T2Nestimation

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1
Topic 2. Distributions, hypothesis testing, and sample size 
determination 
The Student - t distribution (ST&D pg 56 and 77) 
Consider a repeated drawing of samples of size n = 5 from a normal distribution. 
For each sample compute Y , s, sY , and another statistic, t: 
 
t (n-1)= (Y - )/ sY (Remember Z = (Y - )/ Y ) 
 
The t statistics is the number of standard error that separate Y and its hypothesized 
mean µ. 
 
 
Fig. 1. Distribution of t (df=4) compared to Z. The t distribution is symmetric, but wider 
and flatter than the Z distribution, lying under it at the center and above it in the tails. 
df=n-1=4 
Critical values for |t|>|Z| -> less 
sensitivity. This is the price we pay for 
being uncertain about the population 
i
 2
 
When N increases the t distribution tend towards the N distribution
 3
2. 2. Confidence limits based on sample statistics (ST&D p.77) 
 
The general formula for any parameter  is: 
Estimated   Critical value * Standard error of the estimated  
 
So, for a population mean estimated via a sample mean: 
 
Yn
stY 
1,
2
 
The statistic Y is distributed about  according to the t distribution so it satisfies 
 
P{Y - t /2, n-1 sY    Y + t /2, n-1 sY }= 1-  
For a confidence interval of size 1-, use a t value corresponding to  /2. 
Therefore the confidence interval is 
Y - t /2, n-1 sY    Y + t /2, n-1 sY 
 
These two terms represent the lower and upper 1-  confidence limits of the mean. 
The interval between these terms is called confidence interval (CI). 
 
Example: Data Set 1 of Hordeum 14 malt extraction values 
 
Y = 75.94 sY = 1.23 / 14 = 0.3279 A table gives the t0.025,13 value of 2.16 
 
95% CI for  = 75.94 ± 2.160 * 0.3279  [75.23- 76.65] 
 
If we repeatedly obtained samples of size 14 from the population and constructed 
these limits for each, we expect 95% of the intervals to contain the true mean. 
 
 
 True mean 
 
 
Fig. 2 Twenty 95% confidence intervals. One out of 20 intervals does not include 
the true mean. 
 4
28.6
3279.0
00.7894.75 
Ys
Yt 
 
2. 3. Hypothesis testing and power of the test (ST&D 94). 
Example Barley data. Y = 75.94, sY = s n
2 / = 0.3279, t0.025,13 = 2.160, CI: [75.23- 76.65] 
1) Choose a null hypothesis: Test Ho  = 78 against the H1   78. 
2) Choose a significance level: Assign  = 0.05 
3) Calculate the test statistic t: 
 
 
(interpretation: the sample mean is 6.3 SE from the hypothetical mean of 78. Too far!). 
 
4) Compare the absolute value of the test statistic to the critical statistic: 
 | - 6.28 | > 2.16 
 
5) Since the absolute value of the test statistic is larger, we reject H0. 
 
This is equivalent to calculate a 95% confidence interval for the mean. 
Since o (78) is not within the CI [75.23- 76.65] we reject Ho. 
 
is called the significance level of the test (<0.05): probability of incorrectly 
rejecting a true Ho, a Type I error. 
 
is the Type II error: to incorrectly accept Ho when it is false 
 Null hypothesis 
 Accepted Rejected 
True 
Null hypothesis 
False 
Correct decision 
 
Type II error= 
Type I error=  
 
Correct decision= Power= 1-
 
Power of the test: 1- is the power of the test, and represents the probability of 
correctly rejecting a false null hypothesis. It is a measure of the ability of the test to 
detect an alternative mean or a significant difference when it is real 
 
Note that for a given Y and s, if 2 of the 3 quantities , , and n are specified then 
the third one can be determined. 
Choose the right number of replications to keep Type I error  and Type II error 
 under the desired limits (e.g.  <0.05 & <0.20). 
 5
 
Power of a test (ST&D pg 118-119) 
 
What is the power of a test for Ho: = 74.88 in the barley data set against H1:  = 
75.94. Since  = 0.05, n = 14 (t 0.025,13 = 2.160), and sY = 0.32795. 
 
 
The probability of the Type II error  we are looking for is the shaded area to the 
left of the lower curve. 
 
 
Fig. 3. Type I and Type II 
errors in the Barley 
data set. 
 
 
 
 
 
 
 
The area 1- in the 
rejection region = the 
probability that > 75.588 under H1 = power = 
 
P(t>(75.588-75.94)/0.32795)= P(t>-1.072)=0.85 same as above! 
 
