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Unit-VII
T-test One sample & T- test for Dependent
Samples
Muhammad Shahid
MSN,CHQP, BSN, DCN, RN, BSc
Ph D Scholar (Ziauddin University)
Assistant Professor/Principal
School of Nursing
Kharadar General Hospital
1
2
By the end of this session, the students should be able to:
Distinguish the inferential procedures (interval estimation &
hypothesis testing) for a population mean on small samples
(i.e. n<30).
o Identification of t-distribution and its properties
Learn how to test hypothesis which compare the difference
in population means using paired samples
Learn how to construct confidence interval for the
difference in population means using paired samples.
Learning Objectives
3
 If n is large (n30), we can determine confidence intervals for the
population mean using Z, because:
 According to the Central Limit Theorem (CLT) for large
sample size (n) the sampling distribution is normal
 Problem arises: When sample size (n) is small (n<30), because:
 The CLT doesn’t necessarily hold true
 The sampling distribution may not be normal
 Sample standard deviation (s) may provide a poor
approximation of the population standard deviation ().
 Z-distribution probabilities don’t apply
 If “n” is small our confidence interval formula based on Z-scores
doesn’t work
T- Distribution
4
 The solution is to find another curve that accurately characterizes
sampling distribution for small sample size (n<30) called t-
distribution.
 The “t-distribution”
 An alternative that accurately approximates the shape of the
sampling distribution for small sample size (n<30)
 Statisticians have proved this property
 The t distribution is actually a family of distributions with known
probabilities, whose value depends upon sample size.
 Again, we can look up values in a table to determine probabilities
associated with a number of standard deviations away from the
mean for a given sample size (n)
Cont…
5
t-statistic
• T-test is used to test hypothesis about μ when the value for s is
unknown
• The formula for the t statistic is similar in structure to the z-score,
except that t statistic uses estimated standard error from the
sample, not population standard error
• t-statistic is more variable than the z-statistic. The variability
depends upon the sample size, n, or the degrees of freedom (n-1).
• For confidence interval & hypothesis testing, the t-statistic that
we will use will have (n-1) degrees of freedom (df). e.g. If n = 15,
then look at t-distribution for df = 14.
• Degree of freedom: minimum # of known information required to
estimate pop parameter
6
T Statistics Testing
Based on the Degrees of Freedom (df) the critical
value necessary to reject the null hypothesis
changes.
In a t-test, the same t value may correspond to
different significance levels (i.e., critical region
cut-offs) based on the degrees of freedom.
7
How to observe value of “t” from t-Tables?
Choose the
correct df (n-
1)
Choose the
desired
probability
for /2
Find t-value in
correct row
and column
Interpretation is
just like a Z-score.
2.145 = number of
standard errors
away from mean
8
Small-Sample Test of Hypothesis about
a Mean
• If n is small, we cannot assume that the sample
standard deviation will provide a good
approximation of the population standard
deviation. We must use the t-distribution rather
than the z-distribution to make inferences about
the population mean.
• Test Statistic: t
x
s
n

 0
9
Small-Sample Confidence Interval
for population mean ()
• Formula -
• The confidence interval using the t-statistic
is wider than the corresponding confidence
interval using the standard z-statistic.
n
s
t
x df
,
2


10
Hypothesis Test (One sided, Upper tail)
• H0:
• Ha:
• Test statistic:
• Rejection region:
0

 
0

 
df
t
t ,


t
x
s
n

 0
11
Hypothesis Test (One sided, Lower tail)
• H0:
• Ha:
• Test statistic:
• Rejection region:
0

 
0

 
df
t
t ,



t
x
s
n

 0
12
Hypothesis Test (Two sided, Both tails)
• H0:
• Ha:
• Test statistic:
• Rejection region:
0

