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Estimation-Confidence intervalsHypothesis testing
Population and sampling
 Population: is the total set of entities under study.
 example: the height of men, stock prices, all employees of a firm etc.
 Usually it is impossible to survey/measure the entire population
because not all members are observable. If it is possible to measure
the entire population it is often costly to do so and would take a great
deal of time.
 Sample: is a subset of the population. The sample is used to
develop hypotheses about the population under study.
 Researchers focus their analysis on the sample because of the
difficulty of studying the entire population.
 There are many ways to select a sample and the study of this is called
sampling theory. A commonly used method is called Random Sampling
Sampling distribution of the mean
 A sampling distribution of the mean can be thought of as a
relative frequency distribution of sample means
 first randomly select several samples from a population, then
compute the mean of each sample.
 the relative frequency distribution of all the sample means is the
sampling distribution of the mean
Example: let the 5 employees of the firm A be the population under investigation. Each
employee works specific hours per week. The following table demonstrates the hours of
work per week of each employee.
Employees
1.
Dunn
Hours
22
2. Hardy
26
3. Kiers
30
4. Malinowski
26
5. Tillman
22
Sampling distribution of the mean
 Samples of 2 employees are randomly formed . Therefore we
end up with 10 different samples of two employees.
 For each sample we compute the mean of the hours per
week
Samples
Mean of hours of
work per week
{Dunn, Hardy}
48/2 = 24
{Dunn, Kiers}
52/2 = 26
{Dunn, Malinowski}
48/2 = 24
{Dunn, Tillman}
44/2 = 22
{Hardy, Kiers}
56/2 = 28
{Hardy, Malinowski}
52/2 = 26
{Hardy, Tillman}
48/2 = 24
{Kiers, Malinowski}
56/2 = 28
{Kiers, Tillman}
52/2 = 26
{Malinowski,
Tillman}
48/2 = 24
Sampling distribution of the mean
 We construct a relative frequency distribution table of all the
sample means
Sample means
Frequencies
Relative frequencies
- probabilities
22
1
1/10
24
4
4/10
26
3
3/10
28
2
2/10
Graph of the sampling
distribution of the mean of
working hours per week
Relation between the sampling distribution of
the mean and the population distribution
 The mean of the sampling distribution of the mean equals to
the mean of the population from which the scores were
sampled.
 mean of the sampling distribution of the mean:
22 *1  24 * 4  26 * 3  28 * 2
X 
 25.2
10
 Population mean:

22  26  30  26  22
 25.2
5
 The variance of the sampling distribution of the mean
equals to the population variance divided by n (the sample
size)
2
 X2 

n
Central Limit Theorem
 Given a population with mean μ and variance  2, the sampling
distribution of the mean approaches a normal distribution
2
with a mean of μ and a variance of  / n as n, the sample
size, increases.
 How large must n be? Usually n ≥ 30
 Regardless of the shape of the population distribution, the
sampling distribution of the mean approaches a normal
distribution as n increases
X ~ N ( ,
2
n
)
Estimation
 Suppose we have an unknown population parameter
which we'd like to estimate. For instance, we are
interested in estimating a population mean μ or a
population proportion p.
 We can't possibly gather data for the entire population
 So we take a random sample from the population,
and use the resulting data to estimate the value of the
population parameter.
 Point Estimation: we compute a single value from the
sample data to estimate a population parameter.
 For example, the sample mean is a point estimate of
the population mean μ.
1 n
x   xi
n i 1
Interval Estimates
 After we make a point estimate , we are going to draw a
target (an interval) around the point and state the probability
that the real objective is in the target area. Interval estimates
are often desirable because the point estimates vary from
sample to sample.
 The wider the interval, the greater the probability.
 Also, the more accurate the point estimate, the greater the
probability.
 A confidence interval provides a range of values which is
likely to contain the population parameter of interest, usually
either 95% or 99% of the time. These intervals are referred to
as 95% and 99% confidence intervals respectively.
PL1    L2   0.95
where μ denotes the population mean, while
L1 and L2 represent the lower and upper
bounds of the interval
Confidence Intervals
 Confidence intervals are computed at a confidence level,
such as 95 %, selected by the user. A confidence level refers
to the percentage of all possible samples that can be expected
to include the true population parameter.
 (1- α)% denotes the confidence coefficient
 example: (1- 0.05)= 0.95 or 95% confidence interval is
defined as:
70.23 < μ < 100
 a 95% confidence interval is an interval where there is a 0.95
probability of containing the population mean
 the population mean lies between 70.23 and 100
Confidence Intervals
 The sample mean is used as a point estimate of the
population mean; we also use it to construct an interval
estimate for the population mean μ.
 If the population distribution is normal, then the point
estimate x will be approximately normally distributed, with a
mean μ and a standard error  n
 Central Limit Theorem: If the population distribution is not
normal, when sample size n is large, the point estimate x will
be approximately normally distributed, with a mean μ and a
standard error 
n
Confidence Intervals for the mean
 Before we take a random sample from the population, it is
likely (0.95) to find the sample mean in the interval
  1.96

