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Sample Size Needed to Achieve
High Confidence (Means)
Considering estimating X, how many observations n
are needed to obtain a 95% confidence interval for a
particular error tolerance?
The error tolerance E is ½ the
width of the confidence interval
Here,  is a conservative
(high) estimate of the true std
dev X, often gotten by doing
a preliminary small sample
1.960 can be adjusted to get different confidences
Derivation of Formula
• E = z * std. error = z * sigma/sqrt(n)
• Thus, sqrt(n) = z * sigma/E
• So, n = [z * sigma/E] 2
Sample Size Needed to Achieve
High Confidence (Proportions)
Considering estimating pX, how many observations n
are needed to obtain a 95% confidence interval for a
particular error tolerance?
The error tolerance E is ½ the
width of the confidence interval
Here, p is a conservative
(closer to 0.5) estimate of the
true population proportion
pX, often gotten by doing a
preliminary small sample
1.960 can be adjusted to get different confidences
Derivation
• E = z * std. error
• But now, std error = sqrt[p(1-p)/n], so
• E = z * sqrt[p(1-p)]/sqrt(n), and hence
• Sqrt(n) = z * sqrt[p(1-p)]/E,
• => n = [z *sqrt[p(1-p)]/E]2
Election Polling Example:
1500 prospective voters are surveyed. 825 say they will vote for candidate A and 675 say they will vote for B. What is
your estimate of the percentage of voters who will vote for A? Construct a 95% confidence interval.
ˆ 
825
 0.55
1500
To construct the confidence interval, use the normal approximation.
 
0.551  0.55
1500
 0.0128
NORMINV(0.975,0.55,0.0128) = 0.575
The associated confidence interval is, therefore,
55%  2.5%.
The newsmedia would report this result by stating: “The poll has a margin of error of plus or minus 2.5 percentage
points”.
What would happen if the 55% estimate were
based on a sample of size 750?
 
0.551  0.55
750
 0.0182
NORMINV(0.975,0.55,0.0182) = 0.586
The associated confidence interval is,
therefore,
55%  3.6%.
Coke vs. Pepsi Example
From a sample of 500, the number who say they prefer
Coke to Pepsi is 275. Your estimate of the population
proportion who prefer Coke is 275/500 = 55%.
Since n is large, we can apply the CLT and construct a
confidence interval for the population proportion, p:
ˆ  Z
ˆ 1  ˆ 
1 2
n
provides a (1-)100% confidence interval for the
population proportion, p. For a 95% interval in this case,
we would first determine that NORMSINV(0.975) = 1.96.
We would then compute:
0.55  1.96
0.551  0.55
 0.55  0.0436.
500
Hypothesis Testing
• Formulate null hypothesis (action is
associated with alternative)
• Compute Test Statistic
• Determine Acceptance Region
Heart Valve Example
• Original yield = 52%
• Change Process (Sort in batches of 5)
• Sample 100 assemblies: sample yield =
79%
Heart Valve Example: Formulate
Null Hypothesis
• Null Hypothesis: the true process yield is
52% or less ------ Why?????
• Note: Action would be associated with the
alternative----adopt new process only if it
really increases the yield.
Heart Valve Example: Compute
Test Statistic
• We will use the Normal Approximation:
(how many standard deviations to the right of
.52 is .79?)
ˆ 
z
p (1  p )
.52(1  .52)

 .04996
n
100
.79  .52
.27

 5.404
ˆ
.04996
Heart Valve Example: Determine
Acceptance Region
Suppose we want to set the probability of rejecting the null
hypothesis given that it is true (type 1 error) = 0.0001
Compute Z = normsinv(.9999) = 3.71947
Reject the null hypothesis if:
Test statistic > 3.71947
Normal Distribution
-8.00
-6.00
-4.00
-2.00
0.00
2.00
Values of Random Variable X
4.00
6.00
8.00
Heart Valve Example:
conclusion
• Test statistic = 5.404
• Reject null hypothesis if test statistic >
3.71947
• We should REJECT the null hypothesis
Example: 2-Tailed Test on a Single Mean
An auto manufacturer has an old engine that
produces an average of 31.5 mpg. A new engine
is believed to have the same standard deviation in
mpg, 6.6, as the old engine, but it is unknown
whether or not the new engine has the same
average mpg. The sample mean of a random
sample of 100 turns out to be 29.8. Let  = 0.05.
Solution:
1. Formulate Hypothesis:
H0:  = 31.5, H1:   31.5
2. Compute Test Statistic:
z
x   0 29 .8  31.5

 2 .576
 n
6 .6 100
3. Determine Acceptance Region: Fail to reject if
NORMSINV(/2)  z  NORMSINV(1-/2)