 
0

 
df
df
t
t
t
t ,
2
,
2
& 
 


t
x
s
n

 0
13
Sudden Infant Death syndrome
 To investigate whether sudden infant death syndrome (SIDS) might be
related to an imbalance between peptides affecting respiration, the
authors of the paper “Postmortem Analysis of Neuro-peptides in Brains
from Sudden Infant Death Victims” (Brain Research (1984): 279-285)
measured cortex met-enkephalin levels (pmol/ g wet weight) in brain
tissue of 12 SIDS victims. The resulting sample mean and standard
deviation were 7.66 and 3.78, respectively. The mean level for
children who are not victims of SIDS was reported to be 7.48.
 Using a 0.05 significance level, test to determine whether the true
mean met-enkephalin level of SIDS victims differs from that of
children who are not victims of SIDS.
 Construct a 99% confidence interval for mean met-enkephalin level of
SIDS victims using the above information.
14
Test whether the true (population) mean met-enkephalin level of
SIDS victims differs from that of children who are not victims of
SIDS
 H0: µ = 7.84
 Ha: µ ≠ 7.84
 Test statistic:
 Rejection region:
• Conclusion: Fail to reject H0 i.e. the true (population) mean met-
enkephalin level of SIDS victims is same as for children who are not
victims of SIDS
165
.
0
12
78
.
3
48
.
7
66
.
7



t
201
.
2
&
201
.
2 

 t
t
15
99% confidence interval for mean met-
enkephalin level of SIDS victims
 Lower CI:
 Upper CI:
27
.
4
12
78
.
3
)
11
.
3
(
66
.
7 

054
.
11
12
78
.
3
)
11
.
3
(
66
.
7 

16
Example
A researcher believes that children in poverty-stricken
regions are undernourished and underweight. The
population for the weight of 6-year olds is normally
distributed with 20.9 kg.The researcher
collects a sample of 9 children, with a sample mean
of 17.3 kg & SD = 2.51 kg
Using a one-tailed test and a 0.01 level of
significance, determine if this sample is
significantly different from what would be expected
for the population of 6-year olds.
17
303
.
4
9
51
.
2
9
.
20
3
.
17







n
s
X
s
X
t
X


Conclusion: “6-yr old children raised in
poverty-stricken regions weigh significantly
less (X = 17.3 kg) then the general
population of 6-yr olds, t(8) = 2.896,
p < .01.”
18
Example
A teacher was trying to see whether a new teaching
method would increase the Test of English as
Foreign Language (TOFEL) scores of students.
She knew that the students taught by old methods
had  = 580. She tested her method in a class of
20 students and got a mean of 595 and variance of
225. Is this increase statistically significant at the
level of 0.05 in a 2-tailed test?
19
472
4
20
15
580
595
.







n
s
X
s
X
t
X


Conclusion: “The new teaching
method significantly increases TOFEL
scores, t (19) =4.472, p< .05.”
20
20
If………
 Two population or sample is not independent of each other
Paired or Dependent sample method
Assumptions
 The observed data are from the same subject or from a matched
subject and are drawn from a population with a normal
distribution
 does not assume that the variance of both populations are equal
21
21
 You use the paired t-test when there is one measurement variable
and two attribute variables
 Like sugar level before and after intervention ( most common
design)
 Sometimes the pairs are non temporal, such as left vs. right, injured
limb vs. uninjured limb, above a dam vs. below a dam, etc.
 Paired design allows the researchers to detect change more easily
by controlling for extraneous variation among the observations.
 Many biologic measurements exhibits wide variation among
individuals and the use of paired design is thus appropriate in the
health field
22
What is a Paired Difference Observation?
Observation of interest will be the
difference in the readings before and after
intervention called paired difference
observation.
23
Paired Difference Samples “t test” an explanation
• This test is concerned with the pair-wise differences between
sets of data.
• This means that each data point in one group has a related data
point in the other group (groups always have equal numbers).
• This could be data measured before a treatment, and then after.
Or it could be there is a relationship between the groups, such as
identical twins or another relevant matching characteristic (e.g.
Age, Sex, SES or IQ etc.).
24
Reading 1
(Before intervention/Case)
Reading 2 (After
intervention/Matched control)
Difference
X1 Y1 D1=X1-Y1
X2 Y2 D2=X2-Y2
X3 Y3 D3=X3-Y3
. . .
. . .
Xi Yi Di=Xi-Yi
. . .
. . .
Xn Yn Dn=Xn-Yn
_
Where Di is the ith
paired difference observation. XD and SD are the
sample mean and standard deviation of the paired difference respectively.
What is a Paired Difference Observation?
25
Test Statistic and assumptions for Paired Difference Method
• In this method, the test statistic becomes:
Note: Where D0 is mostly zero and the sample size must be small (n<30).
– Assumptions:
• The distribution of differences is approximately normally
distributed.
• The sample differences are random.
