n
 After we take a random sample from the population,
approximately 95% of the intervals
x  1.96

n
will include the population mean μ.
Therefore, x  1.96  is the 95% confidence interval for μ
n
Confidence Intervals for the mean
 When σ is known, or when n > 30, then the (1- α)%
confidence interval for μ is
x  za
2

n
where α is the desired significance level, zα/2 is a value of z
having a tail area of α/2 to its right in the standard normal
distribution.
Critical values of z and levels of confidence
0.99
0.98
0.95
0.90
0.80
2
0.005
0.010
0.025
0.050
0.100
Stand ard N o rm al Distrib utio n
z
0.4
2
2.576
2.326
1.960
1.645
1.282
(1   )
0.3
f(z)
(1   )

0.2
0.1


2
2
0.0
-5
-4
-3
-2
-1
z 
2
0
1
2
Z
z
2
3
4
5
How do we find the critical values from the
standard normal distribution table
For instance assume that α =5%
We find the probability (1- α/2) =
(1- 0.05/2)= 1- 0.025 = 0.9750
Then check the number of z that
corresponds to the specific row and
column of the probability
Confidence Intervals for the mean
 Example 1: European Management Association wishes to
estimate the average income of the administrators of the
European banking sector. The company used a random
sample of 256 administrators and found that the mean
income is 45420 euro. The standard deviation of the sample is
2.05. The company wants to investigate :
 What is the population mean income
 What is the a reasonable range of values for the total
population mean income
Confidence Intervals for the mean: example
 We use the sample mean to estimate the population mean
income. Therefore, the sample mean 45420 is the point
estimate of the unknown population mean parameter.
 We calculate a 95% confidence interval for the population
mean
X  z 0.05
2

n
 45420  1.96
2050
 45420  251  [45169,45671]
256
Confidence Intervals for the mean: example
 Example 2: The president of the MBA department wants to
estimate the weekly homework hours of the department’s
students. He finds that the average weekly hours are 24 with
standard deviation 4, based on a random sample of 49
students.
 What is the population mean?
 Compute a 95% confidence level for the population mean
Confidence Intervals for the mean: example
 We use the sample mean to estimate the population mean.
Therefore, the sample mean of 24 weekly hours is the point
estimate of the unknown population parameter
 The 95% confidence levels for the population mean are:
X  z 0.05
2

n
 24.00  1.96
4
49
 24.00  1.12
Thus, the confidence limits are 22.88 and 25.12 hours
Confidence Intervals for the mean
 When σ is unknown, and n < 30, then the confidence interval
for μ is given by
s
x  ta
, n 1
n
2
where tα/2,n-1 is the critical value of the t distribution at α/2 with
n -1 degrees of freedom, and s is the sample standard deviation
defined as
1 n
2
s
(
x

