Basic Quantitative Methods in the Social Sciences (AKA Intro Stats)

1
Basic Quantitative Methods in the Social
Sciences(AKA Intro Stats)

• 02-250-01
• Lecture 4

2
A Quick Review

• The entire area under the normal curve can be
  considered to be a proportion of 1.00
• A proportion of .50 lies to the left of the mean,
  and a proportion of .50 lies to the right of mean

3
Area Under the Normal Distribution and Z-Scores

• Normal Distribution
• with z-score points
• of reference

4
Properties of Area Under the Normal Distribution

• Since the normal curve is a bell shape, the
  proportion of scores between whole z-scores is
  not equal
• For example, .3413 of the scores lie between the
  z-scores of 0 (the mean) and 1 (or -1), while
  only .1359 of the scores lie between the z-scores
  of 1 and 2 (or -1 and -2)

5
Properties of Area Under the Normal Distribution
.3413
.3413
.1359
.1359
.0215
.0215
.0013
.0013
Z -3 -2 -1 0
1 2 3
6
Properties of Area Under the Normal Distribution

• Z-scores Proportion under the curve
• -1 to 1 .6826 (.3413.3413)
• -2 to 2 .9544
• -3 to 3 .9974
• -4 to 4 1.0000
• Z-scores are expressed in standard deviation
  units, i.e., a z-score of -1 represents one
  standard deviation below (to the left of) the mean

7
Normal Distribution Example

• A study of 2500 University of Windsor students
  showed that the average amount of sleep lost in
  the week prior to writing a statistics exam (in
  hours) was normally distributed with 7.79
  and 1.75 (dont worry, this isnt real
  data!)
• This distribution is shown with the abscissa
  (x-axis) marked in raw score and z-score units

8
Normal Distribution Example
.3413
.3413
.1359
.1359
.0215
.0215
.0013
.0013
X 2.54 4.29 6.04 7.79 9.54
11.29 13.04 Z -3 -2 -1
0 1 2 3
Z -3 -2 -1 0
1 2 3
9
Example cont.

• We can see from this diagram that 34.13 of U of
  W students lost between 6.04 and 7.79 hours of
  sleep in the week prior to a stats test (between
  z-1 and z0)
• 13.59 of students lost between 9.54 and 11.29
  hours of sleep in that week (between z1 and
  z2)
• 49.87 of students lost between 2.54 7.79 hours
  of sleep (between z-3 and z0)
  (.0215.1359.3413 .4987 49.87)

10
Properties of Area Under the Normal Distribution

• The symbol is used to denote the z-score
  having area (alpha) to its right under the
  normal curve
• The proportion of area under the curve between
  the mean and a z-score can be found with the help
  of a table (Table E.10, Howell, p. 452) and a
  little math
• In this example, we want to know the area between
  the mean and z 0.20
• Look under the column mean to z at z0.20
• The proportion 0.0793
• Therefore, .0793 (or almost 8) is the proportion
  of data scores between the mean and the score
  that has a z score of 0.20

11
Example cont.

• This means that the area between the mean and z
  0.20 has an area under the curve of 0.0793

.0793
.4207
Z 0 0.20
12
Example cont.

• Since half of the normal distribution has an area
  of .5000, we can determine the area beyond z
  .20 by subtracting the area from the mean to z
  .20 from .5000
• Area beyond z.20 .5000 - .0793
• Area beyond z.20 .4207
• (Note If you look at the smaller portion in
  the table, you will see its .4207)

13
Example cont.

• Since the normal curve is symmetrical, the area
  between the mean and z -.20 is equal to the
  area between the mean and z .20

.0793
.0793
.4207
.4207
Z -0.20 0 0.20
14
Normal Distribution Table

• Table E.10 has 3 columns
• Mean to z
• Larger portion
• Smaller portion

15
Table Mean to z
16
Table Larger Portion
17
Table Smaller Portion
18
A Couple of Notes

• 1) Always report proportions (area under the
  curve) to four decimal places. This means that if
  you report an area as a percentage, it will have
  two decimal places (e.g., .7943 79.43)
• 2) When using Table E.10, be careful not to
  confuse z.20 with z.02 (this is a common
  mistake)
• 3) Remember that a negative z value has the same
  proportion under the curve as the positive z
  value because the normal distribution is
  symmetrical
• 4) When working on z-score problems, it is highly
  recommended that you draw a normal distribution
  and plot the mean, x, and their corresponding
  z-scores

19
Another Example!

• We often want to know what the area between two
  scores is, as in this example
• Assume that the marks in this class are normally
  distributed with 69.5 and 7.4. What
  proportion of students have marks between 50 and
  80?

