Normal Distribution

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Normal Distribution

• HED 489 Biostatistics

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• Odds are that you've come across the normal
  distribution below. You may know it by it's more
  common namethe "bell-shaped curve or Guassian
  curve (knowing this is very impressiveuse it in
  a conversation with your family todaytheyll be
  impressed). In case you've never seen it before,
  that's the normal distribution on the right. We
  spend time understanding the normal distribution
  for two reasons
• it forms the basis of probability, and
  probability forms the basis of parametric
  statistical tests
• whether the distribution is normal is one of
  theperhaps the primary determinant of which
  family of statistical tests you will apply.

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Formation of The Normal Curve

• Assume for the moment that the data in the slide
  to the right represent ages of a bunch of people.
  As you can see, young people are on the left and
  older people on the right. Most of the people are
  between 9 and 11, right? That's the big bunch in
  the middle.

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• Now, let's say that we draw a line connecting the
  top midpoint of each bar. Here's what that would
  look like

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• See the nice straight lines you get? That's
  because, for purposes of this explanation, I've
  created the distribution such that those straight
  lines would result! Now, how about if we smooth
  the lines so they become a nice curve. That's
  what this next slide shows

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• See how we have the outline of a bell-shaped
  curve. This is an approximation of what would
  happen with these data. The last thing that
  happens when we have a normal distribution is
  that the outline becomes filled in with data.
  That's what this next slide shows.

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• So, if we strip away the unessential information,
  we're left with the first slide you saw in this
  lesson. Remember what it looked like?

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Normal Curve Properties

• Symmetry the mean bisects the distribution
  exactly, such that the two halves of the
  distribution form a mirror image of each other.
• Unimodal one mode
• Standardized characteristics the standard
  deviation of a standardized normal distribution
  will always be 1.0 with a mean of 0.
• Infinite theoretically, the "tails" of the
  distribution never contact the x-axis. In other
  words, the tails are infinite.

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Characteristics of a Normal Distribution

• Bell-shaped examine the shape of the curve
• Points of inflection the point at which a curve
  changes from concave to convex.
• Skewness the ratio of skewness to its standard
  error.
• lt-2 than left or negatively skewed gt2 than right
  or positively skewed.
• Kurtosis the ratio of kurtosis to its standard
  error.
• lt-2 than tails longer than expected gt2 than
  tails shorter than expected.
• Central Tendencies the mean, median, and mode
  have the same point estimate.
• Percentage of Scores for example, 68 of scores
  fall symmetrically within 1 standard deviation
  95 fall within 2, and 99 fall within 3
  standard deviation.

This is REALLY Important to knowLOOK it over and
KNOW these percentages
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Standardized normal distribution, z-scores, and
probability

• By standardizing normally distributed scores, one
  could better understand and compare scores.
• Standardizing is a process through which scores
  are transformed into a common scale (in this
  case, z-scores).
• Probabilities within the normal distribution
  could be represented by z-scores and visa versa
• The standardized normal distribution has a mean
  equal to 0 and a standard deviation equal to 1.

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Moving Towards Probability

• As you can see on the slide to the right, we can
  plot the location of various values by adding or
  subtracting the standard deviation (1.0) to or
  from the mean (0). See where positive and
  negative standard deviations fall on this
  distribution? Typically, we don't go beyond 3.0
  standard deviations, commonly abbreviated "s.d."
  We'll talk about why in just a bit.
• Remember, if we were measuring age, our raw data
  s.d. would be in increments of years. If we were
  measuring water consumption, our raw data
  increment might be ounces. However, both sets of
  data can be transformed into standardized scores
  (z-scores).

