2340 MGMT Section W01
Business Statistics I
Instructor: E. Mark Leany
contact via Blackboard
online.uen.org
alternately:
[e-mail]
Data Distributed by Frequency
Alphabetical
from the Skew
67
Probabilities of Events
166
Areas Under the Normal Curve
This chart is the
same one as
the 3rd page
on your quiz
(and on the
test)
226
Sampling Distribution based on n
272
Confidence Intervals
302
One-Sample Tests of Hypothesis
Chapter 10
326
GOALS
1.
2.
3.
4.
5.
6.
7.
326
Define a hypothesis and hypothesis testing.
Describe the five-step hypothesis-testing procedure.
Distinguish between a one-tailed and a two-tailed test of
hypothesis.
Conduct a test of hypothesis about a population mean.
Conduct a test of hypothesis about a population
proportion.
Define Type I and Type II errors.
Compute the probability of a Type II error.
Hypothesis, Hypothesis and Testing
HYPOTHESIS A statement about the value of a population
parameter developed for the purpose of testing.
HYPOTHESIS TESTING A procedure based on sample
evidence and probability theory to determine whether the
hypothesis is a reasonable statement.
328
5 Steps in Testing a Hypothesis
1.
2.
3.
4.
5.
State the Null Hypothesis and the
Alternate (or Alternative) Hypothesis
Select a Level of Significance ( )
Select the Test Statistic
Formulate the Decision Rule
(Calculate &) Make a Decision
Step 1: State the Null Hypothesis
and the Alternate Hypothesis
NULL HYPOTHESIS A statement about the value of a
population parameter developed for the purpose of
testing numerical evidence.
ALTERNATE HYPOTHESIS A statement that is
concluded if the sample data provide sufficient
evidence that the null hypothesis is false.
329
Important Things to Remember about H0 and H1
329
H0: null hypothesis and H1: alternate hypothesis
H0 and H1 are mutually exclusive and collectively
exhaustive
H0 is always presumed to be true
H1 has the burden of proof
A random sample (n) is used to reject H0
If we conclude 'do not reject H0', this does not
necessarily mean that the null hypothesis is true, it only
suggests that there is not sufficient evidence to reject H0;
rejecting the null hypothesis then, suggests that the
alternative hypothesis may be true.
Equality is always part of H0 (e.g. = , , ).
< and > always part of H1
"Proving" using Hypothesis Tests
If you REJECT the NULL, you conclude that there
is significant evidence that the ALTERNATE is true.
If you DO NOT REJECT the NULL, you conclude
that there is NOT significant evidence that the
ALTERNATE is true.
Technically, you never really accept the NULL, you
just fail to reject it.
This is similar to court where a person is declared "NOT
GUILTY" which is not the same as "INNOCENT".
Note that ALL these "proofs" have a level of
328 possible error associated with them.
How to Set Up a Claim as Hypothesis
In actual practice, the status
quo is set up as H0
If the claim is boastful the
claim is set up as H1 (we
apply the Missouri rule
show me). Remember, H1
has the burden of proof
Inequality
Symbol
Part of:
Larger (or more) than
>
H1
Smaller (or less)
<
H1
Keywords
No more than
H0
At least
H0
Has increased
>
Is there difference?
In problem solving, look for
key words and convert them
into symbols. Some key
words include: improved,
better than, as effective as,
different from, has
changed, etc.
Has not changed
Has improved, is better than. is more
effective
H1
H1
=
See left text
H0
H1
Step 2: Select a Level of
Significance
LEVEL OF SIGNIFICANCE The probability of rejecting
the null hypothesis when it is true
This is known as ALPHA ( )
For 2-tailed test is calculated:
100% - Confidence Level
For 1-tailed test is calculated:
(100% - Confidence Level) / 2
330
Type of Errors in Hypothesis Testing
Type I Error
Defined as the probability of rejecting the null hypothesis
when it is actually true.
This is denoted by the Greek letter
Also known as the Significance Level of a test
Type II Error
Defined as the probability of NOT REJECTING the null
hypothesis when it is actually false.
