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![]() Fall Semester 2002 |
The deVoe Report (June 2, 1980) quoted then U.S. President Jimmy Carter as saying "half the people in this country are living below the median income -- and this is intolerable." What is disputable and what is true in this quote?Here are some of the correct answers:
"President Carter must have meant that whatever the actual value of the median income was, it was intolerable that half of the nation lived below that income."Indeed, that was the answer I was looking for.
The Carter quote was obviously taken out of context. We don't know what he said right before the quoted text, but he might have actually expressed the median income in dollars. Here's an extremely contrived version of this hypothesis, for illustration purposes:
"We have calculated the median yearly income and we found it to be (say) $600. We think this is a problem that needs to be addressed immediately; half the people in this country are living below the median income -- and this is intolerable."There were some answers saying that Carter probably meant the mean. He did not mention the (arithmetical) mean, and that's probably because he did not want to say anything about it. He only wanted to make a comment about the median. He found it too low, and he expressed a concern that the value is too low to be the upper limit of income for half of the nation.
What makes this example intriguing is that, taken out of its context, it drastically polarizes our assumptions about what is said, placing the focus of our understanding on the wrong aspect. To see how this happens in another example (for your enjoyment) witness the following English sentence.
The ship sailed past the harbor sank.How does this sound? Well, here's the same sentence in its original context:
A small part of Napoleon's fleet tried to run the English blockade at the entrance to the harbor. Two ships, a sloop and a frigate, ran straight for the harbor while a third ship tried to sail past the harbor in order to draw enemy fire. The ship sailed past the harbor sank.I hope, perhaps, this makes the point. The question, and the quote, were tricky.
You need to watch for tricks like this in real life too.
Last time we looked at some measures of central tendency.
The homework is asking you to compare them.
The minute paper for today will be communicated in class.
This will be a more mathematical lecture.
1. The Arithmetical Mean.
The arithmetical mean is defined as the sum of the scores divided by the number of scores.
In equation form that is
Properties of the mean:
In other words, this formula (in which zeta is an unknown)
admits a minimum when zeta has this value
We need to verify that.
The median is defined as the scale value below which 50% of the scores (or measurements) fall.
Properties of the median:
The mode is defined as the most frequent score in the distribution.
Usually distributions are unimodal. When a distribution has two modes it is bimodal.
MEASURES OF VARIABILITY
1. The Range.
The range is defined as the difference between the highest and lowest score in the distribution.
2. Deviation Scores.
A deviation score tells how far away the raw score is from the mean of its distribution.
3. The Standard Deviation.
For a population of scores we have:
For a sample we have:
Alternative formula for the standard deviation:
Properties of the standard deviation:
Squared deviations and the mean
We will "prove" today (in lab) that the sum of the squared deviations of all the scores about their mean is a minimum. In other words, the formula below, in which zeta is an unknown (or variable)
admits a minimum for
Let's prove that (and in the process calculate other things as well).
Here are the steps:
That's our data (the scores).
=average(a1:a5)For me that is 2.4 (the arithmetical mean of these 5 numbers).
In B1 write the formula for the first deviation:
=A1-$E$1Notice that the second element has an absolute reference to column E row 1.
When we paste this formula that will become relevant.
The first deviation is -1.4 so the formula works fine.
The deviations are calculated.
=A5-$E$1Excel updated only the relative components of the formula.
=sum(b1:b5)The value should be 0 (zero).
We have just verified that the sum of all the deviations about the mean is zero.
In cell C1 write this formula
=b1^2The value is 1.96 for me, the square of -1.4 as expected.
Select C1. Drag lower right corner to C5. Release mouse button.
Squared deviations have been calculated.
In cell C6 enter the following formula
=sum(c1:c5)That's the sum of the squared deviations about the mean, and it's 5.2 for me.
=count(a1:a5)That's 5. I leave it to you to label your spreadsheet nicely.
=sqrt(c6/e2)
=sqrt(sum(c1:c5)/count(c1:c5))
=stdevp(a1:a5)
=stdev(a1:a5)That should clarify the difference between STDEVP and STDEV.
In A9 through F9 enter 0, 1, 2, 3, 4, 5.
We'll calculate the sum of squared deviations around each of these numbers.
=($A1-A$9)^2
The cells from A10 to F14 should now contain squared deviations.
In cell A15 enter this formula
=sum(a10:a14)
Choose "Line" and click "Finish".
Still another way would be to
There will be a new mean, but that's where the minimum will also be.
Lab notes for tomorrow will be posted in the morning.
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