boys..

> library(alr3)

> attach(BGSgirls)

> names(BGSgirls)

"Sex" "WT2" "HT2" "WT9" "HT9" "LG9" "ST9" "WT18" "HT18" "LG18 "ST18" "Soma"

> pairs(~WT2+HT2+WT9+HT9+LG9+ST9+Soma)

WT2

85 95 125 145 20 60 100

10

13

16

85

95

HT2

WT925

35

45

125

145

HT9

LG924

28

32

20

60

100

ST9

10 13 16 25 35 45 24 28 32 3 5 7

3

5

7

Soma

Abouve scatterplot matrix of Graphs show that the clear relation b/w WT2, HT2, WT9, HT9,

LG9 ,ST9. It means if one of the person growth increase then also others ( variable values )

Increas, So can state that the +ve attitude towards each other variable shown.

> pairs(HT2~HT9+Soma)


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HT2

125 135 145

85

90

95

125

135

145

HT9

85 90 95 3 4 5 6 7

3

4

5

6

7

Soma

Can see the above graph of the matrix of sample correlations b/w the height variables of each ..

3.1.3 For finding a multiple regression model for mean function

> lm(Soma~HT2+WT2+HT9+WT9+ST9)

call

lm(formula = Soma ~ HT2 + WT2 + HT9 + WT9 + ST9)

Coefficients:

(Intercept) HT2 WT2 HT9 WT9 ST9

8.8590417 -0.0792535 -0.0409358 -0.0009613 0.1280506 -0.0092629

> summary(lm(Soma~HT2+WT2+HT9+WT9+ST9))

lm(formula = Soma ~ HT2 + WT2 + HT9 + WT9 + ST9)

the residual values are


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Min 1Q Median 3Q Max

-2.03132 -0.34062 0.01917 0.43939 0.97266

For the coefficient we have to estimate the errors values which is Pr(>|t|)

An Intercept 8.8590417 2.3764431 3.728 0.00041

1) HT2 -0.0792535 0.0354034 -2.239 0.028668

2) WT2 -0.0409358 0.0754343 -0.543 0.589244

3) HT9 -0.0009613 0.0260735 -0.037 0.970704

4) WT9 0.1280506 0.0203544 6.291 3.2e-08.

5) ST9 -0.0092629 0.0060130 -1.540 0.128373

The std Error is 0.5791, 64 degrees, the Multiple R is 0.5211, and by the adjustment of value the

R value would be 0.4837 stated ,the statistics of F would be 13.93, on 5& 64, The P value is

3.309e-09

Yes it fit the multiple linear regression model are good.. The born Childers of girls and boys data.

Executed

Problem 5.6.1 .. .. Jevons gold coins The data

library(alr3)

> attach(jevons)

> names(jevons)

[1] "Age" "n" "Weight" "SD" "Min" "Max"

> plot(Weight~Age)


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The above plot show the values has been decreasing from in order .. but stil its in proper order ..

may not accept the decrees values .. But in its in line. it has more morel ethics to decrees even

Now for the to find SD with Age of Ppl lets plot the value.

> plot(SD~Age)

Fr

The above values shows that (plot or graph) increasing values of Age and in SD (Age is

proportionate S.D)


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Conclusion: So, From the above give data s .. Scatter plot of Weight versus Age. And the SD hasbeen found .. it says the total process of plot and graphs explains the data has some regressiondelay in mean while at initial stage when it compared with SD .. Moved in order by reveres.. I meana Incremental order.. it is Good plot of view . as given standard weight of a gold sovereign wassupposed to be 7.9876 g; the minimum legal weight was 7.9379 g. the final SD shows its in equal to

the actual data provided and summated

The problem 6.17

Continuation with above Jevons The data by using Delta method. The program fallows as

The weight and age are taken in account of SD with Coefficient .