This is denoted by the Greek letter
330
Decisions and Consequences in
Hypothesis Testing
ART - Type I: ALPHA ( ) REJECT when TRUE
BNF - Type II: BETA ( ) "NOT REJECT" when FALSE
331
Step 3: Select the Test Statistic
TEST STATISTIC A value, determined from sample
information, used to determine whether to reject the null
hypothesis.
Example: z, t, F,
2
z
X
Mean - Standard Deviation Known
n
Mean - Standard Deviation Unknown t
X
s
z
Proportion
331
p
(1
n
)
n
Step 4: Formulate the Decision Rule
One-tail vs. Two-tail Test
332
Hypothesis Setups for Testing a Mean ( )
CRITICAL VALUE The dividing point between the
region where the null hypothesis is rejected and the
region where it is not rejected.
Hypothesis Setups for Testing a
Proportion ( )
Step 5: (Calculate &) Make a Decision
Calculate the z or t value
Compare to the z or t value based on your predetermined Level of Significance
Decide whether to REJECT or NOT REJECT
(technically you never ACCEPT) the NULL
hypothesis
332
Additional Clarification of Reject or
NOT with Null vs. Alternate (1of4)
Consider a Court Case as a Hypothesis Test
1.
2.
H0: Defendant is INNOCENT v. HA: Def. is GUILTY
= 10/12 jurors
PRESUMED INNOCENT, but not PROVEN
To taken action (such as imprison), you must
prove GUILT significantly, but you can NOT
EVER prove INNOCENCE
Additional Clarification of Reject or
NOT with Null vs. Alternate (2of4)
Valid conclusions / statements
At "level of 10/12" there is significant evidence to find
the Defendant GUILTY
At "level of 10/12" there is NOT significant evidence to
find Defendant GUILTY; we cannot reject his claim of
INNOCENCE, but neither can we prove it
INVALID conclusions / statements
At "level of 10/12" there is significant evidence to find
the Defendant INNOCENT
At "level of 10/12" there is NOT significant evidence to
find the Defendant INNOCENT
Additional Clarification of Reject or
NOT with Null vs. Alternate (3of4)
Compare to a normal Hypothesis Test
1.
2.
H0: Salary is <= $50,000 v. HA: Salary is > $50,000
= 0.05
PRESUMED <= $50,000, but not PROVEN
To taken action (such as switch jobs), you must
prove > $50,000 significantly, but you can NOT
EVER prove <= $50,000
Additional Clarification of Reject or
NOT with Null vs. Alternate (4of4)
Valid conclusions / statements
At level of 0.05 there is significant evidence to conclude
that the salary is > $50,000
At level of 0.05 there is NOT significant evidence to
conclude that the salary is > $50,000; we cannot reject
the statement <= $50,000, but neither can we prove it
INVALID conclusions / statements
At level of 0.05 there is significant evidence to conclude
that the salary is <= $50,000
At level of 0.05 there is NOT significant evidence to
conclude that the salary is <= $50,000
Level of Significance vs.
% Confident
% confidence = 1 Related to confidence in "Confidence Interval" but
be careful not to confuse them
z is same for confidence interval and 2-tailed test
z for confidence interval does NOT match 1-tailed test,
since you don't have a 1-tailed confidence interval
You can state % confidence interchangeably with
level of significance (remember to fix the numbers)
Level of Significance vs.
% Confident - Examples
When you are rejecting the Null Hypothesis
At level of 0.05 there is significant evidence to conclude
that the salary is > $50,000
We are 95% confident that the salary is > $50,000
When you are NOT rejecting the Null Hypothesis
At level of 0.05 there is NOT significant evidence to
conclude that the salary is > $50,000
We are NOT 95% confident that the salary is > $50,000
IMPORTANT: Either of the following are incorrect:
We are 95% confident that the salary is NOT > $50,000
We are 95% confident that the salary is <= $50,000
"NOT % Confident" vs.