x
)
 i
n  1 i 1
t distribution
 The t distribution is a family of symmetric distributions
 is very similar to the normal distribution when the estimate
of variance is based on many degrees of freedom, but has
relatively more scores in its tails when there are fewer
degrees of freedom.
 As the degrees of freedom increase, Student's t distribution
approaches the normal distribution
How to find critical values from a t distribution
table
df
--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
40
60
120
t0.100
----3.078
1.886
1.638
1.533
1.476
1.440
1.415
1.397
1.383
1.372
1.363
1.356
1.350
1.345
1.341
1.337
1.333
1.330
1.328
1.325
1.323
1.321
1.319
1.318
1.316
1.315
1.314
1.313
1.311
1.310
1.303
1.296
1.289
1.282
t0.050
----6.314
2.920
2.353
2.132
2.015
1.943
1.895
1.860
1.833
1.812
1.796
1.782
1.771
1.761
1.753
1.746
1.740
1.734
1.729
1.725
1.721
1.717
1.714
1.711
1.708
1.706
1.703
1.701
1.699
1.697
1.684
1.671
1.658
1.645
t0.025
-----12.706
4.303
3.182
2.776
2.571
2.447
2.365
2.306
2.262
2.228
2.201
2.179
2.160
2.145
2.131
2.120
2.110
2.101
2.093
2.086
2.080
2.074
2.069
2.064
2.060
2.056
2.052
2.048
2.045
2.042
2.021
2.000
1.980
1.960
t0.010
-----31.821
6.965
4.541
3.747
3.365
3.143
2.998
2.896
2.821
2.764
2.718
2.681
2.650
2.624
2.602
2.583
2.567
2.552
2.539
2.528
2.518
2.508
2.500
2.492
2.485
2.479
2.473
2.467
2.462
2.457
2.423
2.390
2.358
2.326
t0.005
-----63.657
9.925
5.841
4.604
4.032
3.707
3.499
3.355
3.250
3.169
3.106
3.055
3.012
2.977
2.947
2.921
2.898
2.878
2.861
2.845
2.831
2.819
2.807
2.797
2.787
2.779
2.771
2.763
2.756
2.750
2.704
2.660
2.617
2.576
If n = 15, and α = 5%, then for degrees of
freedom (df) = n – 1 = 15 -1 =14, and α/2 =
0.05/2 = 0.025, the critical value is
tα/2, n-1 = 2.145
Confidence Intervals for the mean
 When σ is unknown, and n > 30, then the confidence interval
for μ is
x  za
2
s
n
where zα/2 is the critical value at α/2 of the standard normal
distribution and s is the sample standard deviation
1 n
2
s
(
x

x
)
 i
n  1 i 1
Hypothesis Testing
 Hypothesis testing is a procedure in which a statistical test is used to






