Statistical Techniques I

1

• Statistical Techniques I

EXST7005
The t distribution
2

• The t-test of Hypotheses

• the t distribution is used the same as Z
  distribution, except it is used where sigma (s)
  ,is unknown (or whereY is used instead of m to
  calculate deviations)
• ti (Yi - Y)/S
• t (Y - m0)/SY (Y - m0)/(S/Ön)
• where
• S the sample standard deviation, (calculated
  using Y)
• SY the sample standard error

3

• The t-test of Hypotheses (continued)

• E(t) 0
• The variance of the t distribution is greater
  than that of the Z distribution (except where n
  ), since S2 estimates s2, but is never as good
  (reliability is less)
• Z distribution t distribution
• mean 0 0
• variance 1 ³ 1

4

• CHARACTERISTICS OF THE t DISTRIBUTION

• It is symmetrically distributed about a mean of 0
  
• t ranges to (i.e. - t )
• There is a different t distribution for each
  degree of freedom (df), since the distribution
  changes as the degrees of freedom change.
• It has a broader spread for smaller df, and
  narrows (approaching the Z distribution) as df
  increase

5

• CHARACTERISTICS OF THE t DISTRIBUTION (continued)

• 4) As the df (g, gamma) approaches infinity (),
  the t distribution approaches the Z distribution.
  For example
• Z (no df associated) middle 95 is between
  1.96
• t with 1 df middle 95 is between 12.706
• t with 10 df middle 95 is between 2.228
• t with 30 df middle 95 is between 2.042
• t with df middle 95 is between 1.96

6

• Tables in general

• The tables we will use will ALL be giving the
  area in the tail (a). However, if you examine a
  number of tables you will find that this is not
  always true. Even when it is true, some tables
  will give the value of a as if it were in two
  tails, and some as if it were in one tail.
• For example, we want to conduct a two- tailed Z
  test at the a0.05 level. We happen to know that
  Z1.96.

7

• Tables in general (continued)

• If we look at this value in the Z tables we
  expect to see a value of 0.025, or a/2. But many
  tables would show the probability for 1.96 as
  0.975, and some as 0.05.
• Why the difference? I just depends on how the
  tables are presented.
• Some of the alternatives are shown below.

8

• Tables in general (continued)

• Table gives cumulative distribution starting at -
  infinity. You want to find the probability
  corresponding to 1-a/2.

Table value, 0.975
9

• Tables in general (continued)

• Some tables may also start at zero (0.0) and give
  the cumulative area from this point. This would
  be less common. The value that leaves .025 in
  the upper tail would be 0.475.
• Among the tables like ours, that give the area in
  the tail, some are called two tailed tables and
  some are one tailed tables.

10

• Tables in general (continued)

• One tailed table.

Table value, 0.0.025
11

• Tables in general (continued)

• Two tailed table.

Table value, 0.050
12

• Tables in general (continued)

• Why the extra confusion at this point?
• The Z tables we used gave the area in one tail.
  For a two tailed test you doubled it.
• All our tables will give the area in the tail.
• For the F tables and Chi square tables covered
  later, this area will be a single tail as with
  the Z tables. This is because these
  distributions are not symmetric.

13

• Tables in general (continued)

• Traditionally, many t-tables have given the area
  in TWO TAILS instead of on one tail.
• Most older textbooks have this type of tables.
• SAS will also usually give two-tailed values for
  t-tests
• Our tables will have both two-tailed
  probabilities (top row) and one-tailed
  probabilities (bottom row), so you my use either.
  

14

• Tables in general (continued)

• The same patterns are true for many of the
  computer programs that you may use to get
  probabilities. For example in EXCEL
• If you use the NORMDIST(1.96) function it returns
  a value of 0.975, cumulative from -
• If you enter NORMSINV(0.025) it returns -1.96.
• If you enter TINV(0.05,9999) it returns 1.96, so
  it is two-tailed.
• The TDIST(1.96,9999,1) function allows you to
  specify 1 or 2 tails in the function call.