The magnitude of   depends on 
1. The Type I error rate  
2. The distance between the two means under consideration 
3. The number of observations (n)  n
ssY  
)()(1 012/
01
2/
YY
s
ttPorZZPPower


 

85.0)072.1()
32795.0
88.7494.75
160.2(1  tPtPPower 
Ho is true 
 
 
Ho is false 
/2
 
1-
Acceptance Rejection 
 between 2 
means in SE 
units 
 6
When the distance between the two means is reduced,  increases 
 
Variation of power as a function of the distance between the 
alternative hypotheses (Biometry Sokal and Rohlf) 
 
 
 
 
 
 
 
 
n=5 
SE=1.7 
n=35 
SE=0.7 
 7
2.3.2. Power of the test for the difference between the means of two samples 
 
Two types of alternative hypothesis 
H0: 1 - 2=0 versus H1: 1 - 2  0 (two tail test) -> (t value: top t Table) 
 H1: 1 - 2 >0 (one tail test) -> (t value: bottom t Table) 
 
The general power formula for both equal and unequal sample sizes reads as: 
 
)||()||(
2
21
221
21
2
N
s
ttP
s
ttPPower
pooledYY



 , 
 
where 2pooleds is a weighted variance given by: )1()1(
)1()1(
21
2
22
2
112


nn
snsns pooled 
and 
21
21
nn
nnN  . 
 
When n1 = n2 = n (equal sample sizes) that the formulas reduce to: 
 
2)1(2
))(1(
)1()1(
)1()1( 22
2
1
2
2
2
1
21
2
22
2
112 ss
n
ssn
nn
snsns pooled


 
 
22
2
21
21 n
n
n
nn
nnN  
 
)
2
||()||(
2
21
221
21
2
n
s
ttP
s
ttPPower
pooledYY



 
The variance of the difference between two random variables is the sum of the 
variances (error are always added) (ST&D 113-115). 
 
The degrees of freedom for the critical t /2 are 
General case: (n1-1) + (n2-1) 
Equal sample size: 2*(n-1) 
 8
2. 4. 2 Sample size for estimating µ, when is known. 
Using the z statistic 
If the population variance  is known the Z statistic may be used. 
 Z
Y   so CI = YZY  2/ or nZY  2/ 
The formula for d= half-length of the confidence interval for the mean is 
[ Y ] n
ZZd Y
 
22

 
 
This can be rearranged to estimate the confidence interval in terms of the 
population variance. For =0.05: 
 
n = z 2/2 2 / d2 = z 2/2 (/d)2 = (1.96)2(/d)2= 3.8(/d)2 
 
So if d=  n 4 d= 0.5   n 16 d= 0.25   n 64 
 
 
The equation may be expressed in terms 
of the coefficient of variation 
 
 
CV= s / Y (as a proportion not as a %) 
 d/ is the confidence interval as a fraction of the population mean. 
For example d/ < 0.1 means that the length of the confidence interval should not 
be larger than two tenth of the population mean. 
d/ < 0.1 and so 2d/ < 0.2 
Example: The CVs of yield trials in our experimental station are never higher than 
15%. How many replications are necessary to have a 95% CI for the true mean of 
less than 1/10 of the average yield? 
2d= 0.1 so d= 0.1/2= 0.05 
n= 1.962 0.152/0.052 = 34.6  35 
d d 
2
2
2
2
2
2
2
2 












d
CVZ
d
Zn
 9
2. 4. 3 Sample size for theestimation of the mean 
Unknown 2. Stein's Two-Stage Sample 
 
Consider a (1 - )% confidence interval about some mean µ: 
 
Y - Yn
st
1,
2
    Y + Yn st 1,
2
 
 
The half-length (d) of this confidence interval is therefore: 
 
n
ststd
nYn 1,
2
1,
2

  
This formula can be rearranged to estimate necessary sample size n 
 
2
2
2
2
2
2
2
1,
2 d
Z
d
stn
n

   
 
Stein's Two-Stage procedure involves using a pilot study to estimate s2. 
 
Note that n is now present at both sides of the equation: iterative approach 
 
Example: An experimenter wants to estimate the mean height of certain plants. From a 
pilot study of 5 plants, he finds that s = 10 cm. What is the required sample size, if he 
wants to have the total length of a 95% CI about the mean be no longer than 5 cm? 
 
 Using n = t2 /2,n-1 s2 / d2, n is estimated iteratively, 
 
 initial-n t5%, df n 
 5 2.776 (2.776)2 (10)2 /2.52 = 123 
 123 1.96 (1.96)2 (10)2 / 2.52 = 62 
 62 2.00 64 
 64 2.00 64 
Thus with 64 observations, he could estimate the true mean with a CI no 
longer than 5 cm at =0.05. 
 
