Hypothesis Testing Introduction to Study Skills & Research Methods (HL10040) Dr James Betts.

Hypothesis Testing

Introduction to Study Skills & Research Methods (HL10040)

Dr James Betts

Lecture Outline:

•What is Hypothesis Testing?

•Hypothesis Formulation

•Statistical Errors

•Effect of Study Design

•Test Procedures

•Test Selection.

Statistics

Descriptive Inferential

Correlational

Relationships

GeneralisingOrganising, summarising & describing data

Significance

Sampling Error

Statistics

The dependent variable can be generalised from n to N

Effective sampling is essential to correctly

generalise back to our target population

What is Hypothesis Testing?

A B

A = B

Null Hypothesis

We also need to establish:

1) How unequal are these observations?

2) Are these observations reflective of the general population?

Alternative Hypothesis

Example Hypotheses: Isometric Torque• Is there any difference in the length of time that males and

females can sustain an isometric muscular contraction?

Null Hypothesis Alternative Hypothesis

♂ = ♀ ♂ ♀



Null Hypothesis (H0)

There is not a significant difference in the DV between males and females

Alternative Hypothesis (HA) or experimental (HE)

There is a significant difference in the DV between males and females.

n.b. these are 2-tailed hypotheses. Most common and more recommended.



Useful analogy- the criminal trial

Imagine you are the prosecutor

H0 = Defendant not guilty

HA = Defendant guilty

We must assume that the defendant is innocent until proven guilty.



Energy Intake (calories per day)

1500 2500 3500 4500 5500

Nu

mb

er o

f P

eo

ple

0

20

40

60

80

100

120

140

160

16 17 18 19 20

Sustained Isometric Torque (seconds)

N♂N♀

n♂n♀

n.b. This is why effective sampling is so important...




1500 2500 3500 4500 5500

Nu

mb

er o

f P

eo

ple

0

20

40

60

80

100

120

140

160

16 17 18 19 20


N♂N♀

n♂n♀

…poor/insufficient sampling can lead to errors…

Statistical Errors• Type 1 Errors

-Rejecting H0 when it is actually true

-Concluding a difference when one does not actually exist

• Type 2 Errors

-Accepting H0 when it is actually false (e.g. previous slide)

-Concluding no difference when one does existErrors can occur due to biased/inadequate sampling, poor

experimental design or the use of inappropriate/non-parametric tests.

Back to Study Design• Independent Measures

– Individual scores in each data set are independent of one another

• Repeated Measures– Individual scores in each data set are

dependent/paired/correlated



• Repeated Measures– Individual scores in each data set are

dependent/paired/correlatedTO1 O2

T O1

Oa

P

Pre-Experimental designs.

2 Distinct Groups

Same individuals tested twice



• Repeated Measures

True-Experimental design.

Depends on how equivalent

groups were achieved

O1 T O2

P O4O3

R

Random Group Assignment

Cross-Over Design

Example Hypotheses: Isometric Torque• Is there any difference in the length of time that males and females can sustain an isometric muscular contraction?

• So the above example is an measures design

– Which therefore requires an independent t-test.

Independent

AKA Students’ (Gosset’s) t-test


1500 2500 3500 4500 5500

Nu

mb

er o

f P

eo

ple

0

20

40

60

80

100

120

140

160

16 17 18 19 20


n♂n♀

Independent t-test: Calculation

Mean SD n

♀ 18.5 1.74 25

♂ 17.5 1.72 25

Is this a significant effect?


Mean SD n

♀ 18.5 1.74 25

♂ 17.5 1.72 25

Step 1:

Calculate the Standard Error for Each Mean

SEM♀ = SD/√n = 1.74/5 = 0.348

SEM♂ = SD/√n = 1.72/5 = 0.344


Mean SD n

♀ 18.5 1.74 25

♂ 17.5 1.72 25

Step 2:

Calculate the Standard Error for the difference in means

SEMdiff = √ SEM♀2 + SEM♂2 = √ 0.251 = 0.501


Mean SD n

♀ 18.5 1.74 25

♂ 17.5 1.72 25

Step 3:

Calculate the t statistic

t = (Mean♀ - Mean♂) / SEMdiff = 2.00


Mean SD n

♀ 18.5 1.74 25

♂ 17.5 1.72 25

Step 4:

Calculate the degrees of freedom (df)

df = (n♀ - 1) + (n♂ - 1) = 48


Mean SD n

♀ 18.5 1.74 25

♂ 17.5 1.72 25

Step 5:

Determine the critical value for t using a t-distribution table

Degrees of Freedom Critical t-ratio

44464850

2.0152.0132.0112.009

n.b. Use 0.05 for 2 tailed test


Mean SD n

♀ 18.5 1.74 25

♂ 17.5 1.72 25

Step 6 finished:

Compare t calculated with t critical

Calculated t = 2.00

Critical t = 2.01

Therefore,

t calculated < t critical

Effect size n.s.


Mean SD n

♀ 18.5 1.74 25

♂ 17.5 1.72 25

Interpretation:

P > 0.05 Reject HA & Accept HO

Conclusion:

There is not a significant difference in the DV between males and females.


