Data Sets - The Best Data Sets for Machine Learning

Import the file entitled 101 Dalmations. This is a data set that indicates how friendly people think their dog is on a scale from 1 to 5. Once you have done this, find the mean, median, mode, and normality for the set. Paste the output file below. Interpret the Shapiro-Wilk p-value and indicate whether we can consider the data set to be normal. 

Statistics
FriendlinessOfDogs
N	Valid	101
	Missing	28
Mean		3.5043
Median		3.5200
Mode		3.55

Case Processing Summary
	Cases
	Valid		Missing		Total
	N	Percent	N	Percent	N	Percent
FriendlinessOfDogs	101	78.3%	28	21.7%	129	100.0%

Test for normality 

Descriptives
			Statistic	Std. Error
FriendlinessOfDogs	Mean		3.5043	.05351
	95% Confidence Interval for Mean	Lower Bound	3.3981
		Upper Bound	3.6104
	5% Trimmed Mean		3.5101
	Median		3.5200
	Variance		.289
	Std. Deviation		.53781
	Minimum		1.00
	Maximum		5.00
	Range		4.00
	Interquartile Range		.68
	Skewness		-.714	.240
	Kurtosis		4.236	.476

Tests of Normality
	Kolmogorov-Smirnov a			Shapiro-Wilk
	Statistic	df	Sig.	Statistic	Df	Sig.
FriendlinessOfDogs	.085	101	.069	.945	101	.000
a. Lilliefors Significance Correction

 The data is normally distributed 

 The data is not normally distributed 

In Shapiro-Wilk statistics we reject the null hypothesis if p-value or the significant value is less than 0.05. Since the p-value is less than 0.05 we reject the null hypothesis and conclude that the data is not normally distributed. Hence, we cannot consider the data set to be normal. 

Question 2 

Now, plot a histogram for the dog data. Put a copy below and also identify whether the graph 1) looks normal, and 2) whether you suspect that there might be outliers 

FriendlinessOfDogs 

A visual inspection of the histogram indicates that the friendliness of dogs were not approximately normally distributed, with skewness of -0.714(SE = 0.240) and kurtosis of 4.236(SE = 0.476). there are outliers in the histogram, that is, the two points on the far left on the x-axis position 1.00 and 2.00 and the point on the far right at the x-axis position 5.00. this is because the points falls more than 1.5 times the interquartile range above the third quartile or below the first quartile. 

Question 3 

Now, create a boxplot for the data and paste it below. If you spot any outliers, list them. 

The outliers on a box plot are usually indicated with an asterisk and hence the box plot has outliers at point 99. 

Question 4 

Now, open the “Weights of Cats” file in jamovi. This is a file listing the weights of 201 cats in pounds. Provide the output of the descriptives and tell us the mean, median, mode, standard deviation, and normality. Offer a written interpretation of what you see. Then, provide a frequency histogram and boxplot. Are there outliers? If so, what are they? (Are there “fat cats” or “scrawny cats”? If so, how many and what do they weigh?) 

Descriptives
		WeightsOfCats
N		100
Missing		1
Mean		7.44
Median		7.35
Mode		7.14	ᵃ
Standard deviation		1.05
Minimum		6.38
Maximum		17.0
Skewness		7.80
Std. error skewness		0.241
Kurtosis		71.6
Std. error kurtosis		0.478
Shapiro-Wilk p		< .001
ᵃ More than one mode exists, only the first is reported
From the frequency table the mean is 7.44 and the standard deviation of the weight of the cats is 1.05. The standard deviation indicates that if the average weight of the cats is approximately 7.44 then then most cats in our sample are between 8.49 and 6.39. The standard deviation above and below the mean gives us where most cats tended to be. The data set for the weights of the cats is not normally distributed with skewness of 7.80(SE = 0.241) and kurtosis of 71.6(SE = 0.478). Using the Shapiro-Wilt to explain normality, given the null hypothesis as the data is normally distributed,    The data is normally distributed    The data is not normally distributed

  The Shapiro-Wilt p is less than 0.05, therefore, we reject the null hypothesis and conclude that the data is not normally distributed. 

Plots 

Histogram 

Weight of cats 

A visual inspection of the histogram shows that there are outliers. That is, the point that is far right of the graph. This is because the point falls more than 1.5 times the interquartile range above the third quartile or below the first quartile of the range we expect. 

Box plot 

Inspecting the box plot, there is an indication of the outliers. The outlier is the point that is located above the scale point 15. 

Question 5 

Now, open the “Dog Drool” file in jamovi. This is a file listing the amount dogs drool when they see a bone. Provide the output of the descriptives and tell us the mean, median, mode, standard deviation, and normality. Offer a written interpretation of what you see. Then, provide a frequency histogram and boxplot. Are there outliers? If so, what are they? (Which dogs are too slobbery? Which ones have “drymouth”?) 

Descriptives
		DogDrool
N		95
Missing		0
Mean		18.8
Median		17.9
Mode		14.3	ᵃ
Standard deviation		4.29
Minimum		7.82
Maximum		28.0
Skewness		0.187
Std. error skewness		0.247
Kurtosis		-0.311
Std. error kurtosis		0.490
Shapiro-Wilk p		0.081
ᵃ More than one mode exists, only the first is reported

  The mean of the dog’s drool when they see above is 18.8 and the standard deviation is 4.29. The standard deviation indicates that if the average number of the dog’s drool is approximately 19 then the dog’s drool is between 23 and 15. The standard deviation above and below the mean indicates where the sample of the dog’s drool lies. The dog’s drool data set is approximately normally distributed with skewness of 0.187(SE = 0.247) and kurtosis of -0.3111(SE = 0.490). Given the null hypothesis as the data is normally distributed, 

  The data is normally distributed 

  The data is not normally distributed 

the Shapiro-Wilk = 0.081 which is greater than 0.05 hence we fail to reject the null hypothesis and conclude that the data is normally distributed. 

DogDrool 

Histogram 

The histogram indicates the presence of outliers in the far left of the graph. 

Box plot 

The boxplot does not indicate presence of outliers.