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IQ versus Income
http://www.bus.ucf.edu/mdickie/Research%20Methods/Student%20Papers/Other/MoedingerIQ%20&%20earnings.pdf
"This research examines the extent to which IQ scores affect
earnings. The Bureau of Labor
Statistics’ National Longitudinal Survey of Youth, which tracks
12,686 baby boomers from 1979
on provides the data for this research. Earnings are regressed
against IQ percentile scores in
single regressions as well as in multiple regressions,
controlling for demographic and educational
variables that affect income. The effects of IQ on earnings are
positive and statistically
significant, with and without the control variables."
"Most researchers agree that IQ positively affects earnings.
While they agree about the
direction of the correlation, they disagree about the strength
of the relationship. Previous
research on IQ and earnings has consistently shown a positive
correlation between IQ and
earnings, but this correlation has ranged in strength from .13
to .37 in general samples, with
results typically falling in the mid thirties (Lynn and Vanhaven
2002). The lowest result of .13
came from a study by Brown and Reynolds of 4008 black men
published in 1975 (In Lynn and
Vanhaven 2002). Lynn and Vanhaven conjecture that this low
coefficient of correlation is
caused by the high rates of poverty among blacks, who are less
likely to continue schooling and
enter higherpaying occupations (Lynn and Vanhaven 2002).
Zagorsky examines the effects of
IQ on “wealth, income, and financial distress” in an article for
Intelligence, a peerreviewed
multidisciplinary journal (Zagorsky 2007). He found a
correlation of 0.29 between IQ and
income. Interestingly, he also found that financial distress had
a quadratic relationship to IQ,
with higher IQ scores sometimes leading to higher chance of such
instances of financial distress
as bankruptcy or missed payments of bills (Zagorsky 2007).
Figure 2.1 below shows
Zagorsky’s data plot for income and IQ."
http://www.tutornext.com/ws/examplesofcorrelationresearch
Correlation is the degree to which two variables are related.
Mathematically it is represented by 'r'  Pearson's correlation coefficient. It
is a number that ranges from 1 to +1, where 1 is a perfect negative
correlation (one rises as the other falls), 0 is no correlation, and +1 is a
perfect positive correlation (both rise and fall together). Correlation does NOT
prove causation. Assuming one (at least partially) causes the other, the math
simply provide no way to know. Either variable can be calculated from the other.
For example, there is a correlation between education and income. Does increased
education result in increased salaries (most people would say yes). However, do
people with more money buy more education? That is almost certainly true as
well. There can be very good correlations where neither cause the other. It may
be that both are caused by some third variable. For example, there is a
correlation between the price of Irish Whiskey and Catholic Priests' salara..
http://economix.blogs.nytimes.com/2009/08/27/satscoresandfamilyincome/
August 27, 2009, 1:01 pm SAT Scores and Family Income
By
CATHERINE RAMPELL
Much has been written about the relationship between SAT
scores and testtakers’ family income. Generally speaking, the
wealthier a student’s family is, the higher the SAT score.
Let’s take a look at how income correlated with scores
this year. About twothirds of testtakers voluntarily
report their family incomes when they sit down to take the SAT.
Using this information, the College Board breaks down the
average scores for 10 income groups, each in a $20,000 range.
First, here are the individual test sections:
Source:
College Board
Source:
College Board
Source:
College Board
Here are all three test sections next to each other (zoomed
in on the vertical axis, so you can see the variation among
income groups a little more clearly):
Source:
College Board
A few observations:
 There’s a very strong positive correlation
between income and test scores. (For the math
geeks out there, the R^{2} for each test
average/income range chart is about
0.95.) 
 On every test section, moving up an income
category was associated with an average score
boost of over 12 points. 
 Moving from the secondhighest income group
and the highest income group seemed to show the
biggest score boost. However, keep in mind the
top income category is uncapped, so it includes
a much broader spectrum of families by wealth.

http://answers.yahoo.com/question/index?qid=20090214150019AAtvXkR
Recently, a university surveyed recent graduates of the
English Department for their starting salaries.?Recently, a university surveyed
recent graduates of the English Department for their starting salaries. Four
hundred graduates returned the survey. The average salary was $25,000 with a
standard deviation of $2,500.