20
Example Area Between 2 Scores

• 1) Calculate the z-scores for X values (50 80)
• z (50-69.5)/7.4 -19.5/7.4 -2.64
• z (80-69.5)/7.4 10.5/7.4 1.42
• 2) Find the proportions between the mean and both
  z-scores (consult Table E.10)
• z(-2.64) .4959 is the proportion between the
  mean and z.
• z(1.42) .4222 is the proportion between the
  mean and z.

21
Example Area Between 2 Scores

• Third, add these proportions together to find
  your answer
• .4959 .4222 .9181
• This means that 91.81 of students have Stats
  marks between 50 and 80

22
Smaller and Larger Portions

• Smaller portion proportion in the tail
• Larger portion proportion in the body
• Using the same data ( 69.5 and 7.4) we
  can calculate areas using the Smaller and Larger
  Portions in the Normal Distribution table
• Find the number of students who have stats marks
  of less than 80.6
• z (80.6-69.5)/7.4 1.5

23
Larger Portion

• Area below z 1.5 0.9332
• This means that 93.32 of students had a mark of
  80.6 or less in this class

24
Smaller Portion

• Find the number of students who have marks of
  76.93 or better
• z (76.93-69.5)/7.4 1.00
• Area in smaller portion .1587
• This means that 15.87 of students in this class
  had a mark of 76.93 or better

25
Converting Back to X

• Assume 30 and 5, what raw scores
  correspond to z-1.00 and z1.5?

26
Proportion

• What proportion of scores lie between z-1.00 and
  z1.50?
• Area from mean to z-1.00 .3413
• Area from mean to z1.50 .4332
• Add them together to get the proportion that lies
  between these two z-scores .3413.4332 .7745

27
Finding for Number of Observations

• In this example, if we know the sample size,
  (e.g., n212) we can calculate how many people
  lie between z-1.00 and z1.50
• Area between z-1.00 and z1.50 .7745 (see the
  last slide)
• Multiply the proportion by n
• (.7745)(212) 164.19
• Approximately 164 people

28
And a Little More

• Finally, we can find a z-score from the table if
  we know the proportion of scores (i.e., we can
  work backwards)
• Suppose the birth weight of newborns is normally
  distributed with 7.73 and 0.83
• What birth weight identifies the top (heaviest)
  10 of newborns?

29
Example cont.

• Look at Table E.10 and find the z-score that
  identifies the top proportion of 0.1000 look in
  the smaller portion column (the tail)

.1000
z ?
30
Example cont.

• Looking in the smaller portion column, we find
  that
• z1.28 has an area of .1003
• z1.29 has an area of .0985
• Which do we pick?
• Pick the one that is closest to an area of .1000
  this is z1.28

31
Example cont.

• Now solve for X
• X (1.28)(0.83) 7.73
• 1.06 7.73 8.79
• So any weight equal to or greater than 8.79
  pounds is in the top 10 of birth weights

32
Probability

• Everything that can possibly happen has some
  likelihood of happening probability is a measure
  of that likelihood
• Probability The quantitative expression of
  likelihood of occurrence

33
Probability

• Probability is a ratio of frequencies
• The numerator (top) is the frequency of the
  outcome of interest
• The denominator (bottom) is the frequency of all
  possible outcomes

34
Coin Toss Example

• If a fair coin is tossed in the air, it can land
  on either heads or tails
• This means a coin has 2 possible outcomes
• If we want to know the probability of tossing a
  fair coin and having it land on heads, we
  calculate as follows
• Note fair means a normal coin, one that is not
  weighted differently

