Title: Introduction to Hypothesis Testing
1Introduction to Hypothesis Testing
21 Introduction
- The purpose of hypothesis testing is to determine
whether there is enough statistical evidence in
favor of a certain belief about a parameter. - Examples
- Is there statistical evidence in a random sample
of potential customers, that support the
hypothesis that more than 10 of the potential
customers will purchase a new products? - Is a new drug effective in curing a certain
disease? A sample of patients is randomly
selected. Half of them are given the drug while
the other half are given a placebo. The
improvement in the patients conditions is then
measured and compared.
3What is a hypothesis?
A tentative statement about a population
parameter that might be true or wrong
4Types of hypotheses
- Null hypothesis
- Research/alternative hypothesis
52 Concepts of Hypothesis Testing
- The critical concepts of hypothesis testing.
- Example
- An operation manager wants to determine if the
mean demand during lead time is greater than 350. - If so, the ordering policy should be changed.
- The two hypotheses about a population mean
- H0 The null hypothesis m 350
- H1 The alternative hypothesis m gt 350
This is what you want to prove
62 Concepts of Hypothesis Testing
- Assume the null hypothesis is true (m 350).
m 350
- Sample from the demand population, and build a
test statistic related to the parameter
hypothesized (the sample mean). - Decide, is the value obtained big enough to
reject the null hypothesis?
72 Concepts of Hypothesis Testing
- Assume the null hypothesis is true (m 350).
m 350
- In this case the mean m is not likely to be
greater than 350. Do not reject the null
hypothesis.
8Types of Errors
- Two types of errors may occur when deciding
whether to reject H0 based on the statistic
value. - Type I error Reject H0 while the truth is, it is
true. - Type II error Do not reject H0 while the truth
is, it is false. - Example continued
- Type I error Reject H0 (m 350) in favor of H1
(m gt 350) while the truth is, the real value of m
is 350. - Type II error Do not reject H0 (m 350) while
the truth is, the real value of m is greater than
350.
9Controlling the probability of conducting a type
I error
- Recall
- H0 m 350 and H1 m gt 350.
- H0 is rejected if is sufficiently large
- Thus, a type I error is made if
when m 350. - By properly selecting the critical value we can
limit the probability of conducting a type I
error to an acceptable level.
10What is a critical value?
A value needed to determine weather to reject or
not to reject the null hypothesis.
113 Testing the Population Mean When the
Population Standard Deviation is Known
- The manager of a department store is
thinking about establishing a new billing system
for the stores credit customer. After a through
financial analysis, she determines that the new
system will be cost effective only if the mean
monthly account is more than 170. A random
sample of 400 monthly account is drawn, for which
the sample mean is 178. The manager knows that
the accounts are approximately normally
distributed with standard deviation 65. Can the
manager conclude from this available information
that the new system will be cost-effective?
123 Testing the Population Mean When the
Population Standard Deviation is Known
- Example 1
- A new billing system for a department store will
be cost- effective only if the mean monthly
account is more than 170. - A sample of 400 accounts has a mean of 178.
- If accounts are approximately normally
distributed with s 65, can we conclude that
the new system will be cost effective?
13 Testing the Population Mean (s is Known)
- Example 1 Solution
- The population of interest is the credit accounts
at the store. - We want to know whether the mean account for all
customers is greater than 170.
H1 m gt 170
- The null hypothesis must specify a single value
of the parameter m,
H0 m 170
14Approaches to Testing
- There are two approaches to test whether the
sample mean supports the alternative hypothesis
(H1) - The rejection region method is mandatory for
manual testing (but can be used when testing is
supported by a statistical software) - The p-value method which is mostly used when a
statistical software is available.
15The Rejection Region Method
The rejection region is a range of values such
that if the test statistic falls within that
range, the null hypothesis is rejected in favour
of the alternative hypothesis.
16Steps in rejection region method
- Construct appropriate hypotheses
- Determine a test statistics to be used
- Determine the critical value
- Compare the test statistic with the critical
value. Reject the null hypothesis if the former
is greater than the latter. - Make an appropriate conclusion.
17The Rejection Region Method for a Right - Tail
Test
- Example 1 solution continued
- Recall H0 m 170 H1 m gt 170
therefore, - It seems reasonable to reject the null
hypothesis and believe that m gt 170 if the
sample mean is sufficiently large.
Reject H0 here
Critical value of the sample mean
18The Rejection Region Method for a Right - Tail
Test
- Example 1 solution continued
- Define a critical value for that is
just large enough to reject the null
hypothesis.
19Determining the Critical Value for the Rejection
Region
- Allow the probability of committing a Type I
error be a (also called the significance level). - Find the value of the sample mean that is just
large enough so that the actual probability of
committing a Type I error does not exceed a.
Watch
20Determining the Critical Value for a Right
Tail Test
Example 1 solution continued
P(commit a Type I error) P(reject H0 given
that H0 is true)
is allowed to be a.
