exam interpretation
TRANSCRIPT
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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:
Estimate Std. Error t value Pr(>|t|)
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
> quantile(matrixboot[,1],0.975)
97.5%
231.5225
Can start proceeding with other reaming too becuz 96% above quantities in variable 1. And
conclude the final step.
> quantile(matrixboot[,2],0.025)
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2.5%
0.2520726
> quantile(matrixboot[,2],0.975)
97.5%
0.6915002
> quantile(matrixboot[,3],0.025)
2.5%
-0.3213763
By applying Quantitative to the matrix boot value ..
> quantile(matrixboot[,3],0.975)
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
Number of iterations to convergence: 16
Achieved convergence tolerance: 9.227e-06
> m1 summary(m1)
Formula: LCPUE ~ th1 + th2 * (1 - exp(-th3 * Day))
Parameters:
Estimate Std. Error t value Pr(>|t|)
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
Number of iterations to convergence: 16
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Achieved convergence tolerance: 9.227e-06
> 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"