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• If you collect the ages of, say, 10,000 people,
  and build a frequency distribution, it will
  contain all 10,000 people, right? That goes
  without saying, doesn't it? Well, the same thing
  can be said of the normal distribution. That is,
  all of the data, or 100 of it for a particular
  variablelike agewill be contained within the
  distribution. For any normal distribution, no
  matter what the data are, 99.7 of the data will
  be contained in the space from -3 to 3 s.d. Do
  you see that in the slide on the right? See all
  the blue area between -3 and 3? That's where
  most of the data are. See how little blue there
  is to the left of -3 and to the right of 3?
  There are very few cases there. Combined, in
  fact, only .3 of the data are greater than 3 or
  less than -3.

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• So, if you had those 10,000 ages on slips of
  paper, and you selected one at random, it would
  fall between 3.0 s.d. 99.7 of the time. That is
  to say there is a probability of .997 of
  selecting an age at random that falls between
  3.0 s.d. Because the largest area under the
  curve falls between 1.0 s.d., it stands to
  reason that most of the cases will be contained
  in this area, too. As you can see from the slide,
  about 68.2 of the cases (34.1 2) fall between
  1.0 s.d.
• An additional 14 of the cases are contained
  between 1.0 and 2.0 s.d., and 14 more fall
  between -1.0 and -2.0 s.d. See that? In total,
  95.4 of the cases fall between 2.0 s.d.
• Because there's not much area under the curve
  between 2.0 and 3.0 s.d., only 2.2 of the cases
  will fall between 2.0 and 3.0 s.d., and between
  -2.0 and -3.0 s.d. Remember we said earlier that
  99.7 of the cases fall between 3.0 s.d.? Well,
  this is how we get to 99.7.
• If you didn't follow that, go back and study it
  again until you understand it.

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Taking Another Step

• This slide summarizes what you just learned. That
  is, about 68 of the cases fall between 1.0
  s.d., about 95 between 2.0 s.d., and about 99
  fall between 3.0 s.d. Now, pay close attention
  it's going to get tricky If about 95 of the
  cases fall between 2.0 s.d., then about 5 will
  be greater than -2.0 or 2.0 s.d. Do you see
  that? If all, or 100 of the cases fall somewhere
  along the curve, and you account for 95 of them
  between 2.0 s.d., then 5 are left. About 2.5
  of the cases fall to the left of -2.0 s.d. and
  another 2.5 fall to the right of 2.0 s.d.
• Still with me? Ok, contemplate this if you pick
  a value at random, there is a probability of .05
  that it will be greater than -2.0 or 2.0 s.d.
  Why? Because there's a probability of .95 that it
  will be between 2.0 s.d.

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Calculating the z-score

• Calculating the z-score is relatively simple. 
  Just follow the formula.

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• You will need to know the mean and the standard
  deviation.  Just punch in the score you need and
  you'll get the z-score.  For example, if you
  scored 100 on the test, and the mean is 80 and
  the standard deviation is 10, you'll have the
  formula as such
• z 100 - 80 / 10 z  20 /10z 2
• That is, your score was
• two standard deviations
• above the mean your
• score was higher than
• 97.7 of scores.

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The Normal Distribution What does the z Score
mean?

• It's time for you to open your textbook to the
  inside back cover (to Table A).  Click slide and
  it will appear on-screen (it is also shown on the
  next slide in this program). This table shows the
  area under the curve from zero to any point along
  the curve, out to three decimal points, to 4.0
  s.d.
• This table is also referred to a table of
  "z-scores." See the "z" in the upper left cell?
  For a normal distribution of data, the standard
  deviation is equal to z. More on that in a
  moment. First, I want to get you comfortable with
  this table.
• Remember that about 34 of the cases fall between
  zero and 1 s.d.? In case you forgot that, just
  look at the z-score table, and it will remind
  you. If we selected a case at random from a
  normal distribution of data that the probability
  is about .34 that the value of that case will
  fall between zero and 1.0 s.d.?

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• Look down the left column of Table A (normal
  curve or z-score table), either in the book or on
  the slide at the top. Move your finger one
  column to the right. That's the one labeled ".00"
  The value in the cell where your finger is
  pointing is .3413, right? (next to the z- score
  of 1.0) That's where I found "about" 34 or 34.