n
S
D
x
n
S
x
t
D
D 0
0

26
(1-)% Confidence Interval for Paired Mean Difference
(D)
where “t” has (n-1) degrees of freedom and “n” is
the total number of pairs.
• (1-)% confidence interval for population paired mean
difference (D), when n (number of pairs) is small (n<30):
Formula:
n
S
t
x D
df
D ,
2


27
Hypothesis Test for Paired Mean Difference (D)
• One-tailed Test:
– H0: D  D0 (or 0) vs. Ha: D > D0 (or 0)
– H0: D  D0 (or 0) vs. Ha: D < D0 (or 0)
• Two-tailed Test:
– H0: D = D0 (or 0) vs. Ha: D  D0 (or 0)
• Test statistic:
where “t” has (n-1) degrees of freedom (df) and “n” is the total number of
pairs.
• Critical Region:
i) If Ha: µD > D0 then t > t,df
ii) if Ha: µD < D0 then t < - t,df and
iii) if Ha: µD D0 then t > t/2,df and t < - t/2,df








n
s
D
x
t
D
D 0
28
Example: Trace metals in drinking water
• Trace metals in drinking water wells affect the flavor of the water
and unusually high concentrations can pose a health hazard.
Furthermore, the water in well may vary in the concentration of
the trace metals depending upon from where it is drawn. In the
paper, "Trace Metals of South Indian River Region"
(Environmental Studies, 1982, 62-6), trace metal concentrations
(mg/L) on zinc were found from water drawn from the bottom
and the top of each of 6 wells. Use a 5% level of significance.
H0: D = 0 (There is no difference in mean concentration of Zinc
in water drawn from the bottom & the top of the well)
Ha: D  0 (There is a difference in mean concentration of Zinc
in water drawn from the bottom & the top of the well)
29
Example: Trace metals in drinking water (contd.)
The data is given as follows:
Location Bottom Top Difference
(Xi) (Yi) (Di=Xi-Yi)
1 .430 .415 0.015
2 .266 .238 0.028
3 .567 .390 0.177
4 .531 .410 0.121
5 .707 .605 0.102
6 .716 .609 0.107
30
Choose appropriate
df = (n – 1) =5
Interpretation of
t-score is just
like a Z-score.
i.e. 2.57 means
number of
standard errors
away from mean
What is the critical value using t-Table?
Find t-value in correct
row and column
Tabulated
t =2.57 for
two tailed test
if α/2 = 0.025
or α = 0.05
and df = 5
31
Example: Trace metals in drinking water (contd.)
• Mean difference = = 0.0917
• Standard Deviation of differences = SD = 0.06069
• Test Statistic:
• Critical region: Look up critical t value with “5=(n-1)” d.f. and
/2 = 0.025 to get t.025(5) =2.57 i.e. t<-2.57 & t>2.57
• Conclusion: As tcal lies in the critical region. So, we reject H0
and conclude that trace metal concentrations (mg/L) on zinc
from water drawn from the bottom and the top are
significantly different.
D
x
7
.
3
6
06069
.
0
0
0917
.
0









cal
t
32
Choose appropriate
df = (n – 1) =5
What is the p-value using t-Table?
Find t-value in correct
row and column
Test Statistic
value of
t =3.7 with
α = 0.05
(two tailed)
and df =5
p-value is between
0.01 and 0.02.
33
•Mean difference = = 0.0917
•Standard Deviation of differences = SD = 0.06069
•Number of pairs = n = 6
95% Confidence Interval for the Mean Difference (D) of Zinc
concentration in water drawn from the top and the bottom of the well:
 
)
155400609
.
0
,
02799939
.
0
(
063700609
.
0
0917
.
0
6
06069
.
0
)
57
.
2
(
)
0917
.
0
(




Example: Trace metals in drinking water (contd.)
D
x
34
References
 Fundamentals of Biostatistics, Bernard Rosner,
Wadsworth Publishing Company, USA, 1995, Chapter
8, pages: 253-257 .
 Statistics: The exploration and analysis of data, 3rd
Edition, Jay Devore & Roxy Peck, Brooks/Cole
Publishing Company, 1997, Chapter 10, pages: 373-
384.

