1.
Learning Objectives
• Know how to implement the Hypothesis, Test, Action,
Business (HTAB) system to test hypotheses.
• Understand the logic of Statistical Hypothesis Testing:
– Know how to establish null and alternative hypotheses
– Understand Type I and Type II errors
– Calculate the probability of Type II error when failing to reject the null
hypothesis
• Understand how to test a hypothesis about a single
population parameter:
– Mean
• when
• when
is known (using z-statistic)
is unknown (using t-statistic)
2.
Types of Hypotheses
1. Research Hypothesis
– a statement of what the researcher believes will be
the outcome of an experiment or a study.
2. Statistical Hypothesis
– a more formal structure derived from the scientific
method
– composed of two parts :
• Null hypothesis (Ho) – the assumed value of the parameter if
there is no effect/ impact. We will conclude that this could
be true unless there is a small chance of getting a sample
statistic (mean/proportion/variance) as extreme or more
extreme than from the data (small p-value).
• Alternative hypothesis (Ha) – a statement of whether the
true population parameter is higher, lower, or not equal to
that hypothesized in the null hypothesis.
3.
Types of Hypotheses
3. Substantive Hypothesis - a statistically
significant difference does not imply or mean
a material, substantive difference.
– If the null hypothesis is rejected and the
alternative hypothesis is accepted, then one
can say that a statistically significant result
has been obtained
– With “significant” results, you reject the null
hypothesis
4.
Using the HTAB System
• H – Hypotheses
– Establish the hypothesis
• T – Test
– Conduct the test
• A – Action
– Take statistical action
• B – Business Implications
– Determine the business implications
5.
CPA Salary Example
Example: A survey of CPAs in the U.S., done 15
years ago, found that their average salary was
$74,914. An accounting researcher would like to
test whether this average has changed over the
years. A random sample of 112 CPAs produced a
mean salary of $78,695. Assume that the
population standard deviation of salaries is =
$14,530 (note: this value is typically not known, but
we will assume it for mathematical simplicity. Later,
we will remove this assumption).
6.
Step 1: Hypothesis
Set up the null and alternative hypotheses
Always contains “=“
H0 :
Ha :
$74,914
$74,914
> or < or ≠
7.
Null and Alternative Hypotheses
• The null and alternative hypotheses are mutually
exclusive.
– Only one of them can be selected.
• The null hypothesis is assumed to be true. It is
compared to the observed data via either a
critical value (critical value method) or by
calculating a p-value (p-value method)
• The burden of proof falls on the alternative
hypothesis. Thus, you either reject the null in
favour of the alternative or you fail to reject the
null in favour of the alternative. The latter
statement does not imply that the null is true.
8.
Examples: One-tailed and
Two-tailed Tests
• One-tailed Test
- Means
40
Ha :
- Proportions
H0 :
40
H 0 : p 0.18
H a : p 0.18
• Two-tailed Test
H0 :
12
Ha :
12
9.
Step 2: Determine Appropriate Test
• The z-statistic can be used to test when the
following three conditions are met:
– The data are a random sample from the population
– The sample standard deviation ( ) is known
– At least one of the following conditions are met:
• The sample size (n) is at least 30
• The underlying distribution is normal
• The value of z-statistic can be calculated using the
following formula
z
x
n
10.
Finite Population Correction
• If the sample is drawn from a finite population
of size N, the formula for z-statistic becomes…
z
X
N n
n N 1
11.
Step 3: Set significance level ( )
• Significance level ( ) or Type I error rate
– Committed by rejecting a true null hypothesis
– If the null hypothesis is true, any value that falls in
a rejection region will represent a Type I error.
– The probability of committing a Type I error is
referred to as , the level of significance.
• The significance level is usually set at 0.05.
Other common values are 0.1, 0.01, or 0.001.
12.
Type II Errors
• Type II Error
– Committed when a researcher fails to reject a false null
hypothesis
– The probability of committing a Type II error is referred
to as . Some refer to power, or 1- (the chance of
rejecting the null when it is false), instead.
• In practice, we do not know whether the null
hypothesis is true.
• Type I and Type II error rates are inversely related,
if you reduce one, you increase the other.
• One way of reducing both Type I and Type II error
rates is to increase the sample size, but that
requires more time and money.
13.
