Using Python:
curve80.txt contains the following:
3.4447005e+00 -8.8011696e-01
4.7580645e+00 4.6491228e-01
6.4170507e+00 3.7397661e+00
5.7949309e+00 3.0087719e+00
7.7304147e+00 2.9210526e+00
7.8225806e+00 4.1491228e+00
7.7304147e+00 3.3888889e+00
7.7764977e+00 3.7105263e+00
8.6751152e+00 2.9795322e+00
6.4631336e+00 3.9736842e+00
5.1267281e+00 1.1403509e-01
6.7396313e+00 4.1491228e+00
3.1451613e+00 -5.0000000e-01
9.1589862e+00 4.0906433e+00
8.2373272e+00 2.8040936e+00
4.8041475e+00 -5.0000000e-01
3.5714286e-01 -1.4941520e+00
8.0069124e+00 4.0321637e+00
2.2465438e+00 -7.6315789e-01
6.7626728e+00 4.4122807e+00
5.0115207e+00 1.0204678e+00
8.7211982e+00 3.0087719e+00
1.6935484e+00 -6.7543860e-01
4.8502304e+00 3.7719298e-01
8.6059908e+00 4.6461988e+00
8.2142857e+00 4.1491228e+00
8.1797235e-01 -1.4649123e+00
5.7488479e+00 2.1023392e+00
6.7165899e+00 4.0321637e+00
2.0391705e+00 -9.9707602e-01
5.1036866e+00 1.8976608e+00
4.3433180e+00 5.8187135e-01
4.4815668e+00 -7.6315789e-01
7.3156682e+00 4.9385965e+00
8.5138249e+00 3.4473684e+00
9.0207373e+00 2.8625731e+00
5.4953917e+00 2.1023392e+00
6.0483871e+00 3.5935673e+00
4.5506912e+00 -7.6315789e-01
2.6843318e+00 -6.4619883e-01
6.8087558e+00 4.7046784e+00
1.7857143e+00 -1.3187135e+00
5.4723502e+00 1.7222222e+00
3.3755760e+00 -9.9707602e-01
7.7304147e+00 4.5584795e+00
6.7396313e+00 5.1432749e+00
4.2741935e+00 -1.0263158e+00
4.7811060e+00 1.5467836e+00
5.8870968e+00 2.4532164e+00
8.8133641e+00 4.1783626e+00
5.9101382e+00 3.6228070e+00
4.8502304e+00 7.8654971e-01
6.6013825e+00 4.4707602e+00
1.2557604e+00 -1.2309942e+00
4.1129032e+00 -9.6783626e-01
7.1774194e+00 2.8333333e+00
4.8271889e+00 -2.9532164e-01
2.9147465e+00 -1.0847953e+00
5.1728111e+00 1.6345029e+00
5.8410138e+00 2.8625731e+00
8.4907834e+00 2.3070175e+00
7.4078341e+00 3.7982456e+00
8.1797235e-01 -9.9707602e-01
7.2580645e-01 -4.7076023e-01
7.5921659e+00 4.1491228e+00
8.8133641e+00 3.4766082e+00
2.4769585e+00 -9.3859649e-01
4.5737327e+00 -1.1988304e-01
8.6751152e+00 3.7982456e+00
6.1635945e+00 2.6871345e+00
8.3525346e+00 3.5643275e+00
6.5783410e+00 4.5292398e+00
4.8271889e+00 6.6959064e-01
2.5230415e+00 -1.2309942e+00
2.4193548e-01 3.4795322e-01
6.2327189e+00 4.1783626e+00
8.7903226e+00 3.0380117e+00
2.2695853e+00 -1.0847953e+00
6.3709677e+00 6.1959064e+00
6.0253456e+00 3.0964912e+00

The following has been given:

![In [17]: degrees[1, 3, 5, 7, 10, 18] # Plot settings plt.rcParams[figure.figsize] - (18.0, 7.0) fig, ax- plt.subplots (2,3) axFlat [a for row in ax for a in row] # 2x3 subplots as simple list err train = [] errtest[] for i, d in enumerate (degrees): # Create polynomial features poly = PolynomíalFeatures (d, include-bias-False).fịt (Xtr) # Scale features to standard distribution (mean 0, variance 1) scaler -StandardScaler().fit (XtrP) # Fit the model # Plot the prediction function axFlat [i].plot ( ) #plot the train and test datapoints axissize axF1at [ j . axis ( ) axFlat [i].plot() #plot the fitted curve axFlat[ i].axis (axisSize) # Make predictions YtrP pred - YteP pred .. . # save train and test errors for (b) err_train.append(...) err test.append(. . .) ee 1 Degree 3 Degree 5 VC 18](http://img.homeworklib.com/questions/f9fdb2a0-c171-11eb-bb81-c3becdf2e97b.png?x-oss-process=image/resize,w_560)