"% Confident that something is NOT"
Example:
2 balls in a bag (one is yellow, the other is white)
You pull out one ball without looking
Test the "hypothesis" that, at = 0.05, you will
randomly pulling out the yellow ball
True Statement
You are NOT 95% confident you will pick the yellow ball
False Statement
You are 95% confident you will NOT pick the yellow ball
Testing for a Population Mean with a
Known Population Standard Deviation- Example
Jamestown Steel Company
manufactures and assembles
desks and other office
equipment . The weekly
production of the Model A325
desk at the Fredonia Plant
follows the normal probability
distribution with a mean of 200
and a standard deviation of
16. Recently, new production
methods have been introduced
and new employees hired. The
VP of manufacturing would like
to investigate whether there
has been a change in the
weekly production of the
Model A325 desk.
NOTE: Test this for significance at the 0.01 level.
335
Testing for a Population Mean with a
Known Population Standard Deviation- Example
Step 1: State the null hypothesis and the alternate hypothesis.
H0:
H1:
= 200
200
(note: keyword in the problem has changed)
Step 2: Select the level of significance.
= 0.01 as stated in the problem
Step 3: Select the test statistic.
Use Z-distribution since is known
335
Testing for a Population Mean with a
Known Population Standard Deviation- Example
Step 3: Select the test statistic.
Use Z-distribution since is known
335
Testing for a Population Mean with a
Known Population Standard Deviation- Example
Step 4: Formulate the decision rule.
Reject H0 if |Z| > Z /2
Z
NOTE: z-value to
test against is
actually
2.576.
Z
X
/n
/2
Z
/2
203 . 5 200
Z .01 / 2
16 / 50
1 . 55 is not 2 . 58
Step 5: Make a decision and interpret the result.
Because 1.55 does not fall in the rejection region, H0 is not rejected.
We conclude that the population mean is not different from 200. So we would
report to the vice president of manufacturing that the sample evidence does
not show that the production rate at the plant has changed from 200 per
week.
336
Testing for a Population Mean with a Known
Population Standard Deviation- Another Example
Suppose in the previous problem the vice president
wants to know whether there has been an increase
in the number of units assembled. To put it another
way, can we conclude, because of the improved
production methods, that the mean number of
desks assembled in the last 50 weeks was more
than 200?
Recall: =16, n=200, =.01
338
One-Tailed Test versus Two-Tailed Test
NOTE: z-values to test against are actually 2.576 and 2.326.
338
Testing for a Population Mean with a Known
Population Standard Deviation- Example
Step 1: State the null hypothesis and the alternate hypothesis.
H0:
H1:
200
> 200
(note: keyword in problem the an increase)
Step 2: Select the level of significance.
= 0.01 as stated in the problem
Step 3: Select the test statistic.
Use Z-distribution since is known
338
Testing for a Population Mean with a Known
Population Standard Deviation- Example
Step 4: Formulate the decision rule.
Reject H0 if Z > Z
NOTE: z-value to test
against is actually
2.326.
Step 5: Make a decision and interpret the result.
Because 1.55 does not fall in the rejection region, H0 is not rejected. We
conclude that the average number of desks assembled in the last 50 weeks
is not more than 200
p-Value in Hypothesis Testing
p-VALUE is the probability of observing a sample value as
extreme as, or more extreme than, the value observed,
given that the null hypothesis is true.
This is EASY. It is the probabilities that we learned to calculate back
in chapters 7 and 8.
In testing a hypothesis, we can also compare the p-value to
the significance level ( ).
Decision rule using the p-value:
Reject H0 if p-value < significance level
339
Calculating the p-value
1.
2.
3.
Calculate the z-score
Using the Normal chart, find the probability of
getting (more than / less than) this score, based
on the direction of the equality expression in
the ALTERNATE hypothesis
If it is a two-tailed test, multiple the result by 2.
NOTE: The final calculated p-value can be
between 0.00 and 1 (0% and 100%)
Calculating the p-value - Details
If HA:
Find probability Z > Z-value
If HA:
a.
b.
<#
Find probability Z < Z-value
If HA:
>#
#
If Z-value is positive, Find probability Z > Z-value
If Z-value is negative, Find probability Z < Z-value
Multiply that answer by 2
p-Value in Hypothesis Testing - Example
Recall the last problem where the
hypothesis and decision rules were set
up as:
H 0:
H 1:
200
> 200
Reject H0 if Z > Z
where Z = 1.55 and Z =2.33
Reject H0 if p-value <
0.0606 is not < 0.01
Conclude: Fail to reject H0
339
What does it mean when p-value < ?
