import numpy as np
import scipy as sp
from scipy.stats import normHere are the test scores of 10 students in physics and history:
Physics Scores 15 12 8 8 7 7 7 6 5 3
History Scores 10 25 17 11 13 17 20 13 9 15
Compute Karl Pearson’s coefficient of correlation between these scores. Compute the answer correct to three decimal places.
Output Format
In the text box, enter the floating point/decimal value required. Do not leave any leading or trailing spaces. Your answer may look like: 0.255
This is NOT the actual answer - just the format in which you should provide your answer.
physicsScores=[15, 12, 8, 8, 7, 7, 7, 6, 5, 3]
historyScores=[10, 25, 17, 11, 13, 17, 20, 13, 9, 15]print(np.corrcoef(historyScores,physicsScores)[0][1])0.144998154581
Here are the test scores of 10 students in physics and history:
Physics Scores 15 12 8 8 7 7 7 6 5 3
History Scores 10 25 17 11 13 17 20 13 9 15
Compute the slope of the line of regression obtained while treating Physics as the independent variable. Compute the answer correct to three decimal places.
Output Format
In the text box, enter the floating point/decimal value required. Do not leave any leading or trailing spaces. Your answer may look like: 0.255
This is NOT the actual answer - just the format in which you should provide your answer.
sp.stats.linregress(physicsScores,historyScores).slope0.20833333333333331
Here are the test scores of 10 students in physics and history:
Physics Scores 15 12 8 8 7 7 7 6 5 3
History Scores 10 25 17 11 13 17 20 13 9 15
When a student scores 10 in Physics, what is his probable score in History? Compute the answer correct to one decimal place.
Output Format
In the text box, enter the floating point/decimal value required. Do not leave any leading or trailing spaces. Your answer may look like: 0.255
This is NOT the actual answer - just the format in which you should provide your answer.
def predict(pi,x,y):
slope, intercept, rvalue, pvalue, stderr=sp.stats.linregress(x,y);
return slope*pi+ intercept
predict(10,physicsScores,historyScores)15.458333333333332
The two regression lines of a bivariate distribution are:
4x – 5y + 33 = 0(line of y on x)
20x – 9y – 107 = 0(line of x on y).
Estimate the value of x when y = 7. Compute the correct answer to one decimal place.
Output Format
In the text box, enter the floating point/decimal value required. Do not lead any leading or trailing spaces. Your answer may look like: 7.2
This is NOT the actual answer - just the format in which you should provide your answer.
x=[i for i in range(0,20)]
'''
4x - 5y + 33 = 0
x = ( 5y - 33 ) / 4
y = ( 4x + 33 ) / 5
20x - 9y - 107 = 0
x = (9y + 107)/20
y = (20x - 107)/9
'''
t=7
print( ( 9 * t + 107 ) / 20 )8.5
The two regression lines of a bivariate distribution are:
4x – 5y + 33 = 0(line of y on x)
20x – 9y – 107 = 0(line of x on y).
find the variance of y when σx= 3.
Compute the correct answer to one decimal place.
Output Format
In the text box, enter the floating point/decimal value required. Do not lead any leading or trailing spaces. Your answer may look like: 7.2
This is NOT the actual answer - just the format in which you should provide your answer.
Q.3. If the two regression lines of a bivariate distribution are 4x – 5y + 33 = 0 and 20x – 9y – 107 = 0,
- calculate the arithmetic means of x and y respectively.
- estimate the value of x when y = 7. - find the variance of y when σx = 3.
We have,
4x – 5y + 33 = 0 => y = 4x/5 + 33/5 ————— (i)
And
20x – 9y – 107 = 0 => x = 9y/20 + 107/20 ————- (ii)
(i) Solving (i) and (ii) we get, mean of x = 13 and mean of y = 17.[Ans.]
(ii) Second line is line of x on y
x = (9/20) × 7 + (107/20) = 170/20 = 8.5 [Ans.]
(iii) byx = r(σy/σx) => 4/5 = 0.6 × σy/3 [r = √(byx.bxy) = √{(4/5)(9/20)]= 0.6 => σy = (4/5)(3/0.6) = 4 [Ans.]
variance= σ**2=> 16