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Hypothesis testing
Inferental statistics
Estimation
Hypothesis testing
Estimation of a population parameter
from a sample
- Point
- Interval
Hypothesis testing is a procedure,
based on sample evidence and
probability theory, used to determine
whether the hypothesis is a
reasonable statement and should not
be rejected, or is unreasonable and
should be rejected.
No nothing about the population
parameter
We have a preliminary idea about
the population parameter
Previous class: estimation
Application of Hypothesis testing
1. Is
there a significant
objectives/properites?
difference
between
2. Is there a significant relationship between/among
variables?
Meaning of Hypothesis testing (1)
1. Starting point: we have an idea about the population ->
we expect that the sample shows our idea
Research
question
Statistical
formulation
• Null Hypothesis H0: A
statement about the value
of a population parameter.
It contains „=„
• Alternative Hypothesis H1:
Negotiation of H0
State Hypothesis
Nullhypothesis H0

Negation: alternate hypothesis H1
H0 : No significance relationship
H0 : No significance difference
H0 : Always contains „=„ symbol
Example
Compare the life expectancy of men and Women
State H0 and H1
Meaning of Hypothesis testing (2)
2. Compare the difference between the sample and our
idea with a formula

formula : test function
 If there is no difference the value is 0
 Higher difference means higher value of test function

result: value of test function
3. Make a decision: Is the difference significant? -> make a
decision on our idea
Steps of Hypothesis Testing
Chart Title
Step 1: State null and alternate hypotheses
Step 2: Select a statistical tests
Step 3: Compute the value of test statistic
Step 4: Setup decision rules
Step 5: Make a decision
Do not reject null
Reject null and accept alternate
Errors when we make decision
H0 true
H0 not true
Retain H0
Right decision
(1-a)
Error Type II
( b)
Reject H0
Error Type I
(a)
Right decision
(1-b)
Usage of spss
Search p-value, sig..
If sig<0,05 -> Reject H0
 Else Retain H0
Definitions
Level
of Significance: The probability of
rejecting the null hypothesis when it is actually
true.
Type I Error: Rejecting the null hypothesis
when it is actually true.
Type II Error: Accepting the null hypothesis
when it is actually false
A classification of tests
According the number of samples
One sample
Two samples
More than 2 samples
ONE SAMPLE TESTS
We will examine
Distribution of a variable
Population mean
Population standard. deviation
A proportion in the Population
Testing for the Population Mean
Population standard Population standard
deviation (σ) is
deviation (σ) is not
known
known
X  0
z
/ n
X  0
t
s/ n
When can we use z or t statistics?
If n>=100
 If 30<=n<100 but the distribution is not skewed to the right

strongly (skewness<+1)

If n<30 and there is a normal distribution
EXAMPLE 1
The processors of Fries’ Catsup indicate on
the label that the bottle contains 16 ounces
of catsup. The standard deviation of the
process is 0.5 ounces. A sample of 36
bottles from last hour’s production revealed
a mean weight of 16.12 ounces per bottle.
At the 0.05 significance level can we
conclude that the mean amount per bottle is
different from 16 ounces if the weight
distribution is normal.
EXAMPLE 1
continued
Step 1: State the null and the alternative
hypotheses:
H0:  = 16; H1:  16
Step
2: Identify the test statistic.
Because we know the population
standard
deviation, the test statistic is z.
Step 3: Compute test statistics.
X   0 16.12  16.00
z

 1.44
 n
0.5 36
EXAMPLE 1
continued
Step 4: State the decision rule:
z0.975=1.96
Retain H0 if z is within (-1.96;1.96)
Step 5: Make a decision
Retain the null hypothesis. We cannot conclude
the mean is different from 16 ounces.
EXAMPLE 2
Roder’s Discount Store chain issues its own
credit card. Lisa, the credit manager, wants
to find out if the mean monthly unpaid
balance is more than $400. The level of
significance is set at 0.05. A random check of
172 unpaid balances revealed the sample
mean to be $407 and the sample standard
deviation to be $38. Should Lisa conclude that
the population mean is greater than $400?
EXAMPLE 2
continued
Step 1: H0: µ <400, H1: µ > $400
Step 2: Because standard deviation of the population is
not known we can use the t distribution as the test
statistic.
Step 3: Compute test statisics
X   0 $407  $400
t

 2.42
s n
$38 172
EXAMPLE 2
continued
• Step 4: Decision rule:
t0.95(171)=1.65
Retain H0 if t is within (-∞;1.65)
• Step 5: Make a decision.
H0 is rejected. Lisa can conclude that the mean unpaid
balance is greater than $400.
Example 3
The current rate for producing 5 amp fuses at
Neary Electric Co. is 250 per hour. A new
machine has been purchased and installed that,
according to the supplier, will increase the
production rate. A sample of 10 randomly
selected hours from last month revealed the
mean hourly production on the new machine was
256 units, with a sample standard deviation of 6
per hour. It is also known that the hourly
production follow a normal distribution. At the
0.05 significance level can Neary conclude that
the new machine is faster?
Example 3
continued
Step 1: State the null and the alternate hypothesis.
H0: µ<250; H1: µ > 250
Step 2: Find a test statistic.
It is the t distribution because the population
standard deviation is not known and the sample size
is less than 30, but normal distribution is assumed.
Step 3: Compute a test statistic.
X   0 256  250
t

 3.162
s n
6 10
Example 3
continued
Step 4: State the decision rule.
t0.95(9)=1.833
Retain H0 if t is within (-∞;1.833)
Step 5: Make a decision.
The null hypothesis is rejected. The mean number produced is more
than 250 per hour.
Tests Concerning Proportion
A Proportion is the fraction or percentage that
indicates the part of the population or sample
having a particular trait of interest.
The sample proportion is denoted by p and is
found by:
Number of successesin the sample
p
Number sampled
Test Statistic for Testing a Single
Population Proportion
z
p  0
 0 (1   0 )
n
The sample proportion is p and π is the population proportion.
EXAMPLE 4
In the past, 15% of the mail order solicitations for a
certain charity resulted in a financial contribution. A new
solicitation letter that has been drafted is sent to a
sample of 200 people and 45 responded with a
contribution. At the 0.05 significance level can it be
concluded that the new letter is more effective?
Example
4
continued
Step 1: State the null and the alternate
hypothesis.
H0: π< 0.15 H1: π > 0.15
Step 2: Find a test statistic.
The z distribution is the test statistic.
45statistic.
Step 3: Compute the test
 0.15
p  0
200
z

 2.97
0 (1  0 )
0.15(1  0.15)
200
n
Example 4 continued
• Step 4: State the decision rule.
z0.95=1.65
Retain H0 if z is within (-∞;1.65)
• Step 5: Make a decision.
The null hypothesis is rejected. More than 15 percent are responding with a
pledge. The new letter is more effective.