NORMSINV(0.025)  -2.576  NORMSINV( 0.0975)

-1.959  -2.576  1.959
Since the inequality fails to hold, we
Reject Null Hypothesis!
Our test statistic lies 2.576 standard deviations to the left of the mean of the sampling distribution.
If the null hypothesis were true, we would expect to observe a test statistic this low (or lower) only
about 0.4998% of the time---less than 1/2 of 1 percent.
T-test
 Used when , the standard deviation of the underlying
population is unknown.
 Instead of forming the test statistic with , we substitute an
estimate for , namely the sample deviation:


n
 xi  x 2
S  i 1
.
n 1
 The resulting test statistic is:
t
x  0
.
S n
T-Test
 The test statistic, t, has a t distribution with n-1 degrees of
freedom. In comparison, the test statistic, z, formed using the
population standard deviation, , has a normal distribution .
 The t distribution is shaped like the standard normal distribution
(eg. bell-shaped. but more spread out). Its mean is 0, and its
variance (whenever degrees of freedom > 2) is df/(df-2). As n
(and hence df) increases, the variance approaches 1 & the t
distribution approaches the standard normal distribution.
 When n is large, the z test is sometimes substituted (as a close
approximation) for the t test.
Assumptions: Either n is large enough for the central limit
theorem to hold or the underlying distribution is normal.
T-test & EXCEL
Suppose t has a t distribution with n-1 degrees of freedom.
We define t1 
2 ,n 1
to be the number satisfying:


P t  t1 2 ,n 1  1   2 .
That is, the area under the t distribution (probability) to the left of
t1  2 ,n 1 is 1-/2.
Note: This is the exact analogue of
Z 1  2 for the standard normal
distribution.
To calculate the value for
t1  2 ,n 1 in EXCEL, you use the TINV
function:
t1   2 ,n 1  TINV ( , n  1).
EXAMPLE: 1-Tailed t-Test
Average weekly earnings of full time
employees is reported to be $344. You
believe this value is too low. A random
sample of 1200 employees yields a sample
mean of $361 and a sample deviation of
$110. Formulate the appropriate null
hypothesis and analyze the data:
1. Formulate Null Hypothesis:
H0:  < 344, H1:  > 344
2. Compute Test Statistic:
t
x  0
361  344

 3 .353612
S n 110 1200
3. Analysis: This is an extreme value for the test statistic, more than 3
standard deviations away from the mean (if the null hypotheis were true).
For  = 0.001, we can calculate:
t1    TINV (2  ,1199 )  TINV 0 .002 ,1199   3 .097084 .
Since our test statistic is even larger, we would reject the null hypothesis
at the 99.9% significance level.
The p-value associated with our test statistic is given by:
p-value = TDIST(3.353612, 1199, 1) = 0.000411.
Thus, if the null hypothesis were true, the probability of obtaining a test
statistic as large (or larger) than 3.353612 is only 0.0411%.
Average weekly earnings are almost certainly larger than $344.
p-Values
Definition: The p-value is the smallest level of significance, , for which
a null hypothesis may be rejected using the obtained value of the test
statistic.