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z(score)-(mean of scores)/standard deviation

• The above formula is used to calculate the
  z-score.  What happens if you're interested in
  some s.d. other than 1.0, 2.0, or 3.0? Here's
  where the table really comes in handy! Say you're
  interested in the area under the curve (also
  known as "probability") for a s.d. (or z-score,
  as I want you to begin thinking, as well) of
  1.96? How do you get there. Well, run your finger
  down the first column until you get to 1.9, and
  then run across until you get to .06. What value
  did you find? .4750? If so, you did it just
  right!
• Note that the z-score table just shows half of
  the curve, that is, from zero to the right.
  That's because the curve is symmetrical, so if
  you want to know the area, or probability, for
  the other half of the curve, just double the
  tabled value. For example, if you wanted to know
  the probability for 1.96, multiply .4750 time 2.
  That would give you .9500, and you'd say that the
  probability of selecting a value at random that
  would fall between 1.96 s.d. would be .95.
• Now, I want to give you some practice working
  with the normal curve so I know that you've
  become comfortable with it.

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Calculate the percent under the curve between the
mean and a z-score (or s.d.) of -1.39.

• To do this, use the z-score table. Run your
  finger down the first column until you reach 1.3.
  Run your finger along the 1.3 row until you reach
  the .09 column. The figure in that cell is the
  area under the curve from 0.00 to 1.39 (or 0 to
  1.39). That number is .4177 (or 41.77). This
  is also the probability of selecting a value at
  random and having it fall between 0.00 and 1.39
  (or -1.39).

The probability of selecting a scorebetween the
mean and -1.39 sd or a z-score of -1.39 is .4177
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Calculate the probability of selecting a value at
random greater than a z-score (or s.d.) of 2.01.

• You should be able to find the area under the
  curve, also known as probability, for a value of
  2.01. Run down the first column until you get to
  2.0. Then across the row to the .01 column.
• However, consider the curve at the lower right.
  See the figure .4772? That's the area under the
  curve for 0.00 to 2.00. What you want is the area
  beyond, or to the right of 2.01. To get that, you
  have to subtract the probability for 0.00 to 2.01
  (.4778) from all of the area on the right side of
  the curve. That is, from 0.00 to infinity. What
  is this figure? Can't remember? (remember that
  .50 is to the right of the mean and .50 is to the
  left. Focus on the right .50-.4778 .0222.
  The probability of selecting a value greater than
  a z score of 2.01 is .0222. Look at the z-score
  table.
• Remember, report probability, not percentage.
  Click to get table.

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Calculate the probability of randomly selecting a
value that is greater than a z-score of 2.00 or
less than a z-score of -2.00.

• This assignment requires you to consider both
  sides of the normal curve. What are the chances
  of selecting a z score beyond 2.0 or -2.0?
• Like the last assignment, you need to calculate
  "what's remaining" under the curve from 0.00 to
  -2.00 and 0.00 to 2.00. To do this, you need to
  determine the probability from 0.00 to 2.00 (that
  is .4772 subtract this from .50 to get .0228)
  double that value to get the probability of
  obtaining a z score greater than 2.00 or less
  than -2.0 in this case, .0556.
• Click to get table.

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Assignment 1

• Calculate the z-score equivalents of these
  systolic blood pressure values 100, 120, 130,
  140, and 190, where the mean equals 126.2, and
  the standard deviation equals 18.8.  Click here
  to enter an excel file to do the work.

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Assignment 2

• Calculate the probability of selecting a blood
  pressure value at random greater than 160, where
  the mean is 126.20 and the standard deviation is
  18.80.
• To determine this value, you first need to
  calculate the equivalent z-score, as you did in
  the prior assignment. Then, determine the area
  under the curve represented by that z-score.
  Finally, calculate the area beyond, or greater
  than that value. The resulting value represents
  probability. Click here for an excel file to work
  on.

