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GAINS & LOSSES BY PRESIDENT’S PARTY IN MIDTERM ELECTIONS. MEAN: VARIANCE: ST. DEV.:. What Does it Mean?. On Average, each observation is 17.7 seats away from the mean seat loss (roughly 18 seats). N(-26, 18)--a normal distribution with a mean of negative 26 and standard deviation of 18
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St. Dev. = 18
The Seat Loss Example:
-80 -62 -44 -26 -8 +10 +28
This rule tells us that sixty-eight percent of all observations in a
normally distributed frequency distribution will be +1 to –1
standard deviations from the mean. The standard deviation in our
example is 18.
If the distribution of scores (here seat losses) is more or less
normally distributed, this means that 68% of the cases are + or –
one SD from the mean of 26. So 68% of the cases are between
minus 18 seat losses – negative 26 plus minus 18 - 44 (seat
losses); negative 26 plus 18 = - 8.
-80 -62 -44 -26 -8 +10 +28
The 68-95-99.7 rule allows us to predict that in 95%
of all midterm election the number of seat losses
would be within + 2 sd of the mean -- the mean of
26 minus (18 times 2sd =36 which added to negative
26) equals 62 seats lost. And on the other side 2
standard deviations above the sample mean of -26 is
again 36 which adds up to a gain of 10 seats.
A most important class of frequency distributions is the normal distribution – defined as N [μ, σ] [mu, sigma]. First NDCs describe many natural occurring phenomena, physical, biological, psychological and social. NDCs represent the common case where as people, places and things become more extreme they become less frequent. The average score is most common and the frequency of scores decreases as they diverge from the true score -- the true score being the mean.
All NDCs, not just the ideal bell shape distribution -- is that the area under the curve is the same for all NDCs -- it is100%, meaning that all the cases/observations/people are under the curve. It includes everybody. The frequency distribution contains all the cases or all the people or all scores of the population or sample.
A standard distribution curve can be defined exactly
by its mean and standard deviation
Whatever the mean – as long as the distribution is normally distributed -- and whatever the sd the 68-95-99.7 rule applies -- 68%, two-thirds of all the cases fall within + 1 sd of the mean, 95% of the cases within 2 sd of the mean, and 99.7% of the cases are within + 3 sd of the mean.
In the real world many measures tend to bunch up in the middle. The center is the average, with extreme scores less frequent.
The Standardized Score or Z-score
Z is defined for a population as
and X is any raw score.
But usually, we don’t know the population mean.
Tells us: for any individual (i), how many standard deviations they are from the mean.
We simply take their score (Xi) and subtract the mean, and then divide by the standard deviation.
Doing this allows us to change any value into a standardized value.
Standardized normal distribution has mean of zero and S = 1.
We can create an entire variable that is simply the standardized version of the original variable.
Instead of discussing how many seats above or below the mean were lost in an election we can talk about how many standard deviations above or below the mean the seat loss was.
This allows us to compare one variable to another on the same scale.
We can do this for any interval variable.
Once we have done so, it is easier to make conclusions about the proportion of cases within a given interval.
Which is the same as saying the probability of finding the next case in a given interval.
The total area under the curve is 1.
Half to the right and half to the left.
Again, the mean of the standardized normal is zero.
We want to know what is the probability of a value between Z=0 and Z=a
The area between two points that we want to figure out is our ALPHA-area.
We already know some of these values.
Between 0 and ∞ .5
0 and -∞ .5
1.00 – .84 = .16 => 16% above 1 sd.
Another way : 68% + 1 sd leaves 32% at the tails. Because the NDC is symmetrical , 32/2= 16% above and below at each tail.
From X ~ N with Mean=500 and S=100 what’s the probability of a score between 400 and 600 ?
+ 1 sd either side of the mean = .68 probability
The interpretation of scores in terms of their relative position helps us interpret their relative probability – the relative extremity or typicality – of some event.
Where does a SAT score of 532 stand??
X=532, somewhere between m=500 and +1s, which is at s=600:
500 ——— 532 ——————— 600
m X +1s
Score of 532 is somewhere between 500 and 600, but where precisely? What is the exact probability of this event occurring?
With the Z score statistic we can interpret the probability of any score occurring anywhere within a normal distribution, say a score of 532, not just whole steps of 1 or 2 or 3 sd steps.
In the Z-Score formula, what is the purpose/significance of dividing the deviation between a score [a case or observation] and its distribution mean / by the sd?
Said to "standardize" the distribution. Turns every normal distribution in the family of NDC into a standard shape with the same proportion of cases being within 1, 2, 3 or more standard deviations from the mean.
IQ test scores are normally distributed with a m = 100 and sd=10. What is the Z-scores for an IQ score of …
… 110? Z = (110-100)/10 = 10/10 = Z = 1.0
… 120? (120-100)/10 = 20/10 = Z = 2.0
… 105? (105-100)/10 = 5/10 = Z = 0.50
… 90? (90-100)/10 =-10/10 = Z = -1.0
How do you interpret a score of 109?
Use Z-score = (X-m)/s (109-100)/10=9/10=0.90
What does this Z-score of .90 mean?
Does not mean 90 percent of cases below this score BUT rather that this Z score is .90 standard units above the mean, almost one sd above the mean.
Luckily, statisticians have already calculated the probabilities for every unit of sd. The tables -- called standardized normal probability tables or simply z-tables -- appear in every statistical text. The z table gives the probability of getting a specific score from a normalized distribution.
The z score tells you how distance in units of standard deviation translates into probabilities.
Variance = 100, so S=10
109 - 100
Z = ————= 0.9
What’s the probability of a score above 0.9???
31.59% of cases fall between the mean of this distribution and the Z score of .9.
What percent of scores by chance would be below the Z score of +.90?
50% + 31.59 = 81.59 81.59% of all scores are below a Z-score of .90.What percentage of scores would be beyond a Z-score of .90? That is at or above an X-score of 109?
Because the total area under the curve is 1.0, we can subtract the probability of .8159 from one: (1-.8159) = .1841meaning that 18.41% of all scores are at or above X = 109 or Z = .90.