35
Coin Toss

• Frequency of interest
• Frequency of all possible outcomes
• For a coin toss, this is
• 1
• 2
• The probability of the coin landing on heads is
  p(heads) ½, or p(heads) .5

36
Another Example

• Suppose there are 90 students in a class, 59 of
  them are women and 31 are men
• If one of the students is chosen at random, the
  probability of choosing a woman is
• p(woman) 59/90

37
More Probability

• If the entire class was women (e.g., there were
  no male students), the probability of choosing a
  woman would be 90/90
• If the entire class was men, the probability of
  choosing a woman would be 0/90

38
More Probability

• As a numerical value, probabilities can range
  from 0.00 to 1.00
• The numerator can range from a minimum of 0 to a
  maximum equal to the denominator

39
Express Yourself!

• Probability can be expressed as a fraction, e.g.,
  p(woman) 59/90
• Or as a decimal fraction p(woman)
  .6556
• Although not usually expressed as a percentage
  (e.g., 65.56), they often are in popular media

40
Probability cont.

• Even if we do not know the actual observed
  frequencies (e.g., the number of women),
  probabilities can be determined theoretically
• Without throwing a die, we can deduce the
  probability of landing on a 5

41
Die Example cont.

• We know the die has 6 sides - 6 possible
  outcomes
• We are only interested in one side (the 5), so
  the probability of landing on a 5 is
• p(5) 1/6 0.1667

42
Probability and the Normal Distribution

• The normal distribution can be thought of as a
  probability distribution. Heres how
• We know (from Table E.10) the proportion of
  scores that fall above or below a given z score
• If you were to randomly pick a score from a
  sample of scores, what is the probability that
  you would pick a score that has a corresponding z
  score of .40 or greater?

43
Probability and the Normal Distribution

• The proportion of scores above or below a given z
  score is the same as the probability of selecting
  a score above or below the z score
• e.g., the probability of selecting a score from a
  normal distribution that has a z score of .40 or
  greater is .3446 (the area in the smaller portion
  of z .40)

44
Example 1

• Suppose peoples scores on a personality test are
  normally distributed with a mean of 50 and a
  population standard deviation of 10.
• If you were to pick a person completely at
  random, what is the probability that you would
  pick someone with a score on this personality
  test that is higher than 60?

45
Example 1

• Step 1 Write down what you know
• Step 2 What do you want to find?
• Step 3 Draw the normal distribution, write in
  the mean, standard deviation, and the X and shade
  the area you are looking for

46
Example 1, Step 3
X 20 30 40 50
60 70 80
47
Example 1

• Step 4 Calculate z score(s)
• Step 5 Use Table E.10 to find the probability
  of selecting a score in your shaded area
• Here we want or
• Look up the smaller portion of z1.00

48
Example 1

• Step 6 Interpret
• The probability of picking someone at random who
  has a personality test score of 60 or greater is
  .1587

49
Example 2

• Length of time spent waiting in line to buy
  tickets at the movies is normally distributed
  with a mean of 12 minutes and a population
  standard deviation of 3 minutes.
• If you go to see a movie, what is the probability
  that you will wait in line to buy tickets for
  between 7.5 and 15 minutes?

50
Example 2

• Step 1 Write down what you know
• Step 2 What do you want to find?
• Step 3 Draw the normal distribution, write in
  the mean, standard deviation, and both X scores
  and shade the area you are looking for

51
Example 2, Step 3
X 3 6 7.5 9 12
15 18 21
52
Example 2

• Step 4 Calculate z score(s)
• Step 5 Use Table E.10 to find the probability
  of selecting a score in your shaded area
• Here we want
• or
• Look up the mean to z of z 1.00 .3413
• Look up the mean to z of z -1.50 .4332

53
Example 2

• Add the two areas together! (Each represent the
  mean to z, so adding them together gives you the
  overall shaded area) .3413.4332.7745

54
Example 2

• Step 6 Interpret
• The probability of waiting in line to buy tickets
  at the movie for between 7.5 and 15 minutes is
  .7745. (Note This means that you will wait in
  line for between 7.5 and 15 minutes 77.45 of the
  time).