evaluate whether there is enough evidence in a data sample to infer that a
certain condition is true for the entire population.
The usual process of hypothesis testing consists of five steps:
Formulate the null hypothesis H0 (that the observations are the result of
pure chance) and the alternative hypothesis H1 (that the observations
show a real effect ).
Determine a test statistic (Z) that can be used to assess the validity of
the null hypothesis.
Choose the level of statistical significance α, which controls the
probability of committing Type I error: P(Z  C1 ; H 0 is valid )  
Set the criteria for a decision. In particular, specify two regions, the
acceptance region C0, and the rejection region C1.
Make a decision. We accept the H0 when the value of the test statistic lies
in the acceptance region C0, while we reject the H0 and accept the H1 when
the value of the test statistic lies in the rejection region C1
Hypothesis Testing
 Step 1: Formulate the null hypothesis
 The null hypothesis (H0), stated as the null, is a statement
about a population parameter, such as the population mean,
that is assumed to be true. For example, H0: μ = 0
 An alternative hypothesis (H1) is a statement that directly
contradicts a null hypothesis by stating that the actual value
of a population parameter is less than, greater than, or
different from the value stated in the null hypothesis. For
example: H1: μ < 0 or H1: μ ≠ 16
Hypothesis Testing
 Step 2: Determine a test statistic
 The test statistic is a mathematical formula that allows us to
determine the likelihood of obtaining sample outcomes if the
null hypothesis were true. The value of the test statistic is
used to make a decision regarding the null hypothesis.
 The test statistic is a random variable because is a function of
random variables. For instance, consider hypothesis testing
for the population mean:
2
X 
X ~ N ( , )  Z 
~ N (0,1)
n
/ n
Hypothesis Testing
 Step 3: determine the level of significance (α)
 We decide whether to retain or reject the null hypothesis.
Because we are observing a sample and not an entire
population, it is possible that a conclusion may be wrong.
There are four decision alternatives regarding the truth and
falsity of the decision we make about a null hypothesis:
Hypothesis Testing
 Step 3: determine the level of significance (α)
 Type I error is the probability of rejecting a null hypothesis that is
actually true. Researchers directly control for the probability of
committing this type of error.
 Type II error, is the probability of retaining a null hypothesis that is
actually false.
Since we assume the null hypothesis is true, we control for Type I
error by stating a level of significance. The level we set, called the
alpha level (symbolized as α), is the largest probability of
committing a Type I error that we will allow and still decide to reject
the null hypothesis. This criterion is usually set at 0.05 (α = 0.05)
Hypothesis Testing
 Step 4: Set the criteria for a decision
 We form two regions: the acceptance region C0, and the
rejection region C1.
 The rejection region is the region beyond a critical value in a
hypothesis test. When the value of a test statistic is in the
rejection region, we decide to reject the null hypothesis;
otherwise, we retain the null hypothesis.
 Critical value is the value that separates the rejection region
from the acceptance region
Hypothesis Testing
 The rejection region is defined by the values for which we
reject the null hypothesis. These values which are considered
to be extreme (very large or very low), so that the
probabilities to occur if H0 is valid are minimum
Hypothesis Testing
 Step 5: Make a decision.
 To make a decision, we compare the obtained value of the test
statistic to the critical values. We reject the null hypothesis if the
obtained value exceeds a critical value.
Hypothesis Testing for the mean: twotailed testing
 Two tailed testing is when the alternative hypothesis is stated as
different than (≠).
 Step 1: We define null and alternative hypotheses:
 H0: μ = μ0, Η1: μ ≠ μ0
 Step 2: Choosing a test statistic:
 When the sample size is large (n>30) and the standard deviation of
the population is known, we use the Z statistic:
Z
X  0
~ N (0,1)
/ n
Hypothesis Testing for the mean: twotailed testing
 Step 3: Set the criteria for the decision
 We specify the acceptance region C0, and the rejection region
C1.


C0 : Z * : Z *  c


C1 : Z * : Z *  c
where Z * is the value of the Z test statistic and c denotes the
critical value of the standard normal distribution for a specific
level of significance (α)
Hypothesis Testing for the mean: twotailed testing
 Step 4: determine the level of significance (α):
 The level we set, is the largest probability of committing a Type I error
P(Z  C1 ; H 0 is valid )  
P( Z *  c; H 0 is valid )   
2 * P( Z *  c; H 0 is valid )   
or
P( Z *  c; H 0 is valid )   / 2  c  z / 2
Since we result in c = zα/2, we find the value of c from the standard normal
distribution table
Hypothesis Testing for the mean: twotailed testing
 Step 5: Make a decision
 for example, when α = 5%
Rejection region
Critical value
Rejection region
Critical value
Reject H0 if Z *  1.96 or Z *  1.96
Hypothesis Testing for the mean: σ unknown
and large sample size
 When the sample size is large (n > 30) and the standard
deviation of the population is unknown, we use the Z statistic:
X  0
Z
~ N (0,1)
s/ n
where s is the sample standard deviation, calculated as
1 n
2
s
(
x