15

• The T TABLES

• My t-tables are created in EXCEL, but patterned
  after Steel Torrie, 1980, pg. 577
• df (g) is given on the left side.
• The probability of randomly selecting a larger
  value of t is given at the top (and bottom) of
  the page.
• P(t ³ t0) given at the bottom
• P(t ³ t0) given at the top

16

• The T TABLES (continued)

• Each row represents a different t distribution
  (with different df).
• The whole Z table was a single distribution
• The t table has many different distributions.
• Only the POSITIVE side of the table is given, but
  as with the Z distribution, the t distribution is
  symmetric

17

• The T TABLES (continued)

• The Z table had many probabilities, corresponding
  to Z values of 0.00, 0.01, 0.02, 0.03, etc.
  About 400 in the tables we used. They all fit on
  one page.
• If we are going to give many different
  t-distributions on one page, we lose something.
  We will only give a few selected probabilities,
  the ones we are most likely to use.
• e.g., 0.10, 0.05, 0.025, 0.01, 0.005.

18
Partial t-table - 1 or 2 tails?

• df

0.100
0.050
0.025
0.010
0.005
1
3.078
6.314
12.71
31.82
63.656
2
1.886
2.920
4.303
6.965
9.925
3
1.638
2.353
3.182
4.541
5.841
4
1.533
2.132
2.776
3.747
4.604
5
1.476
2.015
2.571
3.365
4.032
6
1.440
1.943
2.447
3.143
3.707
7
1.415
1.895
2.365
2.998
3.499
8
1.397
1.860
2.306
2.896
3.355
9
1.383
1.833
2.262
2.821
3.250
10
1.372
1.812
2.228
2.764
3.169

1.282
1.645
1.960
2.326
2.576
19
Partial t-table - 1 or 2 tails?

• df

0.200
0.100
0.050
0.020
0.01
1
3.078
6.314
12.71
31.82
63.656
2
1.886
2.920
4.303
6.965
9.925
3
1.638
2.353
3.182
4.541
5.841
4
1.533
2.132
2.776
3.747
4.604
5
1.476
2.015
2.571
3.365
4.032
6
1.440
1.943
2.447
3.143
3.707
7
1.415
1.895
2.365
2.998
3.499
8
1.397
1.860
2.306
2.896
3.355
9
1.383
1.833
2.262
2.821
3.250
10
1.372
1.812
2.228
2.764
3.169

1.282
1.645
1.960
2.326
2.576
20

• Our t-tables

• Selected d.f. on the left side.
• The table stabilizes fairly quickly. Many tables
  don't go over about d.f. 30. The Z tables give
  a good approximation for larger d.f.
• Our tables will give d.f. as follows down the
  leftmost column of the table,
• 1, 2, 3, 4, 5 ,6, 7, 8, 9, 10, 11, 12, 13, 14,
  15, 16, 17, 18, 19 ,20, 21, 22, 23, 24, 25, 26,
  27, 28, 29, 30, 32, 34, 36, 38, 40, 45, 50, 75,
  100,

21

• Our t-tables

• Selected probabilities.
• In the topmost row of the table selected
  probabilities will be given as a for a TWO TAILED
  TEST.
• In the bottommost row of the table selected
  probabilities will be given as a for a ONE TAILED
  TEST.
• Probabilities in out tables are,
• 0.50 0.40 0.30 0.20 0.10 0.050 0.02 0.010 0.002
  0.0010
• 0.25 0.20 0.15 0.10 0.05 0.025 0.01 0.005 0.001
  0.0005

22

• The T TABLES (continued)

• HELPFUL HINT Don't try to memorize "two tail
  top, one tail bottom", just recall the
  characteristics of the distribution when df
  and t 1.96. This leaves 5 in both tails and
  2.5 in one tail.
• So look at any t-table and look to see what
  probability corresponds to df and t 1.96. If
  the value is 0.025, it is the area in one tail.
  If it is 0.050 it is a two tailed table. If the
  area is 0.975, ...