The summer state as

> summary(lm(Weight~Age,weights=(n/SD^2)))$coef

Estimate Std. Error t value Pr(>|t|)

(Intercept) 7.99652179 0.0013219826 6048.88577 9.964249e-12

Age -0.02375617 0.0008797498 -27.00333 1.114497e-04

Then the plot value of weight and age again written

> plot(Weight~Age)


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1 2 3 4 5

7.88

7.9

0

7.9

2

7.9

4

7.96

Age

Weight

> abline(m1)

WITH VALUE COMPARESION OF M1 AS 7.93SO ON

> abline(7.9379,0)

THE ABOVE VAUE SHOWS THAT ALLTHY DO NOT TOUCH EACH OTHER AD DISTICT

IN THE LINE IT MEANS FAIL TO THE OVERALL TREND IN THEGRAPH. IT IS GOOD ..

THE APPLYING DELTA METH NOW DIRECTLY

> delta.method(m1,"(7.9379-b0)/b1")

Estimate SE (7.9379-b0)/b1 0.003317994 1.071017e-05

(7.9379-7.996521)/(-0.002375615)

[1] 24.67614


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> 2.467612-1.96*0.04940154

[1] 2.370785

> 2.467612+1.96*0.04940154

[1] 2.564439

I could notb able to understand way he .. determined a SD error of jevous voind dta where .. The

number in inverse form the plot showed that the data has error when Sd and the delta methods

woudl try to solve the data error more elaborate .. but it has bit on unsolved condition date is good

still my be need more concentration in solving this errors the cut of mark state above as 7.936

around if it would be somewhere 7.91 to 7.915 is good .. good to see it has equal intervals of time

in values.

1)Question)

Anwer) The function (regression) cannot always be written as a linear combination of the terms some times may be bit in order of out flow of data. So weed in condition to estimate the nonlinear

regression

Example: in the turkey diet supplement experiment description.

The Gauss network method is the Best example for nonlinear regression which realise on linear

approximations to the non linear .

> library(alr3)

> attach(lakemary)

> names(lakemary)

[1] "Age" "Length"


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> plot(Length~Age)

Can see the above increasing functional graph as may be bit merged.. but in order and increasing

way. Age Vs Length

> lm(log(1-(Length/200))~Age)

Call:

lm(formula = log(1 - (Length/200)) ~ Age)

Coefficients:

(Intercept) Age

-0.03288 -0.36283

The intercept is beta0,

the age is 1

becuz ( it is lestha .5 , and above the .05 value)

K*=-betahat1

t*=betahat0/-betahat1

t*-> -0.03288/+0.36283

[1] -0.09062095

> nls(Length~Linf*(1-exp(-k*(Age-t0))),start=list(Linf=200,k=0.36283,t0=0.09062095))


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Nonlinear regression model

model: Length ~ Linf * (1 - exp(-k * (Age - t0)))

data: parent.frame()

Linf k t0

192.81044 0.40628 0.08087

residual sum-of-squares: (9014)

Number of iterations to convergence: 4

Achieved convergence tolerance: 1.288e-06

> summary(nls(Length~Linf*(1-exp(-k*(Age-t0))),start=list(Linf=200,k=0.36283,t0=0.09062095)))

Formula: Length ~ Linf * (1 - exp(-k * (Age - t0)))

Parameters:


Linf 192.81044 13.08015 14.741 < 2e-16**

k 0.40628 0.08845 4.593 1.73e-05 **

t0 0.08087 0.24019 0.337 0.737

Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1

Residual standard error: 10.96 on 75 degrees of freedom

Number of iterations 4 for the convergence,So the Convergence tolerance has bees find it conclude

the value as 1.288e-06

With the above value of T0 . we canot control the booster so we have to do boost case with m1 ..

where equal to 9 value.

> bootCase(m1,B=9)

[,1] [,2] [,3]

[1,] 185.2013 0.4560830 0.07057400

[2,] 168.6544 0.6901305 0.48709120

[3,] 202.1633 0.3436856 -0.09846979


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[4,] 194.7148 0.4009924 0.02435563

[5,] 188.7526 0.4779254 0.52709451

[6,] 209.1252 0.3113117 -0.25082608

[7,] 197.3969 0.3947679 0.08624991

[8,] 194.8587 0.3807068 -0.02749190

[9,] 199.5296 0.3629699 -0.07258126

> matrixboot matrixboot

>hist(matrixboot[,1])

Histogramof matrixboot[, 1]

matrixboot[, 1]

Frequency

150 200 250 300

0

100

200

300

400

500

> quantile(matrixboot[,1],0.025)

2.5%

172.4119


97.5%

231.5225

Can start proceeding with other reaming too becuz 96% above quantities in variable 1. And

conclude the final step.