x
)
 i
n  1 i 1
 The other steps of the hypothesis testing procedure
remain the same
Hypothesis Testing for the mean: two tailed
testing with σ unknown and small sample size
 Step 1: We define null and alternative hypotheses:
 H0: μ = μ0, Η1: μ ≠ μ0
 Step 2: Choosing a test statistic:
 When the sample size is small (n < 30) and the standard deviation
of the population is unknown, we use the t statistic:
t
X  0
~ t (n  1)
s/ n
where s is the sample standard deviation, and t is t distribution
with n – 1 degrees of freedom
Hypothesis Testing for the mean: two tailed
testing with σ unknown and small sample size
 Step 3: Set the criteria for the decision
 We specify the acceptance region C0, and the rejection region
C1.


C0 : t * : t *  c


C1 : t * : t *  c
where t * is the value of the t test statistic and c denotes the
critical value of the t distribution for a specific level of
significance (α) with n – 1 degrees of freedom
Hypothesis Testing for the mean: two tailed
testing with σ unknown and small sample size
 Step 4: determine the level of significance (α):
 The level we set, is the largest probability of committing a Type I error
P(t *  C1 ; H 0 is valid )  
P( t *  c; H 0 is valid )   
2 * P(t *  c; H 0 is valid )   
or
P(t *  c; H 0 is valid )   / 2  c  t / 2,n 1
Since we result in c = tα/2, n -1, we find the value of c from the t distribution
table . For example, when the degrees of freedom are 10, we get:
Level of Significance (α)
Critical Value tα/2
0.01
3.169
0.05
2.228
0.1
1.812
Hypothesis Testing for the mean: two tailed
testing with σ unknown and small sample size
 Step 5: Make a decision
 for example, when α = 5%
Rejection region
Rejection region
-2.226
2.226
Critical value
Reject H0 if t *  2.228 or t *  2.228
Critical value
Hypothesis Testing for the mean: one
tailed testing (with σ known)
 One tailed testing is when the alternative hypothesis is stated as
less than (<) or greater than (>).
 Step 1: We define null and alternative hypotheses:
 H0: μ = μ0, Η1: μ < μ0
 Step 2: Choosing a test statistic:
 When the sample size is large (n>30) and the standard deviation of
the population is known, we use the Z statistic:
Z
X  0
~ N (0,1)
/ n
Hypothesis Testing for the mean: one
tailed testing (with σ known)
 Step 3: Set the criteria for the decision
 We specify the acceptance region C0, and the rejection region
C1.
C0 : Z * : Z *  c
C1 : Z * : Z *  c
where Z * is the value of the Z test statistic and c denotes the
critical value of the standard normal distribution for a specific
level of significance (α)
Hypothesis Testing for the mean: one
tailed testing (with σ known)
 Step 4: determine the level of significance (α):
 The level we set, is the largest probability of committing a Type I error
P(Z *  C1 ; H 0 is valid )  
or
P(Z *  c; H 0 is valid )    c  za
Since we result in c = zα, we find the value of c from the standard normal
distribution table
Hypothesis Testing for the mean: one
tailed testing (with σ known)
 Step 5: Make a decision
 for example, when α = 5%
Rejection region
Critical value
Reject H0 if Z *  1.645
Hypothesis Testing for the mean: one tailed
testing (with σ unknown, small sample)
 Step 1: We define null and alternative hypotheses:
 H0: μ = μ0, Η1: μ < μ0
 Step 2: Choosing a test statistic:
 When the sample size is small (n < 30) and the standard deviation
of the population is unknown, we use the t statistic:
t
X  0
~ t (n  1)
s/ n
where s is the sample standard deviation, and t is t distribution
with n – 1 degrees of freedom
Hypothesis Testing for the mean: one tailed testing
(with σ unknown, small sample)
 Step 3: Set the criteria for the decision
 We specify the acceptance region C0, and the rejection region
C1.
C0 : t * : t *  c
C1 : t * : t *  c
where t * is the value of the t test statistic and c denotes the
critical value of the t distribution for a specific level of
significance (α) with n – 1 degrees of freedom
Hypothesis Testing for the mean: one tailed testing
(with σ unknown, small sample)
 Step 4: determine the level of significance (α):
 The level we set, is the largest probability of committing a Type I error
P(t *  C1 ; H 0 is valid )  
or
P(t *  c; H 0 is valid )    c  ta ,n1
Since we result in c = tα, we find the value of c from the t distribution
table. For example, when the degrees of freedom are 10, we get:
Level of Significance (α)
Critical Value tα/2
0.01
-2.764
0.05
-1.812
0.1
-1.372
Hypothesis Testing for the mean: one tailed
testing (with σ unknown, small sample)
 Step 5: Make a decision
 for example, when α = 5%
Rejection region
-1.812
Critical value
Reject H0 if t *  1.812
Hypothesis Testing for the mean: example 1
 Company A sells credit cards. The manager of this company
wants to find out whether the average amount of unpaid
cards overcomes the 400$. A random check of 172 unpaid
cards shows that the sample mean is 407$ while the
population standard deviation is 38$.
Can we infer that average amount of unpaid cards overcomes
the 400$ at level of significance 5%?
Hypothesis Testing for the mean: example