23

• The T TABLES (continued)

• This trick of recalling 1.96 also works for
  different Z tables
• The tables we use give the area in the tail of
  the distribution, Z1.96 corresponds to a
  probability of 0.025
• Some Z tables give the cumulative area under the
  curve starting at -, the probability at Z1.96
  would be 0.975
• Other Z tables give the cumulative area starting
  at 0, the probability at Z1.96 would be 0.475

24

• Working with our t-tables

• Example 1. Let d.f. g 10
• H0 m m0 versus H1 m ¹ m0
• a 0.05
• P(t³t0)0.05 2P(t³t0)0.05 P(t³t0)0.025
  (Probabilities at the top of the table)
• t02.228

25

• Working with our t-tables (continued)

• Example 2. Let d.f. g 10
• H0 m m0 versus H1 m ³ m0
• a 0.05
• P(t³t0)0.05 (probabilities at the bottom of the
  table)
• t01.812

26

• Working with our t-tables (continued)

• Look up the following values.
• Find the t value for H1 m ¹ m0, a0.050, d.f.
  1.960
• Find the t value for H1 m ³ m0, a0.025, d.f.
  1.960
• Find the t value for H1 m ¹ m0, a0.010, d.f.12
  3.055
• Find the t value for H1 m ³ m0, a0.025, d.f.22
  2.074
• Find the t value for H1 m ¹ m0, a0.200, d.f.35
  1.306
• Find the t value for H1 m ¹ m0, a0.002, d.f.5
  5.894
• Find the t value for H1 m m0, a0.100, d.f.8
  -1.397
• Find the t value for H1 m m0, a0.010, d.f.75
  -2.377
• Find the P value for t-1.740, H1 m m0,
  d.f.17 0.050
• Find the P value for t4.587, H1 m ¹ m0, d.f.10
  0.001

27

• t-test of Hypothesis

• We want to determine if a new drug has an effect
  on blood pressure of rhesus monkeys before and
  after treatment. We are looking for a net change
  in pressure, either up or down (two tailed test).
  
• We obtain a random sample of 10 individuals.
  Note n 10, but d.f. g 9
• 1) H0 m m0
• 2) H1 m ¹ m0

28

• t-test of Hypothesis (continued)

• 3) Assume
• Independence (randomly selected sample)
• CHANGE in blood pressure is normally distributed.
  
• 4) a 0.01

29

• t-test of Hypothesis (continued)

• 5) Obtain values from the sample of 10
  individuals (n 10). The values for change in
  blood pressure were
• 0, 4, -3, 2, 0, 1, -4, 5, -1, 4
• SYi 8 SYi2 88 Y SYi /n 8/10 0.8
• S2 (SYi2-(SYi)2)/(n-1) (88-6.4)/9 9.067
• S Ö9.067 3.011
• SY S/Ön 3.011/Ö10 0.952
• t (Y-m0)/SY (0.8-0)/0.952 0.840
• with 9 d.f.

30

• t-test of Hypothesis (continued)

• 6) Get the critical limit and compare to the test
  statistic.
• Critical value of t for
• a 2 tailed test
• with 9 d.f. (n 10, but d.f. g 9)
• a 0.01
• P(t³t0)0.01 2P(t³t0)0.01 P(t³t0)0.005
• t0 3.250
• test statistics t (Y-m0)/SY 0.840 (9 d.f.)

31

• t-test of Hypothesis (continued)

• 6) Get the critical limit and compare to the test
  statistic.
• t0 3.250
• The area leaving 0.005 in each tail is almost too
  small to show on our usual graphs

32

• t-test of Hypothesis (continued)

• 6) Get the critical limit and compare to the test
  statistic.
• t0 3.250
• test statistics t (Y-m0)/SY 0.840 (9 d.f.)
• The test statistic is clearly in the region of
  "acceptance", so we fail to reject the H0.
• 7) Conclude that the new drug does not effect the
  blood pressure of rhesus monkeys.
• Is there an error? Maybe a Type II error.

33

• t-test of H0 Example 2

• A company manufacturing environmental monitoring
  equipment claims that their thermograph (a
  machine that records temperature) requires (on
  the average) no more than 0.8 amps to operate
  under normal conditions. We wish to test this
  claim before buying their equipment. We want to
  reject the equipment if the electricity demand
  exceeds 0.8 amps.

34

• t-test of H0 Example 2 (continued)

• 1) H0 m m0
• 2) H1 m ³ m0, where m0 0.8
• 3) Assume (1) independence and (2) a normal
  distribution of amp values, or at least of the
  mean that we will test.
• 4) a 0.05

35

• t-test of H0 Example 2 (continued)

• 5) Draw a sample. We have 16 machines for
  testing. The values for amp readings were not
  recorded. Summary statistics are given below
• Y 0.96
• S 0.32
• SY S/Ön 0.32/Ö16 0.08
• t (Y-m0)/SY (0.96-0.8)/0.08 2
• with 15 d.f.