The p-value is the probability of obtaining a value of the test statistic as
extreme, or more extreme than, the actual test statistic, when the null
hypothesis is true.
Example:
Your z-statistic in a z-test is 3.162. To
calculate the p-value, use the EXCEL
function NORMSDIST. NORMSDIST(z) is
the probability of obtaining a test statistic
value less than or equal to z. To be more
extreme, the test statistic would have to be
larger than 3.162. Thus,
p-value = 1-NORMSDIST(3.162) = 10.999216 = 0.000784
Example:
Your z-statistic in a z-test is -0.4. To be more
extreme, the test statistic would have to be
less than -0.4. Thus,
p-value = NORMSDIST(-0.4) = 0.3446
Example:
Your t-statistic in a t test with 15 degrees of
freedom is 4.56. To calculate the p-value use
the EXCEL function TDIST. TDIST(t,n-1,1)
= P{test statistic value > t | null hypothesis
true}. (Note the different direction of the
inequality!) Hence
p-value = TDIST(4.56,15,1) = 0.000188
Rules of Thumb for p-values
p-value
< 0.01
interpretation
very significant
between 0.01 & 0.05
significant
between 0.05 & 0.1
marginally significant
> 0.1
not significant
Two-Sample Tests for Means
• Used when we wish to compare the means
of two populations when both means are
unknown
• Tools => Data Analysis => two sample test
• See Two Sample Spreadsheet
Two Sample Tests
• Z-test: two sample for means – when std.
deviations are known for both distributions
• T-test: two sample assuming equal
variances
• T-test: two sample assuming unequal
variances
Usually just use unequal variance test
1.5
1.5
1.5
0.5
0.5
0.5
2.5
2.5
2.5
1.5
1.5
1.5
See spreadsheet "two sample.xls"
hypothesized difference (b column - a column) = 0
Variable 1
Variable 2
Mean
1
2
Variance
0.1875
0.1875
Observations
9
9
Hypothesized Mean Difference
0
df
16
Would we reject or
t Stat
-4.898979486
fail to reject hypothesis
P(T<=t) one-tail
8.02671E-05
that difference = 0?????
t Critical one-tail
1.745884219
P(T<=t) two-tail
0.000160534
t Critical two-tail
2.119904821
t-Test: Two-Sample Assuming Unequal Variances
hypothesized difference (b column - a column) = +1
Variable 1
Variable 2
Mean
2
1
Variance
0.1875
0.1875
Observations
9
9
Hypothesized Mean Difference
1
df
16
t Stat
0
P(T<=t) one-tail
0.5
t Critical one-tail
1.745884219
P(T<=t) two-tail
1
t Critical two-tail
2.119904821
Chi-Square Tests
Like the T-distribution, the chi-square distribution is
defined by its number of degrees of freedom. A chi-square
random variable with k degrees of freedom is normally
2
denoted by the symbol  k , and is defined by the equation:
k
 k2    i2 ,
i 1
That is, the sum of the squares of k standard normal random
variables. Since squares are always non-negative, so is
their sum, and hence a chi-square random variable can only
take on non-negative values. Illustrations of the PDF for 1
and 5 degrees of freedom are shown below.
Probability Density Function
for the Chi-Sq Distribution
1
0.050
3.841
30
25
20
15
10
5
0
Degrees of Freedom:
P-Value:
Chi-Sq Critical Value:
Probability Density Function
for the Chi-Sq Distribution
5
0.050
11.070
30
25
20
15
10
5
0
Degrees of Freedom:
P-Value:
Chi-Sq Critical Value:
Values for the chi-square distribution can be referenced
using the EXCEL functions CHIDIST and CHIINV.
 CHIDIST(x, k) gives the probability that a chisquare random variable with k degrees of freedom
attains a value greater than or equal to x. In other
words, the area under the PDF to the right of x. In
the pictures above, this is reported as the p-value.
 CHIINV(p, k) gives the inverse, or critical value.
That is, if p = CHIDIST(x, k), then CHIINV(p, k) =
x. In the pictures above, this is reported as the chisq critical value.
Examples:
CHIDIST(5,5) = 0.41588
CHIDIST(25,5) = 0.00139
CHIINV(0.41588, 5) = 5
CHIINV(0.00139, 5) = 25
Test for Population Variance
Sometimes it is of interest to draw inferences about the
population variance. The distribution used is the chisquare distribution with n-1 degrees of freedom (where n =
sample size), and
the test statistic is given by:

2

n 1s 2
 02
,
where s2 is the sample variance and the denominator is the
value of the variance stated in the null hypothesis.
Example: Heart Valves.
Without any sorting, the clearance was normally distributed
with mean of 0.005 and standard deviation of 0.000283
(which implies a variance of 8 * 10-8).
One key indicator of process improvement is whether or
not process variability has been reduced. In this case, we
would look to see if the variance of the clearance
dimension has been reduced by sorting.
Heart Valve Example Continued
The null hypothesis in this case is that the variance
has not been reduced:
H0: s2  8 * 10 -8
Suppose that a random sample of size 50 (after
sorting is implemented) yields a sample variance of :
2.308 * 10-9
Computing the test statistic yields:

2
n  1s

 02
2
49  2.308  10 9

 1.414
8  10 8
The critical value is found (for an alpha of 0.001) by:
CHIINV(0.999, 49) = 23.98.
We would reject the null hypothesis for any value of the
test statistic less than the critical value, 23.98.
Example: A machine makes small metal plates used in
batteries. The plate diameter is a random variable with a
mean of 5 mm. As long as the variance is at most 1.0, the
production process is under control & the plates are
acceptable. Otherwise, the machine must be repaired. The
QC engineer wants, therefore, to test the following
hypothesis:
H0: s2 < 1.0
With a random sample of 31 plates, the sample variance is
1.62.
Solution: Computing the test statistic, we see that:
n  1s 2 30 1.62

 