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Assignment 3

• Calculate the probability of selecting a blood
  pressure value at random between 110 and 135
  where the mean is 126.20 and the s.d is 18.80.
  Click here for an excel file to work on.

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• We're about to make the transition from
  descriptive to inferential statistics. The heart
  of inferential statistics is "statistical
  significance." This is the probability that the
  your calculated value was "real" or happened by
  random chance. Here's an example

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• Let's say that you're working with a group of
  people to get their blood pressure down. One of
  the things you do is put them on an exercise
  program to reduce their weight and to strengthen
  their cardiovascular system. All things being
  equal, if your program is successful, their blood
  pressure should come down.
• If you take individuals' blood pressure at the
  beginning of the study, calculate the mean, and
  take it again and average it again at the end of
  the study, the mean blood pressure at the end
  should be lower than at the beginning. However,
  how much does blood pressure have to be lowered
  to assert with some confidence that our program
  was successful? If the beginning group average
  was 128 and the ending was 124, is this
  difference large enough to claim programmatic
  success? What if the ending average was 120? 110?
  This is the issue central to statistical
  significance is the difference between the two
  values so close to the mean of the normal
  distributionzerothat the probability of
  selecting the value at random is too high to
  accept as "real?" Or, is it so far from the mean
  that it's out in one of the tails of the
  distribution, where the chances of pulling out of
  the hat of values randomly is very small, and,
  therefore, more likely "real?"
• Here are some ways that a difference in mean
  values before and after your high blood pressure
  occur by chance alone for any particular group
  of people, their blood pressure might go down for
  reasons other than our exercise program. Maybe
  they had a high salt diet when they started and
  cut down on their sodium intake. Maybe they were
  experiencing a lot of stress and they got it
  under control. Or, maybe their blood pressure
  just went down unexpectedly.
• Conversely, maybe your program actually worked.
  If so, you should be able to select another,
  similar group, conduct the same program, and find
  similar a similar difference between starting and
  ending blood pressure values. Not exactly the
  same, but the difference should be fall in the
  same general area on the normal curve. If you
  conducted this program with 100 similar groups
  and found about the same difference, see how
  you'd be pretty confident that your program
  actually worked? Well, you probably can only run
  it once, so you need that difference value to
  fall a good long way from the mean in order to
  have confidence that your program worked. That's
  what statistical significance does for you.

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• Remember the five blood pressure values that you
  worked with for the last assignment. Pretend
  that you conducted your study five times, and
  those numbers from the assignment represented the
  five studies.
• Many researchers use a statistical significance
  level of .05 as their critical level. This means
  that if the value is statistically-significant,
  it will occur by chance alone, less than 5 times
  in 100.
• Another way of looking at the .05 critical value
  is that it has to fall into one of the two tails
  of the normal distribution, and not from that big
  bunch of scores in the middle. Why? Because there
  are lots of scores in the middle, and you have a
  very good chance of selecting one of them by
  chance alone. So, if the mean difference score is
  in the "bulge" of the distribution, the
  probability that it happened by chance alone is
  too great for you to assume your program worked.

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• So, on the normal curve, in order for a value to
  be statistically-significant at the 95 level or
  greater, you have to have a z-score of greater
  than 1.96 or less than -1.96. That is, the value
  has to come from the area to the left of -1.96 or
  from the area to the right of 1.96. This is the
  only way that you can reduce your odds that the
  value you calculated occurred by random chance
  alone less than 5 times in 100.
• If the value you calculate is statistically-signif
  icant at, let's say the .05 level, you write it
  like this plt.05. "lt" stands for "less than." If
  it's not statistically-significant, you write
  n.s. or ns (meaning that pgt.05. "gt" stands for
  "greater than).
• The reason that we almost always consider both
  sides of the normal curve is because our
  calculated value might be greater or less than
  the mean. For the blood pressure example, our
  program may have failed so badly that we actually
  caused the average ending blood pressure to
  increase! We'll talk a bit more about one-tail vs
  two-tail tests a little later.