Mean SD n

♀ 18.5 1.74 25

♂ 17.5 1.72 25

Evaluation:

The wealth of available literature supports that females can sustain isometric contractions longer than males. This may suggest that the findings of the present study represent a type error

Possible solution: Increase n

Independent t-test: SPSS Output

Independent Samples Test

7.842 .012 -2.333 18 .031 -1.69600 .72710 -3.22358 -.16842

-2.333 15.447 .034 -1.69600 .72710 -3.24188 -.15012

Equal variancesassumed

Equal variancesnot assumed

SwimTime50mF Sig.

Levene's Test forEquality of Variances

t df Sig. (2-tailed)Mean

DifferenceStd. ErrorDifference Lower Upper

95% ConfidenceInterval of the

Difference

t-test for Equality of Means

Group Statistics

10 24.7720 1.25246 .39606

10 26.4680 1.92823 .60976

GroupControl

Visualisation

SwimTime50mN Mean Std. Deviation

Std. ErrorMean

Swim Data from SPSS session 8

Calculated t

df 18 = critical t 2.101

Ignore sign 2.333 > 2.101 So P < 0.05

Repeated Measures Designs• As shown earlier, a repeated measures design

infers that data in each data set can be paired or correlated with one another

• An independent t-test is inappropriate to analyse such data

• Instead, a paired t-test should be used…

Week1 2

Nu

mb

er o

f P

ress

-Up

s

0

20

40

60

80

100

120

140

160

180

200

Advantages of using Paired Data• Data from independent samples is heavily

influenced by variance between subjects

i.e.

This data would have a large SD associated with an

independent t-test simply because some subjects

performed better than others

HOWEVER…

Large SD

(variance)

Week1 2

Nu

mb

er o

f P

ress

-Up

s

0

20

40

60

80

100

120

140

160

180

200

Advantages of using Paired Data• Data from independent samples is heavily

influenced by variance between subjects

…using the same participants on two

occasions allows us to pair up the data…

…now we can remove between subject variance

from subsequent analysis…

Paired t-test: CalculationSubject Week 1 Week 2 Diff (D) Diff2 (D2)

1 10 12

2 50 52

3 20 25

4 8 10

5 115 120

6 75 80

7 45 50

8 170 175

∑D = ∑D2 =Steps 1 & 2: Complete this table

Paired t-test: Calculation

∑D = ∑D2 =

Step 3:


t = n x ∑D2 – (∑D)2 = √ (n - 1)

∑D


∑D = ∑D2 =

Step 3:


t = 8 x 137 – (31)2 = 7.06 √ 7

31


Steps 4 & 5:

Calculate the df and use a t-distribution table to find t critical

Degrees of FreedomCritical t-ratio

(0.05 level)

123456789

12.714.3033.1822.7762.5712.4472.3652.3062.262

df = n -1

Critical t-ratio (0.01

level)63.6579.9255.8414.6044.0323.7073.4993.3553.250

Paired t-test: CalculationStep 6 finished:

Compare t calculated with t critical

Calculated t = 7.06

Critical t = 3.499

Therefore,

t calculated > t critical

Effect size sig.

Mean SD n

Week 1 61.6 56.6 8

Week 2 65.5 57.5 8


Mean SD n

Week 1 61.6 56.6 8

Week 2 65.5 57.5 8

Interpretation:

P < 0.05 Reject H0 & Accept HA

Conclusion:

There is a significant difference in the DV between week 1 and week 2.

Paired Samples Test

-3.87500 1.55265 .54894 -5.17305 -2.57695 -7.059 7 .000VAR00001 - VAR00002Pair 1Mean Std. Deviation

Std. ErrorMean Lower Upper

95% ConfidenceInterval of the

Difference

Paired Differences

t df Sig. (2-tailed)

Paired t-test: SPSS Output

Push-up Data from lecture 3

Calculated t

df 7 = critical t 2.365 (0.05) 3.499 (0.01)

Ignore sign 7.059 > 3.499 So P < 0.01

Paired Samples Statistics

61.6250 8 56.64157 20.02582

65.5000 8 57.54005 20.34348

VAR00001

VAR00002

Pair1

Mean N Std. DeviationStd. Error

Mean

Parametric versus Non-Parametric

• Both the t-tests just shown are parametric tests

• These examine for differences in the mean

• Therefore the mean must be an accurate descriptor

Normal Non-normal?




1500 2500 3500 4500 5500

Nu

mb

er o

f P

eo

ple

0

20

40

60

80

100

120

140

160

16 17 18 19 20


Normal Distribution mean is appropriate t-test

Mean A

Mean B




1500 2500 3500 4500 5500

Nu

mb

er o

f P

eo

ple

0

20

40

60

80

100

120

140

160

16 17 18 19 20


NON-Normal Distribution mean is INappropriate

Mean A

Mean B

Type 2

error

…assumptions of parametric analyses

• All means and paired differences are ND (this is the main consideration)

• N acquired through random sampling

• Data must be of at least the interval LOM

• Data must be Continuous.