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Introduction to t test and types in Nursing.ppt

  • 1. Unit-VII T-test One sample & T- test for Dependent Samples Muhammad Shahid MSN,CHQP, BSN, DCN, RN, BSc Ph D Scholar (Ziauddin University) Assistant Professor/Principal School of Nursing Kharadar General Hospital 1
  • 2. 2 By the end of this session, the students should be able to: Distinguish the inferential procedures (interval estimation & hypothesis testing) for a population mean on small samples (i.e. n<30). o Identification of t-distribution and its properties Learn how to test hypothesis which compare the difference in population means using paired samples Learn how to construct confidence interval for the difference in population means using paired samples. Learning Objectives
  • 3. 3  If n is large (n30), we can determine confidence intervals for the population mean using Z, because:  According to the Central Limit Theorem (CLT) for large sample size (n) the sampling distribution is normal  Problem arises: When sample size (n) is small (n<30), because:  The CLT doesn’t necessarily hold true  The sampling distribution may not be normal  Sample standard deviation (s) may provide a poor approximation of the population standard deviation ().  Z-distribution probabilities don’t apply  If “n” is small our confidence interval formula based on Z-scores doesn’t work T- Distribution
  • 4. 4  The solution is to find another curve that accurately characterizes sampling distribution for small sample size (n<30) called t- distribution.  The “t-distribution”  An alternative that accurately approximates the shape of the sampling distribution for small sample size (n<30)  Statisticians have proved this property  The t distribution is actually a family of distributions with known probabilities, whose value depends upon sample size.  Again, we can look up values in a table to determine probabilities associated with a number of standard deviations away from the mean for a given sample size (n) Cont…
  • 5. 5 t-statistic • T-test is used to test hypothesis about μ when the value for s is unknown • The formula for the t statistic is similar in structure to the z-score, except that t statistic uses estimated standard error from the sample, not population standard error • t-statistic is more variable than the z-statistic. The variability depends upon the sample size, n, or the degrees of freedom (n-1). • For confidence interval & hypothesis testing, the t-statistic that we will use will have (n-1) degrees of freedom (df). e.g. If n = 15, then look at t-distribution for df = 14. • Degree of freedom: minimum # of known information required to estimate pop parameter
  • 6. 6 T Statistics Testing Based on the Degrees of Freedom (df) the critical value necessary to reject the null hypothesis changes. In a t-test, the same t value may correspond to different significance levels (i.e., critical region cut-offs) based on the degrees of freedom.
  • 7. 7 How to observe value of “t” from t-Tables? Choose the correct df (n- 1) Choose the desired probability for /2 Find t-value in correct row and column Interpretation is just like a Z-score. 2.145 = number of standard errors away from mean
  • 8. 8 Small-Sample Test of Hypothesis about a Mean • If n is small, we cannot assume that the sample standard deviation will provide a good approximation of the population standard deviation. We must use the t-distribution rather than the z-distribution to make inferences about the population mean. • Test Statistic: t x s n   0
  • 9. 9 Small-Sample Confidence Interval for population mean () • Formula - • The confidence interval using the t-statistic is wider than the corresponding confidence interval using the standard z-statistic. n s t x df , 2  
  • 10. 10 Hypothesis Test (One sided, Upper tail) • H0: • Ha: • Test statistic: • Rejection region: 0    0    df t t ,   t x s n   0
  • 11. 11 Hypothesis Test (One sided, Lower tail) • H0: • Ha: • Test statistic: • Rejection region: 0    0    df t t ,    t x s n   0
  • 12. 12 Hypothesis Test (Two sided, Both tails) • H0: • Ha: • Test statistic: • Rejection region: 0    0    df df t t t t , 2 , 2 &      t x s n   0
