Homework Help v1.3

[Syfer Van Hyden]

Originally Posted by Ice(v)an

Is anyone familiar with Quartiles?
how do you find Q1 and Q3.
Because the way my professor does it, is he adds up the first two lowest digits, then divides it by 2. and for Q3, he adds the last two digits and divides it by 2.

But on one of the problems she did on board, she didn't do that. So I'm not sure how to do it.

Do u have a graphing calculator like a TI (Texas Instrument)? If so, then put your data in lists. So go to STAT>EDIT>put all your data in L1, and if u have any frequencies put them in L2, if not then dont touch L2. After you're done, go to STAT>scroll to the right column once to CALC and choose 1-Vars-Stats (which is 1variable statistics). Hit enter and u can scroll down for the info u need. Q1 and Q3 r present in there.

It would REALLY help if u post the data here or the question so i can take u through it and find the answer.

[Ice(v)an]

Data
16, 499
11, 990
15, 299
10, 995
13,995
10,949
13,989
10,900
12,999

the professor says that Q1 is 10949 + 10995 /2 and Q3 is 13995 + 15299/2


but her method for this question

136 , 169 , 120, 128 ,129 143, 115, 146 , 96, 86
was different.

she says Q1 = 115 and Q3 was 143
so if I add 86 + 96 /2 I dun get 115

[Syfer Van Hyden]

Originally Posted by Ice(v)an

Data
16, 499
11, 990
15, 299
10, 995
13,995
10,949
13,989
10,900
12,999

the professor says that Q1 is 10949 + 10995 /2 and Q3 is 13995 + 15299/2


but her method for this question

136 , 169 , 120, 128 ,129 143, 115, 146 , 96, 86
was different.

she says Q1 = 115 and Q3 was 143
so if I add 86 + 96 /2 I dun get 115

The reason why the method is different from each other is because ur 1st sets of data had n=9 (9 data sets). And for the other question u have n=10 there (10 sets of data).

As for the answers for the 1st question, iyour right. But then again, for the 2nd question it's not the same method since u have a different number of data.

Ur problem was that u had to rearrange ur data sets, that's what probably confused u, and me.

86, 96, 115, 120, 128, 129, 136, 143, 146, 169.

U take the numbers below the median (in this case median is 128.5= 129+128divided by 2) and those numbers r 86 to 128, the median for that range is 115 which is ur Q1.

Do the same for the upper quartiel: 129 to 169, ur median for that is 143. the middle number.

or u can use the TI way which i presented on the top. Use a graphing calc if u have one. Il teach u all about it.

i hope this helped.

EDIT: Remember that if n (or the number of data sets) changes, then the entire method will change slightly. Keep that in mind. n is key.

[Ice(v)an]

I still dun get how you get 115 from that. And rearranging wasn't my problem, because I had to rearrange it to find the mode.

Anyway, the way I got 115 and 143 was this.

10/2 I get five, then I +1 and divide 2 again, so i get 3. so that means Q1 is three numbers away from the first, and 3 numbers away from the last.

But that didn't work for the first problem. Since I get a decimal number with the whole number.

[Syfer Van Hyden]

^Like i said before, if u have a different number of data sets, then the method will change slightly and using the method u used for the 1st question which had n=10 wont give u the same results u require.

[Ice(v)an]

why do the methods keep changing?
or is the method different when the N is odd or even?

[Syfer Van Hyden]

^Correct, simply when u have an odd number, the MIDDLE number is your median and by that you can find Q1 in the MIDDLE of the 1st half of the data sets.

With even numbers, however, u take the 2 MIDDLE numbers and divide them by 2 to get the median. And by that u can get Q1 by adding the 2 MIDDLE numbers of the 1st half of the data sets.

Note that the same can be done when finding Q3, except this time u will be looking at the UPPER half of the data sets. so with even numbers there is this TINY ADDITION step that u have to go through.
i hope this clears it all.

and remember if u prefer to do this by calc (a TI), i can show u shortcuts.

[Ice(v)an]

So if you were to say, do this in a formula, can you? Cause I'm still not clear on what to do when given a data. When it's even or odd.