21Determining the Critical Value for a Right
Tail Test
Example 1 solution continued
a
0.05
22Determining the Critical value for a Right -
Tail Test
Conclusion Since the sample mean (178) is greater
than the critical value of 175.34, there is
sufficient evidence to infer that the mean
monthly balance is greater than 170 at the 5
significance level.
23The standardized test statistic
- Instead of using the statistic , we can use
the standardized value z. - Then, the rejection region becomes
One tail test
24The standardized test statistic
- Example 1 - continued
- We redo this example using the standardized test
statistic. - Recall H0 m 170
- H1 m gt 170
- Test statistic
- Rejection region z gt z.05 1.645.
25The standardized test statistic
Conclusion Since Z 2.46 gt 1.645, reject the
null hypothesis in favor of the alternative
hypothesis.
26P-value Method
- The p-value provides information about the amount
of statistical evidence that supports the
alternative hypothesis.
27P-value Method
The probability of observing a test statistic at
least as extreme as 178, given that m 170 is
The p-value
28Interpreting the p-value
- Because the probability that the sample mean
will assume a value of more than 178 when m 170
is so small (.0069), there are reasons to believe
that m gt 170.
29Interpreting the p-value
We can conclude that the smaller the p-value the
more statistical evidence exists to support the
alternative hypothesis.
30Interpreting the p-value
- Describing the p-value
- If the p-value is less than 1, there is
overwhelming evidence that supports the
alternative hypothesis.
- If the p-value is between 1 and 5, there is a
strong evidence that supports the alternative
hypothesis. - If the p-value is between 5 and 10 there is a
weak evidence that supports the alternative
hypothesis. - If the p-value exceeds 10, there is no evidence
that supports the alternative hypothesis.
31The p-value and the Rejection Region Methods
- The p-value can be used when making decisions
based on rejection region methods as follows - Define the hypotheses to test, and the required
significance level a. - Perform the sampling procedure, calculate the
test statistic and the p-value associated with
it. - Compare the p-value to a. Reject the null
hypothesis only if p-value lta otherwise, do not
reject the null hypothesis.
32Conclusions of a Test of Hypothesis
- If we reject the null hypothesis, we conclude
that there is enough evidence to infer that the
alternative hypothesis is true. - If we do not reject the null hypothesis, we
conclude that there is not enough statistical
evidence to infer that the alternative hypothesis
is true.
The alternative hypothesis is the more
important one. It represents what we are
investigating.
33A Left - Tail Test
- The SSA Envelop Example.
- The chief financial officer in FedEx believes
that including a stamped self-addressed (SSA)
envelop in the monthly invoice sent to customers
will decrease the amount of time it take for
customers to pay their monthly bills. - Currently, customers return their payments in 24
days on the average, with a standard deviation of
6 days.
34A Left - Tail Test
- The SSA envelop example continued
- It was calculated that an improvement of two days
on the average will cover the costs of the
envelops (checks can be deposited earlier). - A random sample of 220 customers was selected and
SSA envelops were included with their invoice
packs. - The times customers payments were received were
recorded (SSA.xls) - Can the CFO conclude that the plan will be
profitable at 10 significance level? ?
35A Left - Tail Test
- The SSA envelop example Solution
- The parameter tested is the population mean
payment period (m). - The hypotheses areH0 m 22H1 m lt 22 (The
CFO wants to know whether the plan will be
profitable)
36A Left - Tail Test
- The SSA envelop example Solution continued
- The rejection region It makes sense to believe
that m lt 22 if the sample mean is sufficiently
smaller than 22. - Reject the null hypothesis if
37A Left -Tail Test
Left-tail test
- The SSA envelop example Solution continued
- The standardized one tail left hand test is
Define the rejection region
Since -.91 gt 1.28 do not reject the null
hypothesis. The p value P(Zlt-.91)
.1814Since .1814 gt .10, do not reject the null
hypothesis
38A Two - Tail Test
- Example 2
- ATT has been challenged by competitors who
argued that their rates resulted in lower bills. - A statistics practitioner determines that the
mean and standard deviation of monthly
long-distance bills for all ATT residential
customers are 17.09 and 3.87 respectively.
39A Two - Tail Test
- Example 2 - continued
- A random sample of 100 customers is selected and
customers bills recalculated using a leading
competitors rates (see Xm11-02). - Assuming the standard deviation is the same
(3.87), can we infer that there is a difference
between ATTs bills and the competitors bills
(on the average)?
40A Two - Tail Test
- Solution
- Is the mean different from 17.09?
H0 m 17.09
- Define the rejection region
-
41A Two Tail Test
Solution - continued
17.09
We want this erroneous rejection of H0 to be a
rare event, say 5 chance.
42A Two Tail Test
Solution - continued
17.09
43A Two Tail Test
Two-tail test
There is insufficient evidence to infer that
there is a difference between the bills of ATT
and the competitor.
Also, by the p value approach The p-value P(Zlt
-1.19)P(Z gt1.19) 2(.1173) .2346 gt .05
a/2 0.025
a/2 0.025
-1.19
1.19