  • 13. 13 Sudden Infant Death syndrome  To investigate whether sudden infant death syndrome (SIDS) might be related to an imbalance between peptides affecting respiration, the authors of the paper “Postmortem Analysis of Neuro-peptides in Brains from Sudden Infant Death Victims” (Brain Research (1984): 279-285) measured cortex met-enkephalin levels (pmol/ g wet weight) in brain tissue of 12 SIDS victims. The resulting sample mean and standard deviation were 7.66 and 3.78, respectively. The mean level for children who are not victims of SIDS was reported to be 7.48.  Using a 0.05 significance level, test to determine whether the true mean met-enkephalin level of SIDS victims differs from that of children who are not victims of SIDS.  Construct a 99% confidence interval for mean met-enkephalin level of SIDS victims using the above information.
  • 14. 14 Test whether the true (population) mean met-enkephalin level of SIDS victims differs from that of children who are not victims of SIDS  H0: µ = 7.84  Ha: µ ≠ 7.84  Test statistic:  Rejection region: • Conclusion: Fail to reject H0 i.e. the true (population) mean met- enkephalin level of SIDS victims is same as for children who are not victims of SIDS 165 . 0 12 78 . 3 48 . 7 66 . 7    t 201 . 2 & 201 . 2    t t
  • 15. 15 99% confidence interval for mean met- enkephalin level of SIDS victims  Lower CI:  Upper CI: 27 . 4 12 78 . 3 ) 11 . 3 ( 66 . 7   054 . 11 12 78 . 3 ) 11 . 3 ( 66 . 7  
  • 16. 16 Example A researcher believes that children in poverty-stricken regions are undernourished and underweight. The population for the weight of 6-year olds is normally distributed with 20.9 kg.The researcher collects a sample of 9 children, with a sample mean of 17.3 kg & SD = 2.51 kg Using a one-tailed test and a 0.01 level of significance, determine if this sample is significantly different from what would be expected for the population of 6-year olds.
  • 17. 17 303 . 4 9 51 . 2 9 . 20 3 . 17        n s X s X t X   Conclusion: “6-yr old children raised in poverty-stricken regions weigh significantly less (X = 17.3 kg) then the general population of 6-yr olds, t(8) = 2.896, p < .01.”
  • 18. 18 Example A teacher was trying to see whether a new teaching method would increase the Test of English as Foreign Language (TOFEL) scores of students. She knew that the students taught by old methods had  = 580. She tested her method in a class of 20 students and got a mean of 595 and variance of 225. Is this increase statistically significant at the level of 0.05 in a 2-tailed test?
  • 19. 19 472 4 20 15 580 595 .        n s X s X t X   Conclusion: “The new teaching method significantly increases TOFEL scores, t (19) =4.472, p< .05.”
  • 20. 20 20 If………  Two population or sample is not independent of each other Paired or Dependent sample method Assumptions  The observed data are from the same subject or from a matched subject and are drawn from a population with a normal distribution  does not assume that the variance of both populations are equal
  • 21. 21 21  You use the paired t-test when there is one measurement variable and two attribute variables  Like sugar level before and after intervention ( most common design)  Sometimes the pairs are non temporal, such as left vs. right, injured limb vs. uninjured limb, above a dam vs. below a dam, etc.  Paired design allows the researchers to detect change more easily by controlling for extraneous variation among the observations.  Many biologic measurements exhibits wide variation among individuals and the use of paired design is thus appropriate in the health field
  • 22. 22 What is a Paired Difference Observation? Observation of interest will be the difference in the readings before and after intervention called paired difference observation.
  • 23. 23 Paired Difference Samples “t test” an explanation • This test is concerned with the pair-wise differences between sets of data. • This means that each data point in one group has a related data point in the other group (groups always have equal numbers). • This could be data measured before a treatment, and then after. Or it could be there is a relationship between the groups, such as identical twins or another relevant matching characteristic (e.g. Age, Sex, SES or IQ etc.).
