Transcript document

Statistics for Business and
Economics
Chapter 6
Inferences Based on a Single Sample:
Tests of Hypothesis
Learning Objectives
1. Distinguish Types of Hypotheses
2. Describe Hypothesis Testing Process
3. Explain p-Value Concept
4. Solve Hypothesis Testing Problems Based
on a Single Sample
5. Explain Power of a Test
Statistical Methods
Statistical
Methods
Descriptive
Statistics
Inferential
Statistics
Estimation
Hypothesis
Testing
Hypothesis Testing
Concepts
Hypothesis Testing
Population


 


I believe the
population mean
age is 50
(hypothesis).

Random
sample
Mean 
X = 20
Reject
hypothesis!
Not close.
What’s a Hypothesis?
A belief about a population
parameter
I believe the mean GPA of
this class is 3.5!
• Parameter is
population mean,
proportion, variance
• Must be stated
before analysis
© 1984-1994 T/Maker Co.
Null Hypothesis
1. Specified as H0:   Some numeric value
Written with = sign even if , or 
Example, H0:   3
2. Assumed to be true--status quo
3. Gives the sampling distribution of the data
and the test statistic (null distribution)
4. Cannot be proven; only rejected in favor of
the alternative.
Alternative Hypothesis
1. Designated Ha
2. Opposite of null hypothesis
3. Specified Ha:  < or  or > Some value
Example, Ha:  < 3
4. The research hypothesis: what you want
to prove
Identifying Hypotheses
Example
Historically, the average amount spent in the
bookstore has been $50. There is concern that it
may be dropping due to the student use of the
internet. Is the average amount spent in the
bookstore now less than $50?
1. H0:  = 50
2. Ha:  < 50
Basic Idea
Sampling Distribution
It is “impossible”
get a sample
mean of this
value ...
... therefore, we
reject the
hypothesis that
 = 50.
... if in fact this were
the population mean
20
 = 50
H0
Sample Means
Basic Idea
1.
If the sample mean looks as though it could have
come from the sampling distribution given by the null
hypothesis, then we will accept the null hypothesis.
2.
If the sample mean is way out on the tail, or
completely outside the sampling distribution given by
the null hypothesis, we should reject the null
hypothesis.
3.
Only work we have to do: decide what is inside,
and what is outside the distribution!
(Have to DRAW THE LINE!)
Where to Draw the Line
1. Cut point determined by (alpha-probability
of error)
•
Typical values are .01, .05, .10
2. Determines “how far in” to draw the line
•
•
Outside line: rejection region
Inside line: acceptance region
3. Selected by researcher at start
4. Alpha is the “significance level of the test”
Rejection Region
(One-Tail Test)
Sampling Distribution
Fail to reject here— test
statistic in acceptance
region---H0 OK
Rejection
Region
1–

Acceptance
Region
Critical
Value
Ho
Value
Sample Statistic
Observed sample statistic
Rejection Regions
(Two-Tailed Test)
Sampling Distribution
Rejection
Region
Rejection
Region
1–
1/2 
Reject
H0 here!
1/2 
Acceptance
Region
Critical
Value
Ho
Value
Critical
Value
Observed sample statistic
Hypothesis Testing Steps
Hypothesis Testing Steps
1.
2.
3.
State H0, H1, , and n
Collect data and compute test statistic (Xbar)
Draw the line(s)
One tail “<“ alternative:  percentile of Xbar dist’n
One tail “>” alternative: 1 -  percentile
Two tail alternative: /2 and (1 – )/2 percentiles
4.
Draw conclusions.
One-Tailed Z Test Thinking
Challenge
You’re an analyst for Ford. Historically,
Ford Focus models have averaged 32
mpg. You want to test the research
hypothesis (alternative) that the average
miles per gallon of the Ford Focus is now
greater than 32 mpg. Similar models
have a standard deviation of 4.0 mpg.
You take a sample of 64 Focus models
and compute a sample mean of 33.7
mpg. Test at the 0.01 level of
significance. Interpret the results.
Alone
Group Class
One-Tailed Z Test Solution
Step:
1.
2.
3.
4.
Interpretation:
Two-Tailed Z Test Example
Does a cereal production line
produce boxes containing an
average of 368 grams of cereal as
specified on the box---or has the
value changed? To test these
hypotheses, a random sample of
36 boxes showedX = 372.5. The
company has specified  to be 12
grams. Test at the .05 level of
significance. Interpret the results
368 gm.
Two-Tailed Z Test Solution
Step:
1.
2.
3.
4.
Interpretation:
Observed Significance Levels:
p-Values
Short cut:
If you have these, you do not need to
calculate rejection region(s)
p-Value
1.
Probability of obtaining a test statistic more
extreme (or than the actual sample
value given H0 is true
2.
Called observed level of significance
Smallest value of  H0 can be rejected
3.
Used to make rejection decision
If p-value  , do not reject H0
If p-value < , reject H0
Hypothesis Testing Steps Using P-value
(Step 3 reject regions replaced)
1.
2.
3.
State H0, H1, , and n
Collect data and compute test statistic (Xbar)
Compute p-value
One tail “<“ alternative: P(test stat < observed val)
One tail “>” alternative: P(test stat > observed val)
Two tail alternative: Compute one-tail and double
4.
Draw conclusions: If p < alpha, reject null
“If p is low, the null’s gotta go”
One-Tailed Z Test via P-value
You’re an analyst for Ford. Historically,
Ford Focus models have averaged 32
mpg. You want to test the research
hypothesis (alternative) that the average
miles per gallon of the Ford Focus is now
greater than 32 mpg. Similar models
have a standard deviation of 4.0 mpg.
You take a sample of 64 Focus models
and compute a sample mean of 33.7
mpg. Test at the 0.01 level of
significance. Interpret the results.
Alone
Group Class
P-value Solution
Test statistic is Xbar = 33.7 and the alternative is
a “greater than” alternative. So:
p-value = P(Xbar > 33.7| the null is true)
Easy: Just use Minitab to find P(Xbar > 33.7)
using the null distribution:  = 32 and Xbar =
4/8=.5
Two-Tailed p-Value Solution
Just find the one-tail value and double it!
1/2 p-value
1/2 p-value
Z
Xbar (observed)
Two-Tailed P-value Example
You’re a Q/C inspector. You want to find out if
a new machine is making electrical cords to
customer specification: average breaking
strength of 70 lb. with  = 3.5 lb. You take a
sample of 36 cords & compute a sample mean
of 69.7 lb. At the .05 level of significance, is
there evidence that the machine is not meeting
the average breaking strength? Determine the
p-value, draw conclusions.
One Population Tests
One
Population
Mean
Proportion
Variance
(not covered)
Z Test
t Test
Z Test
c2 Test
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
What if  is unknown?
Use the sample standard deviation, s. Then:
1. If the null hypothesis is true and
2. If the raw data are normally distributed, then
The number of number of standard deviations from Xbar to 
follows a t distribution with n-1 degrees of freedom. The pvalue can be obtained using the t distribution.
One-Tailed t Test
Example
Is the average capacity of
batteries at least 140 amperehours? A random sample of 20
batteries had a mean of 138.47
and a standard deviation of
2.66. Assume a normal
distribution. Test at the .05 level
of significance.
One-Tailed t Test
Solution
Step 1:
H0:
Ha:
=
n=
 = 140
 < 140
.05
20
Step 2: Test Statistic:
X   138.47  140
t