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2.5%

0.2520726


97.5%

0.6915002


2.5%

-0.3213763

By applying Quantitative to the matrix boot value ..


97.5%

0.8307757

The confidence intervals are 95 above good

1)variabile (172.4119, 231.5225)

2)variabile (0.2520726, 0.6915002)

3)variabile (-0.3213763, 0.8307757)

Hense from the data ,,we ploted a non linear regggression model..

OR another exaple can be wrote .. as . For non linear modellibrary(car)

> deltaMethod(mols, "-b1/(2*b2)")

Estimate SE

-b1/(2*b2) 183.1104 5.961452

183.1104 is the Days value that maximizes E(LCPUE/Day).

11.4.3


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E(Y/X) = - 2 x +

= = 0.0921151 (intercept)

= - / 2 = - 0.0466052 / 2*-0.0001273 = 183.0526

is the value of the predictor that gives the maximum value of the response,

previously computed with the delta method;

= = -0.0001273 (measure of curvature)

> nls(LCPUE~th1+th2*(1-exp(-th3*Day)),start=list(th1=0.0921151, th2=183, th3=1))

Nonlinear regression model

model: LCPUE ~ th1 + th2 * (1 - exp(-th3 * Day))

data: parent.frame()

th1 th2 th3

-0.03691 3.69560 0.01974

residual sum-of-squares: 10.54



> m1 summary(m1)

Formula: LCPUE ~ th1 + th2 * (1 - exp(-th3 * Day))

Parameters:


th1 -0.036912 0.385318 -0.096 0.924478

th2 3.695597 0.421873 8.760 6.1e-09 ***

th3 0.019743 0.004647 4.249 0.000281 ***

Residual standard error: 0.6626 on 24 degrees of freedom



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> plot(LCPUE~Day)

> lines(Day,predict(m1,data.frame=Day))

0 50 100 150 200 250

0

1

2

3

4

Day

LCPUE

Hens The above grapg is nonlinear regressionmodel graph.where the variables are not in in curvey

of non orders where tried to explain the plot where the days are fell in intervel of LCPUCE..

2)

Answ) Thesimple linear regression modelconsists of the constant function and the variance function , but

binomial regression deals with the number of successes out of M independent trials, each with the sameprobability of success.

In the binomial regression problem, the response counts the number of successes in( Mi )trials, and so (Mi)(Yi) of the trials were failures. In addition, we have p terms or predictors xi possibly including a constant

for the intercept

Yi = response , success : Mi

In linear regression models, the mean function and the variance function generallyhave completely separate parameters, but that is not so for binomial regression.The value of probability (xi) determines both the mean function and the variancefunction, so we need to estimate (xi).

So we canot employee the somple linear regression model into the binary model where theconception is different , and diff in approach in the solution for data. If may be data is samegiven .

Simple linear regression equations:

E(Y |X = x) = 0 + 1x


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Var(Y |X = x) = 2(2.1)The parameters in the mean function are the intercept 0, which is the value ofE(Y |X = x) when x equals zero, and the slope 1, which is the rate of change inE(Y |X = x) for a unit change by varying the parameters

The variance function is assumed to be constant in the Simple linear regression equation , witha positive value 2that is usually unknown. Because variance 2 > 0, the observed value ofthe ith responseyi willtypically not equal its expected value E(Y |X = xi ).

Where as in binary function (Regression)

Binary Equations:

y Bin(m, )

Pr(y = j)= (m j) j (1 )(mj)

Y equals a specific integer j = 0, 1, . . ., m,

(Y |X = xi) Bin(mi, (xi )), i = 1, . . . , n

4) Problem ..

Ans)

d attach(d)

> names(d)

[1] "Country" "Ccode" "dUnRate" "dEmRate" "bbd"

[6] "UnRate94" "EmRate94" "dEmComConst"

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