Question makes no sense to me. How many class intervals are you meant to have?
Hi! I am new here. I have a problem regarding statistics and it's about the Class Interval.. . I don't know how to explain instead I will put my data below:
9, 15, 16, 17, 20, 21, 26, 28, 28, 28, 29, 29, 32, 33, 35, 35, 35, 38, 40, 41, 45, 47, 54, 55, 59, 60, 66, 72, 75
R = 75 - 9 = 66
k = 6
c = 11
The upper limit is only 74 and the 75 cannot be included.. .Meaning, the 75 cannot be included when plotting the Frequency Distribution Column.
Second data
9, 12, 15, 16, 17, 18, 21, 21, 22, 22, 22, 24, 25, 28, 29, 30, 31, 31, 31, 31, 32, 32, 33, 34, 35, 35, 36, 36, 37, 40, 41, 42,
42, 48, 48, 49, 51
The same problem above.. Please Help Me! thanks!...
No. The "K" there is already determined how many intervals, right?? here is the problem.
Class Interval
9 - 19
20 - 30
31 - 41
42 - 52
53 - 63
64 - 74
My data above shows that the LARGEST data is 75, but when determining the Class Interval, the 75 cannot be included because 74 is the last upper limit. Any ways that the 75 will be included??
Tell us what your instructions are for doing this? Are you told to use 6 classes? What rules have you been given (the reality that most of us live in recognise no rules for this, we just do what shows the data to advantage)?
You could just make the class interval length 12 giving intervals:
Which can be shifted about if you wish.Code:9 20 21 32 33 44 45 56 57 68 69 80
CB
We instructed to do the Frequency Distribution Table.
When doing the Frequency Distribution, there were certain rules and steps to follow.
1. Find the range R, where R=highest value - lowest value
R = 75 - 9 = 66
2. Estimate the number of classes K,
K = square root of n, where n = number of observation
K = square root of 29
K = 5.4
K ~ 6
3. Estimate the width c.
c = R / K
c = 66 / 6
c = 11
4. List the lower and upper class limit of the first interval. This interval should contain the smallest observation in the data set. The starting lower limit could be the lowest observation or any number closest to it.
5. List all the class limit by adding the class width to the limits of previous interval. The highest class should contain the largest observation in the data set.
6. Tally the frequency for each class
7. Compute the class marks and class boundaries
Problem??
The class width had already determined by c which is 11.
is it OKIE that the above class interval will not follow the result of the class width which is 11?? The 2nd class interval 21 - 32, the difference is 12 when subtracting the 21 and 9 in the 1st class interval. What is the use of computing the class width if we can adjust the width anytime we want???Code:9 20 21 32 33 44 45 56 57 68 69 80
1. It says estimate the class width in your instructions, you have to adjust it to be compliant with the other conditions. In this case it means you have to increase it slightly.
2. The rule that you are using for the number of cells is not one of the usual rules, I would have expected to see either Sturge's rule or the Freedman–Diaconis rule. Not that they make that much difference in this case.
The table should have a fixed class width, if it does not fix it. Other than that the class intervals are arbitary. They are chosen to suite that data. Here we want 5 or 6 classes with 9 in the first and 75 in the last.is it OKIE that the above class interval will not follow the result of the class width which is 11?? The 2nd class interval 21 - 32, the difference is 12 when subtracting the 21 and 9 in the 1st class interval. What is the use of computing the class width if we can adjust the width anytime we want???
CB
So what should I do???
Take the estimate of 11 for the class width and increase it by a convient small amount so that all of the data fall in the cells. I would increase it to 12, but 11.5 would do but it would make the arithmetic inconvienient.
(or scrutinise your notes or instructions closely to see if they tell you what to do in this situation)
CB