NONPARAMETRIC STATISTICS

1
NONPARAMETRIC STATISTICS
Vonnet Estaris, RND
2
Nonparametric Statistics

• or distribution-free statistics is used when the
  population from which the samples are selected is
  not normally distributed.
• This can also be used to test hypotheses that do
  not involve specific population parameters such
  as µ, s, or ?.

3
Advantages

• There are five advantages that nonparametric
  methods have over parametric methods
• They can be used to test population parameters
  when the variable is not normally distributed.
• They can be used when the data is nominal or
  ordinal.
• They can be used to test hypotheses that do not
  involved population parameters.
• In most cases, the computations are much easier
  than those for the parametric counterparts.
• 5. They are easy to understand.

4
Disadvantages

• There are three disadvantages of nonparametric
  methods
• They are less sensitive than the parametric
  counterparts when the assumptions of the
  parametric methods are met.
• They tend to use less information than the
  parametric tests.
• 3. They are less efficient than their parametric
  counterparts when the assumptions of the
  parametric methods are met.

5
Ranking

• Many nonparametric tests involve the ranking of
  data, that is, the positioning of a data value in
  a data array according to some rating scale.
  Ranking is an ordinal variable.

6

• For example, suppose a judge decides to rate five
  speakers on an ascending scale to 1 to 10, with 1
  being the best and 10 being the worst, for
  categories such as voice, gestures, logical
  presentation and platform personality.
• The ratings are shown in the chart.

7
The rankings are shown next.
What happens if two or more speakers receive
the same number of points? Suppose the judge
awards points as follows
8
The speakers are then ranked as follows
When there is a tie for two or more places, the
average of the ranks must be used. In this case,
each would be ranked as
2 3
5
2.5


2
2
9
Sign Test

• The sign test is a nonparametric test that can be
  used with a single group using the median rather
  than the mean.
• For example, we can ask Did the children in the
  study by Dennison and colleagues have the same
  median level energy intake as the 1286 kcal
  reported in the NHANES III study? (Because we do
  not know the median in the NHANES data, we assume
  for this illustration that the mean and median
  values are the same.

10
(No Transcript)
11

• If the median of the population of 2-year-old
  children is 1286, the probability is 0.50 that
  the any observation is less than 1286. (The
  probability is also 0.50 that any observation is
  greater than 1286). We count the number of
  observations less than 1286 and can use binomial
  distribution with p 0.50. The table contains
  the data on the energy level in 2-year-olds
  ranked from lowest to highest. Fifty-seven
  2-year-olds have energy lower than 1286 and 37
  have higher energy levels. The probability of
  observing X 57 out of n 94 values less than
  1286 using the binomial distribution.

12

• Step 1. The null alternative hypothesis are
• H0 The population median energy intake level
  in 2- year old children is 1286 kcal, MD 1286.
• H1 The population median energy intake level
  in 2- year-old children is not 1286 kcal, or MD ?
  1286.

• Step 2. Assuming energy intake is not normally
  distributed,
• the appropriate test is the sign test and
  because the
• sample size is large, we can use z
  distribution.

z X np - (1/2)
v np (1-p)
13

• X number of children with energy levels less
  than 1286
• n total number of children
• p probability is 0.5, to reflect the 50 chance
  that the observation is less than (or greater
  than) the median.
• Step 3. We use a 0.05 so we can compare the
  results with those found in the t test.
• Step 4. The critical value of the z distribution
  for a 0.05 is 1.96. So, if there is less than
  1.96 or greater than 1.96, we will reject the
  null hypothesis of no difference in median levels
  of energy intake.

14

• Step 5. The calculations are
• z 57 94(0.5) - 0.5
• v 94(0.5) (1-0.5)
• 57 47 - 0.5
• 4.85
• 9.5
• 4.85
• 1.96

• Step 6. The value of the sign test is 1.96 and
  is right on
• the line with 1.96. It is traditional that
  we do not reject
• the null hypothesis unless the value of the
  test statistics
• exceeds the critical value.

15
Wilcoxon Signed-Rank Test

• When the samples are dependent, as they would be
  in before-and-after test using the same subjects,
  the Wilcoxon signed-rank test can be used in
  place of the t test for dependent samples.

16

• Ex. In a large department store, the owner wishes
  to see whether the number of shoplifting
  incidents per day will change if the number of
  uniformed security officers is doubled. A sample
  of 7 days before security is increased and 7 days
  after the increase shows the number of
  shoplifting incidents.

17

• Is there enough evidence to support the claim, at
  a 0.05, that there is a difference in the
  number of shoplifting incidents before and after
  the increase in security?
• Solution
• Step 1. State the hypotheses and identify the
  claim.
• H0 There is no difference in the number of
  shoplifting incidents before and after the
  increase in security.
• H1 There is a difference in the number of
  shoplifting incidents before and after the
  increase in security. (claim)
• Step 2. The critical value of the z distribution
  for a 0.05 is 2. So, if there is less than 2
  or greater than 2, we will reject the null
  hypothesis of no difference in the number of
  shoplifting incidents before and after the
  increase in security.

18

• a. Make a table as shown here.

b. Find the differences (before and after), and
place the values in the Difference column.
19

• c. Find the absolute value of each difference,
  and place the results in the Absolute value
  column. (The absolute value of any number except
  0 is the positive value of the number. Any
  differences of 0 should be ignored.)

20

• d. Rank each absolute value from the lowest to
  highest, and place the rankings in the Rank
  column. In case of a tie, assign the values that
  rank plus 0.5.

21

• e. Give each column plus or minus sign,
  according to the sign in the Difference column.

22

• f. find the sum pf the positive ranks and the sum
  of the negative ranks separately.
• Positive rank sum (3.5) (5) (6)
  (3.5) (7) 25
• Negative rank sum (-1.5) (-1.5)
  -3
• g. Select the smaller of the absolute values of
  the sums (-3), and use this absolute value as
  the test value ws. In this case, ws. -3 3
• Step 4. Make the decision. Reject the null
  hypothesis if the test value is less than or
  equal to the critical value. In this case, 3 gt 2
  hence, the decision is not to reject null
  hypothesis.
• Step 5. Summarize the results. There is not
  enough evidence to support the claim that there
  is difference in the number of shoplifting
  incidents. Hence, the security increase probably
  made no difference in the number of shoplifting
  incidents.

23

• The rationale behind the signed-rank test can
  be explained by a diet sample. If the diet is
  working, then the majority of the postweights
  will be smaller then the preweights. When the
  postweights are subtracted from the preweights,
  the majority of the sign will be positive, and
  the absolute value of the sum of the negative
  will be small. This sum will probably be smaller
  than the critical value obtained, and the null
  hypothesis will be rejected. On the other hand,
  if the diet does not work, some people will gain
  weight, other will lose some weight and still
  other people will remain about the same weight.
  In this case, the sum of the positive ranks and
  the absolute value of the negative ranks will
  approximately equal and will be about one-half of
  the sum of the absolute value of all the ranks.
  In this case, the smaller of the absolute values
  of the two sums will still be larger than the
  critical value obtained and the null hypothesis
  will not be rejected.