Above is the expected plot output of part a
Above is the expected plot
output of part b
import numpy as np
from sklearn.model_selection import train_test_split, KFold
from sklearn.linear_model import LinearRegression
from sklearn.preprocessing import PolynomialFeatures,
StandardScaler
import matplotlib.pyplot as plt
from sklearn.pipeline import make_pipeline
data = np.loadtxt('curve80.txt')
X = data[:,0]
X = np.atleast_2d(X).T
Y = data[:,1]
Xtr, Xte, Ytr, Yte =
train_test_split(X,Y,test_size=0.25,shuffle=False)
degrees = [1,3,5,7,10,18]
plt.rcParams['figure.figsize']=(18.0,7.0)
fig, ax = plt.subplots(2,3)
axFlat = [a for row in ax for a in row]
err_train = []
err_test = []
for i, d in enumerate(degrees):
poly = PolynomialFeatures(d,
include_bias=False).fit(Xtr)
XtrP = poly.transform(Xtr)
XteP = poly.transform(Xte)
scaler = StandardScaler().fit(XtrP)
XtrP = scaler.transform(XtrP)
XteP = scaler.transform(XteP)
lrP = LinearRegression().fit(XtrP, Ytr)
xsP =
np.sort(np.random.uniform(0.0,X.max()+1,(1000,1)),axis=0)
ysP =
lrP.predict(scaler.transform(poly.transform(xsP)))
axFlat[i].scatter(Xte,Yte,c='green') # plotting
the train and test data points
axFlat[i].scatter(Xtr,Ytr,c='red')
axisSize = axFlat[i].axis()
axFlat[i].plot(xsP,ysP,c='black') # plotting the
fitted curve
axFlat[i].axis(axisSize)
# make predictions
YtrP_pred = lrP.predict(XtrP)
YteP_pred = lrP.predict(XteP)
# Save train and test errors
err_train.append(np.mean((Ytr-YtrP_pred)**2))
err_test.append(np.mean((Yte-YteP_pred)**2))
plt.show()
plt.rcParams['figure.figsize']=(15.0,8.0)
plt.semilogy([1,2,5,7,10,18],err_train, c='orange',label='Train
Error')
plt.semilogy([1,2,5,7,10,18],err_test,c='green', label='Test
Error')
plt.xticks([1,2,5,7,10,18])
plt.legend()
plt.show()
![D matplotlib.pyplot scatte × Polynom al fit Matpl × ■ Plotting Cross Validated × . matí tib.p plotse mil × ● matplotlib.p plot legen X + ⓘ localhost: Jupyter HomeworkLib Last Checkpoint. 2 hours ago (unsaved changes) Logout File Edit View Insert Cell Kemel Help TrustedP Python 3 GO In [11] import numpy as np from sklearn.model_sclection import train test_split, KFold from sklearn.linear model import LinearRegression from sklearn.preprocessing import PolynomialFeatures, Standardscaler import matplotlib.pyplot as plt from sklearn.pipeline import make_pipeline In [2]: datanp.loadtxt(curve80.txt) In [28]: X= data [:,0] xp.atleast_2d (x) .T Y-data [ 1,1] Xtr, Xte, Ytr, Yte-train test split (X,Y, test size=0.25,shuffle-Palse) - - - In [681 degrees [,3,5,7,10,18] plt.rcParamsfigure.figsize fig, ax = plt . subplots (2,3) axFlat = [a for row in ax for a in row] -(18.0,7.0 crr train [1 err test = [] for i, d in enumerate (degrees): poly PolynomialFeatures (d, include bias-False) .fit(xtr) XtrP poly.transform (Xtr) Xtep-poly. transform (Xte) scaler- Standardscaler).fit (XtrP) XtrP scaler.transform (XtrP)](http://img.homeworklib.com/questions/fb3ea310-c171-11eb-899d-1f9200515a9a.png?x-oss-process=image/resize,w_560)