To accelerate the iteration you can start with Z: 
 
 n = z2 s2 / d2 = (1.96)2 (10)2 / 2.52 = 62 
 
 10
2. 4. 4 Sample size estimation for a comparison of two means 
When testing the hypothesis Ho: o, we can take into account the possibility of 
a Type I and Type II error simultaneously. 
To calculate n we need to known the alternative 1 or at least the minimum 
difference we wish to detect between the means  = |o - 1
The formula for computing n, the number of observations on each treatment, is: 
n = 2 ( / (Z/2 + Z)2 
 
For = 0.05 and = 0.20, z= 0.8416 and z /2 = 1.96, (Z/2 + Z)2=7.85 8 
We can define in terms of 
 If δ = 2σ, n ≈ 4 
 If δ = 1σ, n ≈ 16 
 If δ = 0.5σ, n ≈ 64 
 
We rarely know 2 and must estimate it via sample variances: 
 
2
221,221,
2
2
2 


 


  nnnn
pooled tt
s
n  , where 2
2
2
2
1 sss pooled
 
 
n is estimated iteratively. If no estimate of s is available, the equation may be 
expressed in terms of the CV, and  as a proportion of the mean: 
 
 
Example: Two varieties are compared for yield, with a previously estimated s
2
 = 
2.25 (s=1.5). How many replications are needed to detect a difference of 1.5 
tons/acre with a  = 5%, and  = 20%? 
Approximate: n  2 (/(Z/2 + Z)2 = 2 (1.5/1.5)2(1.96+0.8416)= 15.7 
Then use n = 2 (s / (t/2 + t)2 to estimate the sample size iteratively. 
 
guesstimate n df = 2(n - 1) t0.025 t0.20 estimated n 
16 30 2.0423 0.8538 16.8 
17 32 2.0369 0.8530 16.7 
The answer is that there should be 17 replications of each variety. 
 
n  2 [(/) / ((Z/2 + Z)2  2(CV/%)2(Z/2 + Z)2 
 11
2. 4. 5. Sample size to estimate population standard deviation 
 
The chi-squared (2) distribution is used to establish confidence intervals around the 
sample variance as a way of estimating the true, unknown population variance. 
 
2. 4. 5. 1. The Chi- square distribution (ST&D p. 55) 
654321
0 . 5 
0 . 4 
0 . 3 
0 . 2 
0 . 1
0 . 0 
C h i - s q u a r e 
2 df 
4 df 
6 df 
 
Distribution of 2 , for 2, 4, and 6 degrees of freedom. 
 
Relation between the normal and chi-square distributions. 
The 2 distribution with df = n is defined as the sum of squares of n independent, 
normally distributed variables with zero means and unit variances. 
2α, df=1= Z2(0,1) α/2 2α/2, df= 
 21, 0.05 = 3.84 and Z2(0,1), 0.025 = t2, 0.025 = 1.962 = 3.84 
Note: Z values from both tails go into the upper tail of the χ2 because of the disappearance of the minus sign in the 
squaring. For this reason we use  for the 2 and /2 for Z and t. 
Z Y Yi
i
n
i
i
2
1
2
2 2
21

     ( ) ( )   
 
If we estimate the parametric mean  with a sample mean, we obtain: 
1 1
2
2
2
2 ( )
( )Y Y n si    …due to: 
 

n
i
i
n
YYs
1
2
2
1
)(
  2
1
2 )1()( snYY
n
i
i 

 
 
This expression, which has a 2n-1 distribution, provides a relationship between the 
sample variance and the parametric variance. 
 12
2. 4. 5. 2. Confidence interval for 2 
We can make the following statement about the ratio (n-1) s2/2 that has2n-1 
distribution, 
P { 21-/2, n-1  (n-1) s2/2  2/2, n-1} = 1 -  
 
Simple algebraic manipulation of the quantities in the inequality yields 
 
P { 21-/2, n-1 /(n-1)  s2/2  2/2, n-1/(n-1)} = 1 -  
which is useful when the precision of s2 can be expressed in terms of the % of 2. 
 
Or inverting the ratio and moving (n-1): 
 P {(n-1) s2/ 2/2, n-1  2  (n-1) s2/ 21-/2, n-1} = 1 -  
which is useful to construct 95% confidence intervals for 2. 
 
Example: What sample size is required to obtain an estimate of  that 
deviates no more than 20% from the true value of  with 90% confidence? 
 
 Pr {0.8 < s/ < 1.2} = 0.90 = Pr {0.64 < s2/2 < 1.44} = 0.90 
 
thus 
 2(1 - /2, n-1) / (n-1)= 0.64 and 2 (/2, n-1) / (n-1)= 1.44 
 
Since 2 is not symmetrical, the above two solutions may not identical for 
small n. The computation involves an arbitrary initial n and an iterative 
process: 
 
 df 1 - /2 = 95% /2 = 5% 
n (n-1) 2(n-1) 2(n-1) /(n-1) 2(n-1) 2(n-1) /(n-1) 
21 20 10.90 0.545 31.4 1.57 
31 30 18.50 0.616 43.8 1.46 
41 40 26.50 0.662 55.8 1.40 
36 35 22.46 0.642 49.8 1.42 
35 35 21.66 0.637 48.6 1.43 
 
Thus a rough estimate of the required sample size is ~ 36.