In the given training dataset:
Number of images with (Label=Tree) = 4
Number of images with (Label=not Tree) = 6
Probability of value of R being 0-100 when (Label=Tree) is given = P(R=0-100 | Label=Tree) = =2/4 = 1/2
Probability of value of R being 0-100 when (Label=not Tree) is given = P(R=0-100 | Label = not Tree) = =1/6
Similarly,
P(R=171-255 | Label=Tree) = 2/4 = 1/2
P(R=171-255 | Label=not Tree) = 1/6 = 1/6
P(G=0-100 | Label=Tree) = 0/4 = 0
P(G=0-100 | Label=not Tree) = 3/6 = 1/2
P(G=171-255 | Label=Tree) = 1/4
P(G=171-255 | Label=not Tree) = 1/6
P(G=101-170 | Label=Tree) = 3/4
P(G=101-170 | Label=not Tree) = 2/6 = 1/3
P(B=0-150 | Label=Tree) = 1/4
P(B=0-150 | Label=not Tree) = 3/6 = 1/2
P(B=151-255 | Label=Tree) = 3/4
P(B=151-255 | Label=not Tree) = 3/6 = 1/2
P(3D=Yes | Label=Tree) = 3/4
P(3D=Yes | Label=not Tree) = 2/6 = 1/3
P(3D=No | Label=Tree) = 1/4
P(3D=No | Label=not Tree) = 4/6 = 2/3
For the testing set:
1) For Image A,
P(Image A | Label=Tree) = P(R=0-100 | Label=Tree)*P(G=0-100 | Label=Tree)*P(B=0-150 | Label=Tree)*P(3D=Yes | Label=Tree) = (1/2)*0*(1/4)*(3/4) = 0
P(Image A | Label=not Tree) = P(R=0-100 | Label=not Tree)*P(G=0-100 | Label=not Tree)*P(B=0-150 | Label=not Tree)*P(3D=Yes | Label=not Tree) = (1/6)*(1/2)*(1/2)*(1/3) = 1/72
According to Naive Bayes classifier, Label's output will be according to whichever one's value is higher. We can see that (1/72) > 0. So, Label = not Tree for Image A.
2) For Image B,
P(Image B | Label=Tree) = P(R=171-255 | Label=Tree)*P(G=171-255 | Label=Tree)*P(B=0-150 | Label=Tree)*P(3D=No | Label=Tree) = (1/2)*(1/4)*(1/4)*(1/4) = 1/128
P(Image B | Label=not Tree) = P(R=171-255 | Label=not Tree)*P(G=171-255 | Label=not Tree)*P(B=0-150 | Label=not Tree)*P(3D=No | Label=not Tree) = (1/6)*(1/6)*(1/2)*(2/3) = 1/108
According to Naive Bayes classifier, Label's output will be according to whichever one's value is higher. We can see that (1/108) > (1/128). So, Label = not Tree for Image B.
3) For Image C,
P(Image C | Label=Tree) = P(R=171-255 | Label=Tree)*P(G=101-170 | Label=Tree)*P(B=151-255 | Label=Tree)*P(3D=Yes | Label=Tree) = (1/2)*(3/4)*(3/4)*(3/4) = 27/128 = 0.2109
P(Image C | Label=not Tree) = P(R=171-255 | Label=not Tree)*P(G=101-170 | Label=not Tree)*P(B=151-255 | Label=not Tree)*P(3D=Yes | Label=not Tree) = (1/6)*(1/3)*(1/2)*(1/3) = 1/108 = 0.009
According to Naive Bayes classifier, Label's output will be according to whichever one's value is higher. We can see that (0.2109) > (0.009). So, Label = Tree for Image C.
2. Giving the following training dataset: B Label Image 1 0-100 171-255 0-150 Yes Tree Image...
Download the hypothesis testing dataset from the class website. 1. This is case 1, in which the value of population standard deviation is assumed to be known. a. Compute the standard deviation of the textbook price and assume this to be the population standard deviation. Type this value in E5. b. Pick a random sample of size 15. c. Compute the mean of your random sample in E4 using the AVERAGE function. d. Calculate the standard error in cell E9...