36

• t-test of H0 Example 2 (continued)

• 6) Get the critical limit and compare to the test
  statistic.
• Critical value of t for
• a 1 tailed test (see H1)
• with 15 d.f. (n 16, but d.f. g 15)
• a 0.05
• P(t³t0)0.05 t0 1.753

37

• t-test of H0 Example 2 (continued)

• 6 continued) Get the critical limit and compare
  to the test statistic.
• t0 1.753
• test statistics t (Y-m0)/SY 2 (15 d.f.)

t0 1.753
38

• t-test of H0 Example 2 (continued)

• 6 continued) Get the critical limit and compare
  to the test statistic.
• The test statistic falls in the area of
  rejection.
• 7) Conclusion We would conclude that the
  machines require more electricity than the
  claimed 0.8 amperes.
• Of course, there is a possibility of a Type I
  error.

39

• t test with SAS

• Recall our test of blood pressure change of
  Rhesus monkeys. We can take the values of blood
  pressure change, and enter them in SAS PROC
  UNIVARIATE.
• Values 0, 4, -3, 2, 0, 1, -4, 5, -1, 4
• SAS PROGRAM DATA step
• OPTIONS NOCENTER NODATE NONUMBER LS78 PS61
• TITLE1 't-tests with SAS PROC UNIVARIATE'
• DATA monkeys INFILE CARDS MISSOVER
• TITLE2 'Analysis of Blood Pressure change in
  Rhesus Monkeys'
• INPUT BPChange
• CARDS RUN

40

• t test with SAS (continued)

• SAS PROGRAM procedures
• PROC PRINT DATAmonkeys RUN
• PROC UNIVARIATE DATAmonkeys PLOT VAR BPChange
• TITLE2 'PROC Univariate on Blood Pressure
  Change' RUN

41

• t test with SAS (continued)

• SAS PROGRAM output
• Moments
• N 10 Sum Wgts 10
• Mean 0.8 Sum 8
• Std Dev 3.011091 Variance 9.066667
• Skewness -0.15751 Kurtosis -0.95777
• USS 88 CSS 81.6
• CV 376.3863 Std Mean 0.95219
• TMean0 0.840168 PrgtT 0.4226
• Num 0 8 Num gt 0 5
• M(Sign) 1 PrgtM 0.7266
• Sgn Rank 6.5 PrgtS 0.3984

42

• Notes on SAS PROC Univariate

• Note that all values we calculated match the
  values given is SAS.
• Note that the standard error is called the "Std
  Mean". This is unusual, it is called the "Std
  Error" in most other SAS procedures.
• The test statistic value matches our calculated
  value (0.840).
• SAS also provides a "PrgtT 0.4226"

43

• Notes on SAS PROC Univariate (continued)

• The value provided by SAS is a P value (PrgtT
  0.4226)
• It indicates that the calculated value of t0.840
  would leave 0.4226 (or 42.46 percent) of the
  distribution in the tails. Note that it is for
  the absolute value of t, so it leaves 0.2113 (or
  21.13 ) in each tail.

44

• Notes on SAS PROC Univariate (continued)

• The P-value indicates our calculated value would
  leave 21.13 in each tail, our critical region
  has only 0.5 in each tail. Clearly we are in
  the region of "acceptance".

Calculated value
Upper Critical region
Lower Critical region
45

• Example 2 with SAS

• Testing the thermographs using SAS PROC
  UNIVARIATE. We didn't have data, so we cannot
  test with SAS.
• A NOTE. SAS automatically tests the mean of the
  values in PROC UNIVARIATE against 0. In the
  thermograph example our hypothesized value was
  0.8, not 0.0.
• But from what we know of transformations, we can
  subtract 0.8 from each value without changing the
  characteristics of the distribution.

46

• Example 2 with SAS (continued)

• Another example .
• Freund Wilson (1993) Example 4.2
• We receive a shipment of apples that are supposed
  to be "premium apples", with a diameter of at
  least 2.5 inches. We will take a sample of 12
  apples, and test the hypothesis that the mean
  size is equal 2.5 inches, and thus qualify as
  premium apples. If LESS THAN 2.5 inches, we
  reject.