(a) .10, we have some evidence that H0 is not true.
(b) .05, we have strong evidence that H0 is not true.
(c) .01, we have very strong evidence that H0 is not true.
(d) .001, we have extremely strong evidence that H0 is not
true.
339
LYING with Statistics
?
Create Hypothesis at 0.05
Run the test and find INSIGNIFICANT
Find p-value = 0.07
Rerun at 0.10 to state SIGNIFICANT
THIS is BAD statistics
NOTE: Not bad to run a test and state
p-value (even if you do not reject Null)
Testing for the Population Mean:
Population Standard Deviation Unknown
When the population standard deviation ( ) is
unknown, the sample standard deviation (s) is used in
its place
The t-distribution is used as test statistic, which is
computed using the formula:
341
Testing for the Population Mean: Population
Standard Deviation Unknown - Example
The McFarland Insurance Company Claims Department reports the mean cost to
process a claim is $60. An industry comparison showed this amount to be larger
than most other insurance companies, so the company instituted cost-cutting
measures. To evaluate the effect of the cost-cutting measures, the Supervisor
of the Claims Department selected a random sample of 26 claims processed
last month. The sample information is reported below.
At the 0.01 significance level is it reasonable a claim is now less than $60?
342
Testing for a Population Mean with a
Known Population Standard Deviation- Example
Step 1: State the null hypothesis and the alternate hypothesis.
H0:
H1:
$60
< $60
(note: keyword in the problem now less than)
Step 2: Select the level of significance.
= 0.01 as stated in the problem
Step 3: Select the test statistic.
Use t-distribution since is unknown
342
t-Distribution Table (portion)
343
Testing for a Population Mean with a
Known Population Standard Deviation- Example
Step 4: Formulate the decision rule.
Reject H0 if t < -t ,n-1
NOTE: Calculation
discrepancy in
z-value
343
Step 5: Make a decision and interpret the result.
Because -1.818 does not fall in the rejection region, H0 is NOT
REJECTED at the .01 significance level. We have NOT demonstrated
that the cost-cutting measures reduced the mean cost per claim to less
than $60. The difference of $3.58 ($56.42 - $60) between the sample
mean and the population mean could be due to sampling error.
Testing for a Population Mean with an Unknown
Population Standard Deviation- Example
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.
At the .05 significance level can Neary conclude that the new
machine is faster?
Testing for a Population Mean with an Unknown
Population Standard Deviation- Example
Step 1: State the null and the alternate hypothesis.
H0: 250
H1: > 250
Step 2: Select the level of significance.
It is .05.
Step 3: Find a test statistic.
Use the t distribution because the population standard deviation is not
known and the sample size is less than 30.
Testing for a Population Mean with an Unknown
Population Standard Deviation- Example
Step 4: State the decision rule.
There are 10 1 = 9 degrees of freedom. The null
hypothesis is rejected if t > 1.833.
t
X
s
256 250
n
6
10
3.162
Step 5: Make a decision and interpret the results.
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 x/n
The test statistic is computed as follows:
350
Assumptions in Testing a Population Proportion using
the z-Distribution
A random sample is chosen from the population.
It is assumed that the binomial assumptions discussed in Chapter 6 are
met:
(1) the sample data collected are the result of counts;
(2) the outcome of an experiment is classified into one of two mutually
exclusive categoriesa success or a failure;
(3) the probability of a success is the same for each trial;
(4) the trials are independent
The test we will conduct shortly is appropriate when both n and n(1- )
are at least 5.
When the above conditions are met, the normal distribution can be used
as an approximation to the binomial distribution
349
Assumptions when using a Normal
approximation for hypothesis testing
with a proportion (short version)
1.
It meets the conditions of a BINOMIAL
distribution
2.