What is the best point estimate of the population mean?
A. $25,000
B. $2,500
C. 400
D. $62.5
What is the 95% confidence interval for the mean salary of all graduates from
the English Department?
A. [$22,500, $27,500]
B. [$24,755, $25,245]
C. [$24,988, $25,012]
D. [$24,600, $25,600]2 years ago
Report Abuse
.apc_128
Best Answer  Chosen by VotersFirst question: A. $25,000
You are trying to estimate the mean, or average, of the population. The average
for the sample is $25,000. For this problem, we should assume that the sample is
representative of the population. The answer A, $25,000 out of the four choices
is closest to the sample mean value $25,000.
Second question: B. [$24,755, $25,245]
Without a graphing calculator:
To find confidence intervals without a graphing calculator, you do have to plug
it in the formula. The formula of a confidence interval for a mean where the
sample standard deviation is:
sample mean ± (tcritical value) [sample standard deviation/√(sample size)]
Note that we are missing a piece of information: the tcritical value. To obtain
the tcritical value, you must check a table. Your teacher will tell you where
to find this table.
The tcritical value we will use for this problem is 1.649. The rest of the
problem is just plugging your values in the formula. Note that you will get 2
values.
25,000 ± (1.649) [2500/√(400)]≈24754 or 25246
Since the boundaries of the confidence interval are 24754 and 25246, the closest
answer is B, [$24,755, $25,245].
With a graphing calculator:
To obtain your confidence interval on a TI83 graphing calculator, go to STATS,
then to TESTS. Scroll down to TInterval. Enter your data, and press Enter on
Calculate. The calculator will give you the interval of (24754, 25246). The
closest answer is B, [$24,755, $25,245].
Note: If your teacher asks you to show the formula and its values despite the
fact that you have a graphing calculator, you must use the invT( function, which
is not present in all TI graphing calculators, or the table mentioned above. I
have a TI84+ Silver Edition, which came with this function. If you have this
function in your TI graphing calculator, you may find it by going 2nd►DISTR and
scrolling down. When you input your values, you use confidence level and degrees
of freedom. To find the degrees of freedom, subtract 1 from the sample size.
invT(.95, 399)=1.648681473
http://www.brainmass.com/homeworkhelp/statistics/alltopics/133795
What is the probability that the mean salary offer for these
80 students is 24,250 or less? The mean salary offered to students who are
graduating from Coastal State University this year is $24,230, with a standard
deviation of $3,712. A random sample of 80 Coastal State students graduating
this year has been selected. What is the probability that the mean salary offer
for these 80 students is 24,250 or less?
Carry your intermediate computations to at least four decimal places. Round your
answer to at least three decimal places.
Can you explain this how to do this?
http://www.tutorvista.com/math/confidenceintervalstandarddeviation
Examples of Confidence Interval Standard Deviation :A Random
sample of 100 school teachers in a particular state has a mean salary of 31,578.
It is knwon from the previous data that the standard deviation of the salaries
of the teachers in the state is 4,415. Construct a 99% confidence interval
estimate for the true mean salary for public school teachers for a given state.
Solution: Given α = 0.01, Zα/2 = 2.576, x¨= 31578, n=100 , σ = 4,415 and σx = σ
/ √n = 441.5
Thus the 99 percent confidence interval estimate for the mean salary, using the
formula, is 31,578 ± 2.576 x 441.5 = 31,578 ± 1,137.3. That is we are 99 percent
confident that the average salary of the public school teachers for the given
state will be between 30,440.9 and 32,715.3.
http://answers.yahoo.com/question/index?qid=20080714153120AAKU6My
The weekly salaries of
teachers in one state are normally distributed
w/ a mean of $490 and a Std. dvtn of $45
What is the probability that a
randomly selected teacher earns
more than $525 a week?
by
pepsi4me...
 Member since:
 June 11, 2008
 Total points:
 10011 (Level
6)
Best Answer  Chosen by Asker
Convert to zscore (standard
deviations):
$525$490 = $35
$35/$45 = Z of 0.78
From Z score table:
The value from 0.78 to zero =
.2823
The value above 0.78 is .50 (the
entire positive side)  .2823 =
.2177 = 21.77%
 