47

• Example 2 with SAS (continued)

• 1) H0 m m0
• 2) H1 m m0
• 3) Assume
• Independence (randomly selected sample)
• Apple size is normally distributed.
• 4) a 0.05
• 5) Draw a sample. We will take 12 apples, and let
  SAS do the calculations.

48

• Example 2 with SAS (continued)

• The sample values for the 12 apples are
• 2.9, 2.1, 2.4, 2.8, 3.1, 2.8, 2.7, 3.0, 2.4, 3.2,
  2.3, 3.4
• As mentioned, SAS automatically tests against
  zero, and we want to test against 2.5. So, we
  subtract 2.5 from each value and test against
  zero. The test should give the same results.

49

• Example 2 with SAS (continued)

• SAS Program data step
• options ps61 ls78 nocenter nodate nonumber
• data apples infile cards missover
• TITLE1 'Test the diameter of apples against 2.5
  inches'
• LABEL diam 'Diameter of the apple'
• input diam diff diam - 2.5
• cards run
• SAS Program procedures
• proc print dataapples var diam diff run
• proc univariate dataapples plot var diff run

50

• Example 2 with SAS (continued)

• SAS PROC UNIVARIATE Output
• Moments
• N 12 Sum Wgts 12
• Mean 0.258333 Sum 3.1
• Std Dev 0.394181 Variance 0.155379
• Skewness -0.11842 Kurtosis -0.8353
• USS 2.51 CSS 1.709167
• CV 152.5863 Std Mean 0.11379
• TMean0 2.270258 PrgtT 0.0443
• Num 0 12 Num gt 0 8
• M(Sign) 2 PrgtM 0.3877
• Sgn Rank 25 PrgtS 0.0493

51

• Example 2 with SAS (continued)

• 6) Now we want compare the observed value to the
  calculated value. This case is a little tricky.
• We have a one tailed test (H1 m m0)
• And we chose a 0.05
• The usual critical limit would be a t value with
  11 d.f. This value is -1.796.
• SAS gives us a t value of 2.27.

52

• Example 2 with SAS (continued)

• Reject? No, it is a positive 2.27, not negative.
  So we would not reject the hypothesis.
• 7) Conclude the size of the apples is not
  significantly below the 2.5 inch diameter we
  required.

53

• Example 2 with SAS (continued)

• I used the t values here, not the SAS provided P
  values. Why? Because we were doing a one tailed
  test and the SAS P values are 2 tailed.
• However, we can use them if we understand them.

Our critical limit was here, t-1.796.
54

• Example 2 with SAS (continued)

• The two tailed P value provided by SAS showed
  that the area in two tails was 0.0443, so the
  area in each tail was 0.02215.

Our calculated value was here, t2.27.
Our critical limit was here, t-1.796.
55

• Example 2 with SAS (continued)

• If the values were not in different tails, we can
  see that the size of the calculated value tails
  (0.02215 on each side) is well within the region
  of rejection. So if they had been on the same
  side, they would have been significantly
  different.
• Because they were in different tails they not
  cause rejection of the null hypothesis.

56

• The Null hypothesis

• We could not reject the apples as too small. Had
  we noticed that the mean was greater than 2.5, we
  would not even have had to conduct the test and
  do the calculations.
• But other hypotheses could have been tested.
  Maybe "Prime apples" are supposed to have a mean
  size greater than 2.5. We could reject the
  apples if we could not prove that the size was
  greater than 2.5.

57

• The Null hypothesis (continued)

• Previously we tested,
• 1) H0 m m0
• 2) H1 m m0
• But we could have tested
• 1) H0 m m0
• 2) H1 m ³ m0
• In this case if we set a to a one sided 0.05 we
  would have rejected the H0 since the tail was
  0.0443/2 0.02215. We would still have taken
  the apple shipment.

58

• The Null hypothesis (continued)

• But which is the right test?
• It depends on what is important to you. Do you
  loose your job for sending back apples that were
  not really too small, or do you loose your job
  for accepting apples that did not meet the
  criteria. The correct alternative depends on
  what you have to prove (with an a100 chance of
  error) and what is important to you.

59

• One more example in SAS

• Test for differences in seed production at two
  levels on a plant (top and bottom). We have ten
  vigorous plants bearing Lucerne flowers, each of
  which has flowers at the top bottom. We want
  to test for differences in the number of seeds
  for the average of two pods in each position.
• For each plant take two pods from the top and get
  and average, and two from the bottom for an
  average.