[Syfer Van Hyden]

^Yes u can do it in a formula, but the odd n has a separate formula than the even n.

And i dont think that i can explain it simpler than i already have, sorry man. Just a quick question because u havent answered me this, do u have a TI calc?

[ahme]

Can I jump in the middle of you?

I'll try to simply put how to calculate the quartiles but let's first define quartiles. Quartiles are, as defined in wikipedia, is any of the three values which divide the sorted data set into four equal parts, so that each part represents 1/4th of the sampled population. In other words, it the data is sorted, there are 25% of the numbers less than Q1, 50% of the numbers less that Q2 and 75% less that Q3.

How to calculate q quartile?
Let's start with Q2 since it is the easiest. Q2 is simply the median. The easiest way to find the median is to do the following:
1. sort the numbers.
2. count the numbers:
if odd, pick the middle number
if even, pick the two middle numbers and average them

ex, 1,3,5,7,9 (5 numbers -> median is middle number which is 5)
1,3,5,7,9,10 (6 numbers -> median is average of 5 and 7 = 6)

What about Q1?
By now we know how to find the Q1 which cuts the data into two halves. The easiest way to find Q2 is to find the number that would cut the lower half (all numbers lower that the median into two halves) into two halves.
Q3 is the same as Q1 except that we use the upper half of the data (those numbers that are greater than the median).

ex,
136 , 169 , 120, 128 ,129 143, 115, 146 , 96, 86

1. sort: 86, 96, 115, 120, 128, 129, 136, 143, 146, 169
2. 10 numbers -> median is avg. of the to middle numbers (128 and 129 = 128.5 = Q2)
Q2 (median of lower half):
1. Take lower half (all numbers STRICTLY less than 128.5): 86, 96, 115, 120, 128
2. median (we have five numbers so median is the middle number) = 115 = Q2

If I'm to add three more number to the set of numbers:
86, 96, 115, 120, 128, 129, 136, 143, 146, 169, 190, 191,192

Q1:
1. the data is sorted
2. Q1 = median of the data (13 numbers so median is the middle number) = 136
Q2:
1. take the lower half of the numbers (all numbers STRICTLY less than 136):
86, 96, 115, 120, 128, 129
2. Q2 = median (we have six numbers so median is avg of the two middle number) = (115+120)/2 = 117.5

[Syfer Van Hyden]

^Someone has explained it much much better than i have, u ahme. But i skipped the whole intro part of Q1 and Q3 because i assumed Ice(v)an knew what they were and thought that it wasn't necessary. I thought i could help but im not cut out for explaining things really, i just do them, period. Oh well.

[]MpC[hebe]

Originally Posted by ahme

Can I jump in the middle of you?

I'll try to simply put how to calculate the quartiles but let's first define quartiles. Quartiles are, as defined in wikipedia, is any of the three values which divide the sorted dataset into four equal parts, so that each part represents 1/4th of the sampled population. In other words, it the data is sorted, there are 25% of the numbers less than Q1, 50% of the numbers less that Q2 and 75% less that Q3.

How to calculate q quartile?
Let's start with Q2 since it is the easiest. Q2 is simply the median. The easiest way to find the median is to do the following:
1. sort the numbers.
2. count the numbers:
if odd, pick the middle number
if even, pick the two middle numbers and average them

ex, 1,3,5,7,9 (5 numbers -> median is middle number which is 5)
1,3,5,7,9,10 (6 numbers -> median is average of 5 and 7 = 6)

What about Q1?
By now we know how to find the Q2 which cuts the data into two halves. The easiest way to find Q1 is to find the number that would cut the lower half (all numbers lower that the median into two halves) into two halves.
Q3 is the same as Q1 except that we use the upper half of the data (those numbers that are greater than the median).

ex,
136 , 169 , 120, 128 ,129 143, 115, 146 , 96, 86

1. sort: 86, 96, 115, 120, 128, 129, 136, 143, 146, 169
2. 10 numbers -> median is avg. of the to middle numbers (128 and 129 = 128.5 = Q2)
Q1 (median of lower half):
1. Take lower half (all numbers STRICTLY less than 128.5): 86, 96, 115, 120, 128
2. median (we have five numbers so median is the middle number) = 115 = Q2

