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 |
Delegate your assignment to our experts and they will do the rest.
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.