…but see Norman (2010) Adv. Health Sci. Educ.

Non-Parametric Tests

• These tests use the median and do not assume anything about distribution, i.e. ‘distribution free’

• Mathematically, value is ignored (i.e. the magnitude of differences are not compared)

• Instead, data is analysed simply according to rank.


• Independent Measures

– Mann-Whitney Test


– Wilcoxon Test

e.g. Exam grades (ordinal) from 14 students in 2 separate schools

Mann-Whitney U: CalculationStep 1:

Rank all the data from both groups in one series, then total each

Student

School A School B

StudentGrade GradeRank Rank

J. S. L. D. H. L. M. J. T. M. T. S. P. H.

T. J. M. M. K. S. P. S. R. M. P. W. A. F.

B- B- A+ D- B+ A- F

D C+ C+ B- E C-

A- Median = B-; Median = C+;∑RA = ∑RB =


Calculate two versions of the U statistic using:

Median = B-; Median = C+;∑RA = ∑RB =

U1 = (nA x nB) + 2

(nA + 1) x nA - ∑RA

AND…

U2 = (nA x nB) + 2

(nB + 1) x nB - ∑RB


Calculate two versions of the U statistic using:

Median = B-; Median = C+;∑RA = ∑RB =

U1 = (nA x nB) + 2

(nA + 1) x nA - ∑RA

…OR to save time you can calculate U1 and then U2 as follows

U2 = (nA x nB) - U1

Mann-Whitney U: CalculationStep 3 finished:

Select the smaller of the two U statistics (U1 = 17.5; U2 = 31.5)

…now consult a table of critical values for the Mann-Whitney test

n

0.05

0.01

6

5

2

7

8

4

8

13

7

9

17

11

Calculated U must be less than critical U to conclude a significant difference

Conclusion

Median A = Median B

Test Statisticsb

17.500

45.500

-.900

.368

.383a

Mann-Whitney U

Wilcoxon W

Z

Asymp. Sig. (2-tailed)

Exact Sig. [2*(1-tailedSig.)]

VAR00001

Not corrected for ties.a.

Grouping Variable: VAR00002b.

Mann-Whitney U: SPSS Output

Calculated U (lower value)

17.5 > 8

So P > 0.05 n.s.

Ranks

7 8.50 59.50

7 6.50 45.50

14

VAR000021.00

2.00

Total

VAR00001N Mean Rank Sum of Ranks


• Independent Measures

– Mann-Whitney Test


– Wilcoxon Test

e.g. One group pre-test post-test, assumed non-normal

Wilcoxon Signed Ranks: CalculationStep 1:

Rank all the differences in one series (ignoring signs), then total each

AthletePre-training OBLA (kph)

Rank

J. S. L. D. H. L. M. J. T. M. T. S. P. H.

15.6 17.2 17.7 16.5 15.9 16.7

17.0

0.5 0.3 -1 0.3

0.1 -0.2 0.1 ∑Signed Ranks =

Post-training OBLA (kph)

Diff. Signed Ranks

16.1 17.5 16.7 16.8 16.0 16.5

17.1

6 4.5 -7 4.5 1.5 -3

1.5

- +

-7

-3

6 4.5

4.5

1.5

1.5

Medians = 16.7 16.7

Wilcoxon Signed Ranks: CalculationStep 2:

The smaller of the T values is our test statistic (T+ = 18; T- = 10)

…now consult a table of critical values for the Wilcoxon test

n

0.05

6

0

7

2

8

3

9

5

Calculated T must be less than critical T to conclude a significant difference

Conclusion

Median A = Median B

Test Statisticsb

-1.364a

.172

Z

Asymp. Sig. (2-tailed)

VAR00002 -VAR00001

Based on negative ranks.a.

Wilcoxon Signed Ranks Testb.

Wilcoxon Signed Ranks: SPSS Output

10 > 2

So P > 0.05 n.s.

Ranks

2a 3.00 6.00

5b 4.40 22.00

0c

7

Negative Ranks

Positive Ranks

Ties

Total

VAR00002 - VAR00001N Mean Rank Sum of Ranks

VAR00002 < VAR00001a.

VAR00002 > VAR00001b.

VAR00002 = VAR00001c.

So which stats test should you use?

Q1. What is the LOM?

Ordinal

Nominal Interval/Ratio

Q2. Are the data ND?NoYes

Q3. Are the data paired

or independent?

Why do we use Hypothesis Testing?

• It is easy (i.e. data in P value out)

• It provides the ‘Illusion of Scientific Objectivity’

• Everybody else does it.

Problems with Hypothesis Testing?

• P<0.05 is an arbitrary probability (P<0.06?)

• The size of the effect is not expressed

• The variability of this effect is not expressed

• Overall, hypothesis testing ignores ‘judgement’.

[email protected]

Hypothesis Testing Introduction to Study Skills & Research Methods (HL10040) Dr James Betts.

Documents

Transcript of Hypothesis Testing Introduction to Study Skills & Research Methods (HL10040) Dr James Betts.