Both n*
349
and n*(1- ) are at least 5
Test Statistic for Testing a Single
Population Proportion - Example
Suppose prior elections in a certain state indicated it is
necessary for a candidate for governor to receive at
least 80 percent of the vote in the northern section of
the state to be elected. The incumbent governor is
interested in assessing his chances of returning to
office and plans to conduct a survey of 2,000
registered voters in the northern section of the state.
Using the hypothesis-testing procedure, assess the
governors chances of reelection.
NOTE: Test this for significance at the 0.01 level.
349
Test Statistic for Testing a Single
Population Proportion - Example
Step 1: State the null hypothesis and the alternate hypothesis.
H0:
H1:
.80
< .80
(note: keyword in the problem at least)
Step 2: Select the level of significance.
= 0.01 as stated in the problem
Step 3: Select the test statistic.
Use Z-distribution since the assumptions are met and
n and n(1- ) 5
350
Testing for a Population Proportion - Example
Step 4: Formulate the decision rule.
Reject H0 if Z < -Z
351
Step 5: Make a decision and interpret the result.
The computed value of z (-2.80) is in the rejection region, so the null hypothesis is
rejected at the .05 level. The difference of 2.5 percentage points between the sample
percent (77.5 percent) and the hypothesized population percent (80) is statistically
significant. The evidence at this point does not support the claim that the incumbent
governor will return to the governors mansion for another four years.
Type II Error
Recall Type I Error, the level of significance, denoted
by the Greek letter , is defined as the probability
of rejecting the null hypothesis when it is actually
true.
Type II Error, denoted by the Greek letter ,is defined as
the probability of FAILING TO REJECT the null hypothesis
when it is actually false.
352
Type II Error - Example
A manufacturer purchases steel bars to make cotter pins. Past
experience indicates that the mean tensile strength of all
incoming shipments is 10,000 psi and that the standard
deviation, , is 400 psi.
You are going to create a hypothesis test sampling incoming
shipments of steel bars at the .05 significance level. If the
tensile strength of the incoming bars is significantly different
than 10,000 psi, you will reject the shipment.
Suppose the population mean of an incoming shipment is
really 9,900 psi. What is the probability of a Type II Error?
352
H0 :
10,000 psi vs. H A :
10,000 psi
Type I and Type II Errors Illustrated
353
Calculation of a "Critical X-Bar"
used with a Type II Error
Xc
zc
zc
n
n
zc
Xc
Xc
Xc
10,000 1.96 400
Xc
10,000 1.96 400
100
100
n
zc
Xc
n
10,000 78.4 10,078.4
10,000 78.4 9,921.6
RESTATE decision based on Critical X-Bar: "At the 0.05 significance
level if the sample mean strength falls between 9,921.6 psi and
10,078.4 psi, accept the lot. Otherwise the lot is to be rejected."
Calculation of a Type II Error based on
Critical X-Bar and Population Mean
z
Xc
1
n
Critical X-Bar varies based on the Significance ( )
Value of
1 varies based on the (supposed)
Actual Population Mean
z
z at 0.54 = 0.2054;
352
9,921.6 9,900
400
100
0.54
= 0.5000 - 0.2054 = 0.2946
IMPORTANT: This is more accurate than the Book.
Calculation of a Type II Error in ONE
step
Given:
z
z
Xc
zc
zc
n
n
1
zc
n
10,000 9,900
1.96
400
100
z at 0.54 = 0.2054;
z
and
Xc
1
n
1
n
1.96 2.5 0.54
= 0.5000 - 0.2054 = 0.2946
Limitations of Type II Errors
Type II Errors only exist if you "Fail to Reject" the
Null Hypothesis when it is False
If (supposed) Actual Population Mean fits the
parameters of the Null Hypothesis then, by
definition, = 0.00 (regardless of your sample)
Reason: If the Actual Population Mean is within
the Null Hypothesis, then it is impossible to
Falsely Accept (i.e. "Fail to Reject") it.
Limitations of Type II Errors Examples
H0 :
100 vs. H A :
For
1
100,
For
1
110,
For
1
85,
100
H0 :
0.00
can exist (and will be
can exist (and will be
100 vs. H A :
For
1
100,
For
1
110,
For
1
85,
0.00)
0.00)
100
0.00
can exist (and will be
0.00
0.00)