If I'm to add three more number to the set of numbers:
86, 96, 115, 120, 128, 129, 136, 143, 146, 169, 190, 191,192

Q2:
1. the data is sorted
2. Q2 = median of the data (13 numbers so median is the middle number) = 136
Q1:
1. take the lower half of the numbers (all numbers STRICTLY less than 136):
86, 96, 115, 120, 128, 129
2. Q1 = median (we have six numbers so median is avg of the two middle number) = (115+120)/2 = 117.5

I think you got Q1 and Q2 mixed up in the second half of your post. I corrected it in the quote above. Otherwise, looks like pretty solid info.

Please note that a definition that states that Q1, Q2, and Q3 actually have to be part of the dataset would be mistaken. They are simply the values which have either "at most" or "at least" 25%, 50%, and 75% of the sample data below them. Because of that "at least" or "at most" distinction, there is a fair degree of leeway when actually calculating the quartile values by hand and in computer software. That decision usually comes down to what you are testing, and whether overestimating would be appropriate. The actual mathematical distinction between “at most” and “at least” comes in deciding whether to include Q2 (assuming it is in the dataset) when finding both Q1 and Q3, or whether to take only the numbers below and above Q2 when finding Q1 and Q3 respectively. Just say consistent…

Based on your n = 13 sample data, here are the two methods described:

86, 96, 115, 120, 128, 129, 136, 143, 146, 169, 190, 191,192

The “at most” method:

Q2 = middle element (because there is one this time) = 136

The “at most” method excludes 136 when finding Q1 and Q3:

Q1 dataset: 86, 96, 115, 120, 128, 129
Q3 dataset: 143, 146, 169, 190, 191,192

Because there is no middle element in each smaller dataset, we follow the averaging method between the two middle elements. Why do we do that? Why not just pick any element between the two middle elements? Well, you can, but it will generally suck when trying to use such values for developing statistical inference for a large population.

Q1 = (115 + 120) / 2 = 117.5
Q3 = (169 + 190) / 2 = 179.5

So, our original data is broken up as follows:

86, 96, 115, |, 120, 128, 129, 136, 143, 146, 169, |, 190, 191,192

Notice that at most 25% (in this case, at most 3) of the 13 elements is in each grouping and that 136 is excluded.

The quartile sets would probably be listed as follows:
86, 96, 115
120, 128, 129
143, 146, 169
190, 191,192

Now lets have a go with “at least” method. This method includes Q1, Q2, and Q3 where applicable. Lets start again with our fresh dataset:

86, 96, 115, 120, 128, 129, 136, 143, 146, 169, 190, 191,192

Exactly the same as above, Q2 is the middle element (since there is one here):

Q2 = 136

In the “at least” method, we look at the following Q1 and Q3 datasets

Q1 dataset: 86, 96, 115, 120, 128, 129, 136
Q3 dataset: 136, 143, 146, 169, 190, 191,192

This time, Q1 and Q3 are very obvious, because each dataset has a clear middle value:

Q1 = 120
Q3 = 169

So, our original data is broken up as follows:

86, 96, 115, 120, 128, 129, 136, 143, 146, 169, 190, 191,192

Notice that at least 25% (in this case, at least 4 since 3 < 13/4) of the 13 elements is in each grouping

The quartile sets would probably be listed as follows:
86, 96, 115, 120
120, 128, 129, 136
136, 143, 146, 169
169, 190, 191,192

Personally, I have no preference between these methods, but I typically teach the first one.

[Ice(v)an]

Sorry Wesker, for not seeing what you were trying to say. It might not be what you were saying, it might be me not seeing it. I don't know, but from ahme post, I just saw it. I saw that he found the median, then the lower number, the median of that range, is Q1, and for Q3, it's the median of those range of numbers.

I believe this is it, if it's not, then sigh...

[Syfer Van Hyden]

^That's absolutely fine, i'm not any good at explaining things. But i thought i would be able to explain this correctly because it was a very simple stats question. Oh well. I tried.