60

• One more example in SAS (continued)

• Calculate the difference between the mean for the
  top and mean for the bottom and test to see if
  the difference is zero (i.e. no difference).
• 1) H0 m m0
• 2) H1 m ¹ m0
• 3) Assume
• Independence (randomly selected sample)
• Number per pod is normally distributed.
• 4) a 0.05

61

• One more example in SAS (continued)

• 5) Take a sample. We have 10 plants, so n 10
  and d.f. 9.
• TOP BOTTOM
• 4.0 4.4
• 5.2 3.7
• 5.7 4.7
• 4.2 2.8
• 4.8 4.2
• 3.9 4.3
• 4.1 3.5
• 3.0 3.7
• 4.6 3.1
• 6.8 1.9

62

• One more example in SAS (continued)

• SAS PROGRAM DATA step
• options ps61 ls78 nocenter nodate nonumber
• data flowers infile cards missover
• TITLE1 'Seed production for top and bottom
  flowers'
• LABEL top 'Flowers from the top of the
  plant'
• LABEL bottom 'Flowers from the bottom of
  the plant'
• LABEL diff 'Difference between top and
  bottom'
• input top bottom
• diff top - bottom
• cards run
• SAS PROGRAM procedures
• proc print dataflowers var top bottom diff
  run
• proc univariate dataflowers plot var diff run

63

• One more example in SAS (continued)

• SAS Output
• OBS TOP BOTTOM DIFF
• 1 4.0 4.4 -0.4
• 2 5.2 3.7 1.5
• 3 5.7 4.7 1.0
• 4 4.2 2.8 1.4
• 5 4.8 4.2 0.6
• 6 3.9 4.3 -0.4
• 7 4.1 3.5 0.6
• 8 3.0 3.7 -0.7
• 9 4.6 3.1 1.5
• 10 6.8 1.9 4.9

64

• One more example in SAS (continued)

• SAS Output
• Moments
• N 10 Sum Wgts 10
• Mean 1 Sum 10
• Std Dev 1.598611 Variance 2.555556
• Skewness 1.669385 Kurtosis 3.934593
• USS 33 CSS 23
• CV 159.8611 Std Mean 0.505525
• TMean0 1.978141 PrgtT 0.0793
• Num 0 10 Num gt 0 7
• M(Sign) 2 PrgtM 0.3438
• Sgn Rank 19.5 PrgtS 0.0469

65

• One more example in SAS (continued)

• 6) Compare the test statistic to the critical
  limits. With 9 d.f. we know that our critical
  limit for a two tailed test at an a 0.05 would
  be t2.262.

66

• One more example in SAS (continued)

• SAS reports
• Mean 1
• t 1.978
• P(gtt) 0.0793. This area leaves almost 4 in
  each tail (0.0793/20.03965) and our critical
  region includes only 2.5 in each tail.
  Therefore, the observed value falls in the area
  of "acceptance".
• We fail to reject the null hypothesis.

67

• One more example in SAS (continued)

• 7) Conclude the number of seeds does not differ
  between the top and bottom of the plant.
• Of course, we may have made a Type II error.
• SAS program. log and output for all examples is
  available on line as both raw text (ASCII code)
  and as HTML.

68

• Summary

• The t distribution is similar the Z
  distribution, but it is used where the value of
  s2 is not known, and is estimated from the
  sample. This is a much more common case.
• The t distribution is very similar to the Z
  distribution in that it is a bell shaped curve,
  centered on zero and ranges from .

69

• Summary (continued)

• It can be also be used with observations or
  samples, the formulas are
• ti (Yi - Y)/S
• t (Y - m0)/SY (Y - m0)/(S/Ön)
• Not all tables or computer algorithms give the
  area in the tail. Some give the cumulative
  frequency starting at -.

70

• Summary (continued)

• In the t-tables
• Each row represents a different t distribution
  (with different df).
• The t table has many different distributions.
  Only the POSITIVE side of the table is given,
  since the t distribution is symmetric.
• Only selected probabilities were provided in the
  t-table,
• The t-test of hypothesis was very similar to the
  Z test we had done before.

71

• Summary (continued)

• SAS PROC UNIVARIATE will do t-test, but it only
  tests the hypothesized value of zero.