![D matplotlib.pyplot scatte × Polynom al fit Matpl × ■ Plotting Cross Validated × . mat lotlib.p plotse mil × ● matplotlib.p plot legen X + ⓘ localhost: Jupyter HomeworkLib Last Checkpoint. 2 hours ago (unsaved changes) Logout File Edit View Insert Cell Kemel Help TrustedP Python 3 GO plt.rcParamsfigure.figsize 1-(18.0,7.0) fig, ax = plt . subplots (2,3) axFlat = [a for row in ax for a in row] err train[l err test[1 for i, d in enumerate (degrees): poly PolynomialFeatures (d, include bias-False) .fit(xtr) XtrP poly.transform (Xtr) Xtep-poly. transform (Xte) scaler-Standardscaler ).fit (XtrP) XtrP scaler.transform (XtrP) Xtep = 3caler. transform (Xtep) lrP LinearRegression ).fit(XtrP, Ytr) xsP ysp np.sort (np. randan. uniform (O.0, X. max () +1, (1000, 1)), axis-0) 1rp . predict (sealer, transform(poły. transform(X8P))) = = axPlat[1],scatter (Xte, Yte, c- green) # plotting the train and axFlat [i].scatter (Xtr, Ytr,c-red) axisSize axFlat [1] . axis () axFlat [1] .plot (X3P,ysp, c-black) # plotting the fitted curve axFlat[i] axis (axisSizc) test data points # make predictions YtrP pred lrP.predict (XtrP) YtcP_pred lrP.predict (XteP) # save train and test errors err_train.append (np.mean( (Ytr-YtrF_pred) 2)) err test.append (np.mean ((Yte-YteP pred)*2)) plt.show()](http://img.homeworklib.com/questions/fbc87c40-c171-11eb-8300-c7405117d94e.png?x-oss-process=image/resize,w_560)
![2 HomeworkLib matplotlib.pyplot scattexm 2 New Messages! × ■Plotting Cross Validated× . mat lotlib.plotse k× matplotlib.pyplotlegenX+ ⓘ localhost: Jupyter HomeworkLib Last Checkpoint. 2 hours ago (unsaved changes) Logout File Edit View Insert Cell Kemel Help TrustedPython 3 O plt.show) -2 -2 In [691 plt.rcParamsfigure.figsize 1-(15.0,8.0) plt.semilogy(Il,2,5,7,10,181,err train, -orange ,label- Train Error) plt.semilogy([1,2,5,7,10,18],err test, -green label-Test Error) plt.xticks(1,2,5,7,10,18]) plt.legend() plt.show )](http://img.homeworklib.com/questions/fc5318d0-c171-11eb-bbff-5dcfeeadf712.png?x-oss-process=image/resize,w_560)
![HomeworkLib matplotlib.pyplot scattexm 2 New Messages! × ■Plotting Cross Validated× . mat lotlib.plotse k× matplotlib.pyplotlegenX+ ⓘ localhost: jupyter HomeworkLib Last Checkpoint 2 hours ago (unsaved changes) Logout File Edit View Insert Cell Kemel Help Trusted Python 3 O +鈊 In 169]: 龟 pit .rcparameltigure.士1431ze|-(15.0, 8.U) plt.semilogy(I1,2,5,7,10,181,err_train, c-orange, label-Train Error plt.semilogy(I,2,5,7,10,18),err test,c-green, label-Test EEO plt.xticks (,2,5,7,10,181) plt.legend() pit·show ( ) ㅑ역 Train Error Test Error 10 101 100 18](http://img.homeworklib.com/questions/fcdda8b0-c171-11eb-9584-d52684791a03.png?x-oss-process=image/resize,w_560)
from the above experiments degree 10 has the lowest test error because it gives us the best fit of our data.
