What if you have an outlier? My text specifically told me not to use my outlier when deciding class width. I have 45 data points with 7 classes. If I include my outlier I will have 18 classes. Isn't that large for a small data set?
You should only have 7 classes, because that is what was specified. Your class width would just be wider. If it asked you to leave the outlier, then you need to leave it in there.
If your data set contains values that are greater than 0, you round up to the next whole number. If you don't, the maximum value is missed. Even if it's a whole number, you round up to the next whole number to guarantee that the max is included in the last class. If your values are all between 0 and 1, then you would round up to the nearest decimal place that matches your data set. For example, if all data is rounded to 2 decimal places, you would round up the the nearest hundredth.
The number of classes really is open to interpretation. Typically, we use anywhere between 5-20. It depends on the range of the data and the number of data points give. In the homework platform, my students are told how many classes to use for each problem. If they used a different number of classes than specified, they wouldn't match the homework platform answer. In the real world, it's open to interpretation of the person analyzing the data.
If it's for homework, it's typically given in the problem, so that anyone doing the question would end with the same thing. In the real world, you decide based on the range of the data. You can play with different number of classes to see what works best. 5-20 is the typical range for classes.
![Understand Frequency Tables, Cumulative & Relative Frequency in Statistics - [7-7-3]](http://i.ytimg.com/vi/Md2_E_hTvdQ/mqdefault.jpg)
6 years later and this video is very helpful!! Thank you for this!
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helpful!
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Great help
super helpful, helped me with my psych stats project
Glad it was helpful to you!
Very helpful video. I am taking graduate statistics. Beginner friendly.
Thank you. Good luck in your studies!
When I did the tally’s for my chart the numbers were too large for the classes ?
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Glad it helped
How do you find the class boundaries?
What if you have an outlier? My text specifically told me not to use my outlier when deciding class width. I have 45 data points with 7 classes. If I include my outlier I will have 18 classes. Isn't that large for a small data set?
You should only have 7 classes, because that is what was specified. Your class width would just be wider. If it asked you to leave the outlier, then you need to leave it in there.
how do you round up 8 to 9? just go up by one whole number every time?
If your data set contains values that are greater than 0, you round up to the next whole number. If you don't, the maximum value is missed. Even if it's a whole number, you round up to the next whole number to guarantee that the max is included in the last class. If your values are all between 0 and 1, then you would round up to the nearest decimal place that matches your data set. For example, if all data is rounded to 2 decimal places, you would round up the the nearest hundredth.
why use 6 classes if I use 7 or 8 classes is it wrong?
The number of classes really is open to interpretation. Typically, we use anywhere between 5-20. It depends on the range of the data and the number of data points give. In the homework platform, my students are told how many classes to use for each problem. If they used a different number of classes than specified, they wouldn't match the homework platform answer. In the real world, it's open to interpretation of the person analyzing the data.
How is the number of class determined? Or is given as apart of the question?
If it's for homework, it's typically given in the problem, so that anyone doing the question would end with the same thing. In the real world, you decide based on the range of the data. You can play with different number of classes to see what works best. 5-20 is the typical range for classes.
why is she choosing 6 and not 7😂😂😂....
Formula for class interval is
CI= 1+3.322LogN
The answer will be 5.91 = 6