Decision Table for
Hypothesis Testing
State of Null (Truth)
Decision:
Null True
Null False
Fail to
reject null
Correct
Decision
Type II error
( )
Reject null
Type I error
( )
Correct Decision
(Power)
14.
Step 4: Decision Rule
• A decision rule has to be made about when the
difference between the sample and hypothesized
population mean (under the null hypothesis) is small
or large.
• The rejection region is the area on the curve where
the null hypothesis is rejected. Here the value of the
sample mean is too far from the hypothesized
population mean to conclude that they are the same.
• The nonrejection region is the area where the null
hypothesis is not rejected. Here the sample mean is
close enough to the hypothesized population mean
to conclude that the null hypothesis could be true.
15.
Rejection and Non-rejection Regions
Rejection Region
Rejection Region
Non Rejection Region
=$74,914
Critical Value
Critical Value
16.
Decision rule – CPA Example
Rejection Region
Rejection Region
Non Rejection Region
=$74,914
Z
/2
= Z0.025 = -1.96
Z1-
Z=0
/2
= Z0.975 = +1.96
17.
Critical Z values:
Level of significance = 0.05
Non-rejection Region
95%
.025
.4750
.025
.4750
X
-1.96
0
1.96
Z
18.
Values of z for common
Levels of Confidence
Confidence Level
90%
95%
98%
99%
Z Value
1.645
1.96
2.33
2.575
19.
Decision Rule – CPA Example
Thus, we will reject H0 if Z > 1.96 or Z < -1.96
or reject H0 if |Z| > 1.96
20.
Critical Value Based on
Sample Mean
Alternatively, you could calculate a critical value based
not on Z, but on x-bar.
zc xc
or
xc
so
n
1.96
xc 74,914
14,530 112
14,530 74,914 2,691
74,914 1.96
112
lower xc 72,223 and upper xc 77,605
Thus, we would reject the null if the sample mean is
above $77,605 or below $72,223.
21.
One-tailed Tests
Depending on the problem, one-tailed tests are sometimes
appropriate.
H0 :
40
H0 :
40
Ha :
40
Ha :
40
Rejection Region
Non Rejection Region
=40 oz
Critical Value
Rejection Region
Non Rejection Region
=40 oz
Critical Value
22.
Step 5 (Gather Data) and
Step 6 (Compute Test Statistic)
z
X
78695 74914 2.75
/ n 14530 / 112
24.
Step 8: Business Decision
Statistically, the researcher has enough evidence
to reject the figure of $74,914 as the true average
salary for CPAs. Based on the evidence gathered,
it suggests that the average has increased over
the 15-year period.
25.
Alternative Method: the p-value
• p-value – another way to reach statistical conclusion in
hypothesis testing
If the null hypothesis is true, the p value is the probability of
getting a sample mean as extreme or more extreme than
what you observed.
If the sample mean is in the rejection region, the p-value will
be small. These two methods are always consistent.
• p-value <
reject H0, p-value
do not reject H0
• For two tailed test, /2 is used in each region
The p value is then compared to α/2 instead of to determine
statistical significance.
Some statisticians (and packages) double the p-value for a two
sided test instead and compare to .
26.
p-value for CPA Example
Px
$78,695 |
$74,914
P
x
n
P( Z
2.75) 0.003
$78,695 $74,914
$14,530
112
27.
Review of 8 Steps
of Hypothesis Testing
1 – Establish the null and alternative hypotheses
2 – Determine the appropriate statistical test
3 – Set , the Type I error rate / significance level
4 – Establish the decision rule
5 – Gather sample data
6 – Analyze the data
7 – Reach a statistical conclusion
8 – Make a business decision
28.
The U.S. Farmers’ Production Company (USFPC) builds
large harvesters. For a harvester to be properly
balanced when operating, a 25-pound plate is
installed on its side. The machine that produces these
plates is set to yield plates that average 25 pounds.
The distribution of plates produced from the machine
is normal. However, the shop supervisor is worried
that the machine is out of adjustment and is
producing plates that do not average 25 pounds. To
test this, he randomly selects 20 of the plates from
the day before and weighs them.
29.
• Establish the null and alternative hypotheses
H0: =25 pounds (where
mean weight of all plates)
Ha: ≠25
• Determine the appropriate statistical test.