  • 24. 24 Reading 1 (Before intervention/Case) Reading 2 (After intervention/Matched control) Difference X1 Y1 D1=X1-Y1 X2 Y2 D2=X2-Y2 X3 Y3 D3=X3-Y3 . . . . . . Xi Yi Di=Xi-Yi . . . . . . Xn Yn Dn=Xn-Yn _ Where Di is the ith paired difference observation. XD and SD are the sample mean and standard deviation of the paired difference respectively. What is a Paired Difference Observation?
  • 25. 25 Test Statistic and assumptions for Paired Difference Method • In this method, the test statistic becomes: Note: Where D0 is mostly zero and the sample size must be small (n<30). – Assumptions: • The distribution of differences is approximately normally distributed. • The sample differences are random.                 n S D x n S x t D D 0 0 
  • 26. 26 (1-)% Confidence Interval for Paired Mean Difference (D) where “t” has (n-1) degrees of freedom and “n” is the total number of pairs. • (1-)% confidence interval for population paired mean difference (D), when n (number of pairs) is small (n<30): Formula: n S t x D df D , 2  
  • 27. 27 Hypothesis Test for Paired Mean Difference (D) • One-tailed Test: – H0: D  D0 (or 0) vs. Ha: D > D0 (or 0) – H0: D  D0 (or 0) vs. Ha: D < D0 (or 0) • Two-tailed Test: – H0: D = D0 (or 0) vs. Ha: D  D0 (or 0) • Test statistic: where “t” has (n-1) degrees of freedom (df) and “n” is the total number of pairs. • Critical Region: i) If Ha: µD > D0 then t > t,df ii) if Ha: µD < D0 then t < - t,df and iii) if Ha: µD D0 then t > t/2,df and t < - t/2,df         n s D x t D D 0
  • 28. 28 Example: Trace metals in drinking water • Trace metals in drinking water wells affect the flavor of the water and unusually high concentrations can pose a health hazard. Furthermore, the water in well may vary in the concentration of the trace metals depending upon from where it is drawn. In the paper, "Trace Metals of South Indian River Region" (Environmental Studies, 1982, 62-6), trace metal concentrations (mg/L) on zinc were found from water drawn from the bottom and the top of each of 6 wells. Use a 5% level of significance. H0: D = 0 (There is no difference in mean concentration of Zinc in water drawn from the bottom & the top of the well) Ha: D  0 (There is a difference in mean concentration of Zinc in water drawn from the bottom & the top of the well)
  • 29. 29 Example: Trace metals in drinking water (contd.) The data is given as follows: Location Bottom Top Difference (Xi) (Yi) (Di=Xi-Yi) 1 .430 .415 0.015 2 .266 .238 0.028 3 .567 .390 0.177 4 .531 .410 0.121 5 .707 .605 0.102 6 .716 .609 0.107
  • 30. 30 Choose appropriate df = (n – 1) =5 Interpretation of t-score is just like a Z-score. i.e. 2.57 means number of standard errors away from mean What is the critical value using t-Table? Find t-value in correct row and column Tabulated t =2.57 for two tailed test if α/2 = 0.025 or α = 0.05 and df = 5
  • 31. 31 Example: Trace metals in drinking water (contd.) • Mean difference = = 0.0917 • Standard Deviation of differences = SD = 0.06069 • Test Statistic: • Critical region: Look up critical t value with “5=(n-1)” d.f. and /2 = 0.025 to get t.025(5) =2.57 i.e. t<-2.57 & t>2.57 • Conclusion: As tcal lies in the critical region. So, we reject H0 and conclude that trace metal concentrations (mg/L) on zinc from water drawn from the bottom and the top are significantly different. D x 7 . 3 6 06069 . 0 0 0917 . 0          cal t
  • 32. 32 Choose appropriate df = (n – 1) =5 What is the p-value using t-Table? Find t-value in correct row and column Test Statistic value of t =3.7 with α = 0.05 (two tailed) and df =5 p-value is between 0.01 and 0.02.
  • 33. 33 •Mean difference = = 0.0917 •Standard Deviation of differences = SD = 0.06069 •Number of pairs = n = 6 95% Confidence Interval for the Mean Difference (D) of Zinc concentration in water drawn from the top and the bottom of the well:   ) 155400609 . 0 , 02799939 . 0 ( 063700609 . 0 0917 . 0 6 06069 . 0 ) 57 . 2 ( ) 0917 . 0 (     Example: Trace metals in drinking water (contd.) D x
  • 34. 34 References  Fundamentals of Biostatistics, Bernard Rosner, Wadsworth Publishing Company, USA, 1995, Chapter 8, pages: 253-257 .  Statistics: The exploration and analysis of data, 3rd Edition, Jay Devore & Roxy Peck, Brooks/Cole Publishing Company, 1997, Chapter 10, pages: 373- 384.