Step 1: H0: µ = $400, H1: µ > $400
Step 2: the level of significance is equal to 0.05
Step 3: since the sample size is large, and the population standard
deviation is known, we use the Z test statistic
Step 4: the null hypothesis H0 is rejected when Ζ >1.65
Z

X  0 $407  $400

 2.42
 n
$38 172
Step 5: Since 2.42 > 1.65, we reject the null hypothesis.
Therefore, we conclude that the average amount of unpaid cards overcomes
the 400$ at level of significance 5%
Hypothesis Testing for the mean: example 2
 Canon, Inc., launches a new machine that makes paper copies
faster than the previous copy machines. The company
selected randomly 24 new machines and estimated that the
average speed of the new machine is 24.7 paper copies per
second with standard deviation 7.4 seconds/copy. The older
models produced 27 papers per second.
 The firm wants to check whether the new machine has the
same average paper copying speed with the previous models
at level of significance 5%.
Hypothesis Testing for the mean: example 2



Step 1: H0: µ = 27, H1: µ ≠ 27
Step 2: the level of significance is equal to 0.05
Step 3: since the sample size is small, and the population standard
deviation is unknown, we use the t test statistic
 Step 4: the null hypothesis H0 is rejected when [t < -2.069] or t  2.069]
(for α/2 = 0.05/2 = 0.025 and n -1 = 24 -1 =23 degrees of freedom: tα/2 = 2.069)
t 

x  0
24.6 - 27

 1.59
s
7.4/ 24
n
Step 5: Since -1.59 > -2.069 , we accept the null hypothesis.
Therefore, we conclude that the new machine has the same average speed of
paper copying with the older models
The notion of the p-value
 The p-value represents the probability of the occurrence of a
specific value of the test statistic
 The p-value is used as an alternative to rejection points to provide
the smallest level of significance at which the null
hypothesis would be rejected.
 The smaller the p-value, the stronger the evidence is in favor of the
alternative hypothesis.
 Calculation of the p-value:

one-sided testing:
 Two-sided testing:

p  value  P Z  Z *


p  value  2  P Z  Z *

If p-value ≥ 0.05 , we have strong support of the null
hypothesis (accept H0) ; otherwise we reject the null
hypothesis
The notion of the p-value
 previous example:
 We found that the value of the test statistic is Z = 1.35. In this
case we have two-sided testing, so


p  value  2  P Z  Z *  2 PZ  1.35   2 PZ  1.35 
 2(1  PZ  1.35) 
 2(1  0.911492)  0.177016
 Since the p-value > 0.05, we find strong evidence in support of
the null hypothesis. Therefore, we accept H0