Using Python: curve80.txt contains the following: 3.4447005e+00 -8.8011696e-01 4.7580645e+00 4.6491228e-01 6.4170507e+00 3.7397661e+00 5.7949309e+00 3.0087719e+00 7.7304147e+00 2.9210526e+00...
Classification in Python: Classification In this assignment, you will practice using the kNN (k-Nearest Neighbors) algorithm to solve a classification problem. The kNN is a simple and robust classifier, which is used in different applications. The goal is to train kNN algorithm to distinguish the species from one another. The dataset can be downloaded from UCI Machine Learning Repository: https://archive.ics.uci.edu/ml/machine-learning-databases/iris/ (Links to an external site.)Links to an external site.. Download `iris.data` file from the Data Folder. The Data Set description...
Regression Analysis Problem #1 The following table shows the sales (in $100,000) of a certain product as a function of the past 10 months and the level of advertisement (in S1,000) for the corresponding months Sales 100 110 107 110 118 120 117122 120 125 Month 2 4 5 7 8 10 Advertisement 1.5 2.1 2,6 2.8 3.5 3.7 3.6 3.4 3.4 2.9 Part A: Considering a simple linear regression of sales vs, time (months), perform the following analysis manually,...
Need help with stats true or false questions
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Q1 (30 points) Consider Problem 11.45, Page 637. Please note that for this problem the data will be entered in R as follows: #Enter data on x = Dose Level of Drug, and y = Potency of Drug (Problem 11.45, page 637) x<-c(2, 2, 2, 4, 4, 8, 8, 16, 16, 16, 32, 32, 64, 64, 64) y<-c(5, 7, 3, 10, 14, 15, 17, 20, 21, 19, 23, 29, 28, 31, 30) For this problem, answer the following questions. In...
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3. [25 marks] Some female psychology students were investigating
whether intelligence depends on brain size. They each took a
standard test that measured verbal IQ and also underwent an MRI
scan to measure their brain size. The resulting data is below, file
named IQBrain.csv.
IQ
BrainV
132
816.932
132
951.545
90
928.799
136
991.305
90
854.258
129
833.868
120
856.472
100
878.897
71
865.363
132
852.244
112
808.02
129
790.619
86
831.772
90
798.612
83
793.549
126
866.662
126
857.782...
3. [25 marks] Some female psychology students were investigating
whether intelligence depends on brain size. They each took a
standard test that measured verbal IQ and also underwent an MRI
scan to measure their brain size. The resulting data is below, file
named IQBrain.csv.
IQ
BrainV
132
816.932
132
951.545
90
928.799
136
991.305
90
854.258
129
833.868
120
856.472
100
878.897
71
865.363
132
852.244
112
808.02
129
790.619
86
831.772
90
798.612
83
793.549
126
866.662
126
857.782...
I need help interpreting logistic regression results to answer
the following question: Does GRE scores, undergraduate GPA and the
prestige (yes or no) of their undergraduate program effect
admission (yes or no) into graduate school?
Fit Group 4 Logistic Fit of ADMIT 2 By GRE 1.00 Contingency Analysis of ADMIT 2 By TOPNOTCH 2 4 Mosaic Plot Logistic Fit of ADMIT 2 By GPA 1.00 1.00 0.75 0.75 No 0.75 No No ADMIT 2 0.50 N 0.50 ADMIT 2 ADMIT...
The first photo is the data I
had collected in Minitab.I am confused on what the b1= to then get
the degree of freedom. I need this information to answer question
16 to plug in the right information in minitab to get t*multiplier.
Overall need help with getting the answer to #16 so then I can
continue the rest of the problems. Thanks! (also for 17 what is
S.E.)
Regression: icu versus age Simple Analysis of Variance Source DF Adj...