 48 .6
2
2
2
0
1.00
 For a critical value of  = 0.05, the critical value is
found by:
CHIINV(0.05, 30)=43.77.
Since our test statistic lies to the right of the critical
value, we would reject the null hypothesis.
 The p-value is given by:
CHIDIST(48.6, 30) = 0.017257.
Thus, we would reject the null hypothesis for any
value of  > 0.017257, and fail to reject the null
hypothesis for smaller values.
Important Note: The use of the
chi-square test on variance
requires that the underlying
population be normally
distributed.
Chi-Square Test for Independence
It is often useful to have a statistical test that helps us to determine
whether or not two classification criteria, such as age and job performance
are independent of each other. The technique uses contingency tables,
which are tables with cells corresponding to cross-classifications of
attributes.
In marketing research, one place where the chi-square test for
independence is frequently used, such tables are called cross-tabs. You
will recall that we have previously used the pivot-table facility within
EXCEL to produce contingency or cross-tabs tables from more unwieldy
tabulations of raw data.
Example: A random sample of 100 firms is taken. For each firm, we
record whether the company made or lost money in its most recent fiscal
year, and whether the firm is a service or non-service company. A 2 X 2
contingency table summarizes the data.
Profit
Loss
Total
Industry Type
Service
Non-service
42
18
6
34
48
52
Total
60
40
100
Using the information in the table, we want to investigate
whether the two events:
 the company made a profit in its most recent fiscal
year, and
 the company is in the service sector
are independent of each other.
Before stating the test, we need to develop a little bit of
notation:
r = number of rows in the table
c = number of columns in the table
Oij = observed count of elements in cell (i, j)
Eij = expected count of elements in cell (i, j) assuming
that the two variables are independent
Ri = total count for row i
Cj = total count for column j
The expected number of items in a cell is equal to the
sample size, n, times the probability of the event signified
by the particular cell. In the context of a contingency table,
the probability associated with cell (i, j) is the joint
probability of occurrence of both events. That is,
E  n  Pi  j.
ij
From the definition of independence, it follows that
E  n  Pi  P j.
ij
From the row and column totals, we can estimate:
Pi 
R
,
n
P j 
C
i
j
n
Substituting in these estimates, we get:
The expected count in cell (i, j) is
E 
ij
RC
i
n
j
Example: Using the data from the contingency table of the previous
example, we can calculate:
E 
RC
60  48

 28.8,
n
100
E 
RC
60  52

 31.2,
n
100
E 
RC
40  48

 19.2,
n
100
E 
RC
40  52

 20.8.
n
100
11
12
21
22
1
1
2
2
1
2
1
2
The resulting table of expected counts is shown below:
Profit
Loss
Industry Type
Service
Non-service
28.8
31.2
19.2
20.8
Expected Counts
Profit
Loss
Industry Type
Service
Nonservice
28.8
31.2
19.2
20.8
Actual Counts
Profit
Loss
Total
Industry Type
Service
Nonservice
42
18
6
34
48
52
Total
60
40
100
The chi-square test statistic for independence is given by:
  
r
c
2
i 1
With degrees of freedom:
j 1
O
ij
E
E
ij

2
ij
 r  1  c  1
Note that the chi-square test statistic is always nonnegative. If the observed counts exactly equal the expected
(under the hypothesis of independence) counts, then the
value of the test statistic would be zero. The greater the
difference between observed and expected counts, the
larger the test statistic becomes.
We would reject the null hypothesis (of independence) only
when we obtain a sufficiently large value of the test
statistic.
We can compute the chi-square test statistic for our
example as follows:

2
42  28.8

28.8
2
18  31.2 

31.2
2
6  19.2 

19.2
2
34  20.8

20.8
2
 29.09 .
The number of degrees of freedom is 1 degree of freedom.
Using the CHIINV function to compute a critical value for
 = 0.01, we see that CHIINV(0.01, 1) = 6.63. The
rejection region (for a confidence level of 99%) is any test
statistic larger than 6.63.
Since our computed value of the statistic is much larger, we
would reject the null hypothesis. Alternatively, we could
compute the p-value by using the CHIDIST function:
CHIDIST(29.09, 1) = 6.91 * 10-8
from which we see that we would reject the null hypothesis
for any reasonable value of .
Using the CHITEST function:
An easier way to do this is to use the EXCEL CHITEST
function. To do this, you need to have two tables in your
spreadsheet
 one containing the original contingency table data of
actual observed counts, and
 one containing the expected counts
The following spreadsheet information contains:
 the original counts in range A1:B2,
 the expected counts in range D1:E2,
 the chitest function formula in cell F1, and
 the value returned by the chitest function formula in
cell F2.
42
6
18
34
2 8 .8
1 9 .2
3 1 .2
2 0 .8
= C H IT E S T (A 1 :B 2 ,D 1 :E 2 )
6 .9 2 1 6 2 6 1 2 2 2 0 6 2 3 E -0 8
Note that the value returned by the chi-test function
formula is the p-value.
Chi-Test Examples
• See 95 Murders.xls spreadsheet