 2.57
S
2.66
n
20
Step 3:
P-value = P(t < -2.57) = .0093 (using Minitab for t with 19 df)
Step 4:
Conclude: Reject H0 because p =.0093 < .05
Minitab: Calc >> Probability Distributions >> t……
One-Tailed t Test
Thinking Challenge
You’re a marketing analyst for Wal-Mart. WalMart had teddy bears on sale last week. The
weekly sales ($ 00) of bears sold in 10 stores
was:
8 11 0 4 7 8 10 5 8 3
At the .05 level of significance, is there evidence
that the average bear sales per store is more than
5 ($ 00)? (Find the p-value where the alternative
is “ > 5”)
a) Do by hand the hard way!
b) Do using Minitab: Stat >> Basic Statistics >>
One Sample t
Z Test of Proportion
One Population Tests
One
Population
Small
Sample
Proportion
Z Test
t Test
Z Test
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
Large
Sample
Mean
One-Sample Z Test
for Proportion
• 1. Assumptions
– Two categorical outcomes
– Population follows binomial distribution
– Normal approximation can be used
• np  3 np 1  p
does not contain 0 or n
b g
• 2. Z-test statistic for proportion
Z
p  p0
p0  (1  p0 )
n
Hypothesized
population proportion
One-Proportion Z Test Thinking
Challenge
• You’re an accounting manager.
A year-end audit showed 4%
of transactions had errors. You
implement new procedures. A
random sample of 500
transactions had 25 errors.
Has the proportion of
incorrect transactions changed
at the .05 level?
Alone
Group Class
One-Proportion Z Test Solution
Step 1: (H0, Ha, n, )
Step 2: (Z test statistic)
Step 3: P-value
Step 4: Conclusion
Decision Making Risks
Errors in
Making Decision
1. Type I Error
•
•
•
Reject true null hypothesis
Has serious consequences
Probability of Type I Error is (alpha)
— Called level of significance
2. Type II Error
•
•
Do not reject false null hypothesis
Probability of Type II Error is (beta)
Decision Results
H0: Innocent
Jury Trial
Actual Situation
Verdict
Innocent
Guilty
Innocent
Guilty
H0 Test
Actual Situation
Decision
Correct
Error
Accept
H0
Error
Correct
Reject
H0
H0 True
H0
False
1–
Type II
Error
()
Type I Power
Error () (1 – )
 &  Have an
Inverse Relationship
You can’t reduce both
errors simultaneously!


Factors Affecting 
1. True value of population parameter
•
Increases when difference with hypothesized
parameter decreases
2. Significance level, 
•
Increases when decreases
3. Population standard deviation, 
•
Increases when  increases
4. Sample size, n
•
Increases when n decreases
Conclusion
1. Distinguished Types of Hypotheses
2. Described Hypothesis Testing Process
3. Explained p-Value Concept
4. Solved Hypothesis Testing Problems Based
on a Single Sample
5. Discussed (briefly!) Type I and Type II
errors