Recall from earlier session, the conditions for the tdistribution:
1. The sample was randomly selected from the population
2. The population standard deviation ( ) is unknown
3. One of these conditions are met:
The sample size (n) is at least 30 OR
the underlying distribution is normal
These conditions are met!
The degrees of freedom are n-1 = 20-1 = 19 in this example
30.
Recap: t Distribution
A family of distributions - a unique distribution for each
value of its parameter using degrees of freedom (d.f.),
every sample size having a different distribution
t
x
s
n
31.
t19
t
/2
t19
t0.025
t19
2.093
X
25.51 25 1.04
t
n 1 s / n 2.1933/ 20
32.
7 – Reach a statistical conclusion
Since |t| = 1.04 < tc = 2.093, do not reject H0
8 – Make a business decision
There is not enough evidence to show that the
plates are different from 25 pounds. (Note: Is this
because the true population mean is close to 25
pounds, or is there a large chance that we have
suffered from a Type II error? Good question –
more on calculating type II error rates later.)
33.
R: Lower Tail Test of Population Mean
when Population Variance is known.
Problem: Suppose a manufacturer claims that the mean
lifetime of a light bulb is more than 10,000 hours. In a sample
of 30 light bulbs, it was found that they only last 9,900 hours
on average. Assume the population standard deviation is 120
hours. At .05 significance level, can we reject the claim by the
manufacturer?
Go to
http://www.r-tutor.com/elementary-statistics/hypothesis-testing
34.
R: Upper Tail Test of Population Mean
when Population Variance is known.
Problem: Suppose the food label on a cookie bag states that
there is at most 2 grams of saturated fat in a single cookie. In a
sample of 35 cookies, it is found that the mean amount of
saturated fat per cookie is 2.1 grams. Assume that the
population standard deviation is 0.25 grams. At .05
significance level, can we reject the claim on food label?
Go to
http://www.r-tutor.com/elementary-statistics/hypothesis-testing
35.
R: Two-tailed Test of Population Mean
when Population Variance is known.
Problem: Suppose the mean weight of King Penguins found in
an Antarctic colony last year was 15.4 kg. In a sample of 35
penguins same time this year in the same colony, the mean
penguin weight is 14.6 kg. Assume the population standard
deviation is 2.5 kg. At .05 significance level, can we reject the
null hypothesis that the mean penguin weight does not differ
from last year?
Go to
http://www.r-tutor.com/elementary-statistics/hypothesis-testing
36.
R: Lower Tail Test of Population Mean
when Population Variance is unknown
Problem: Suppose the same manufacturer claims that the mean
lifetime of a light bulb is more than 10,000 hours. In a sample of
30 light bulbs, it was found that they only last 9,900 hours on
average. Assume the sample standard deviation is 125 hours.
At 0.05 significance level, can we reject the claim by the
manufacturer?
Go to
http://www.r-tutor.com/elementary-statistics/hypothesis-testing
37.
R: Upper Tail Test of Population Mean
when Population Variance is unknown
Problem: Suppose the same food label on a cookie bag states
that there is at most 2 grams of saturated fat in a single cookie.
In a sample of 35 cookies, it is found that the mean amount of
saturated fat per cookie is 2.1 grams. Assume that the sample
standard deviation is 0.3 gram. At 0.05 significance level, can
we reject the claim on food label?
Solution
Go to
http://www.r-tutor.com/elementary-statistics/hypothesis-testing
38.
R: Two Tail Test of Population Mean
when Population Variance is unknown
Problem: Suppose the mean weight of King Penguins found in
an Antarctic colony last year was 15.4 kg. In a sample of 35
penguins same time this year in the same colony, the mean
penguin weight is 14.6 kg. Assume the sample standard
deviation is 2.5 kg. At 0.05 significance level, can we reject the
null hypothesis that the mean penguin weight does not differ
from last year?
Go to
http://www.r-tutor.com/elementary-statistics/hypothesis-testing
39.
R: Two Tail Test of Population Mean
when Population Variance is unknown
Go to http://ww2.coastal.edu/kingw/statistics/Rtutorials/index.html
Enter Single Sample t test and practice the t.test() command
with the temperature data given
A particular slide catching your eye?
Clipping is a handy way to collect important slides you want to go back to later.
Be the first to comment