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```One Sample Tests of
Hypothesis
Chapter 10
McGraw-Hill/Irwin
Learning Objectives
Chapter 10
1 Define a hypothesis.
3 Describe Type I and Type II errors.
5 Distinguish between a one-tailed and two-tailed hypothesis
6 Apply the five-step procedure to the test of hypothesis about a
population mean.
7 Compute and interpret a p-value.
8 Conduct a test of hypothesis about a population proportion.
9-2
Hypothesis and Hypothesis 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.
10-3
Null and 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 accepted if the sample
data provide sufficient evidence that the null hypothesis is false.
10-4
Hypotheses: H0 and H1

H0: null hypothesis and H1: alternate hypothesis






Examples: H0: defendant is innocent; H1: defendant is guilty
H0: µ =10; H1: µ >10
H0 is presumed to be true, but in business analysis we
usually seek to reject it, that is, to prove it is not true.
H1 has the burden of proof
We collect evidence from the sample.
If our decision is '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 used in H0; “≠” “<” and “>” is used in H1
10-5
Decisions and Errors in Hypothesis Testing
(Acquaint)
(Defendant
innocent)
Correct decision
I error
(Defendant Type
Type II error
guilty)
P(Type
P(Type IIII error)=β
error)=β
(Convict)
Type I error
P(Type I error)=α
P(Type I
Also called
error)=α
significance level
Correct decision
10-6
Test Statistic versus Critical Value
TEST STATISTIC A value, determined from sample information, used to
determine whether to reject the null hypothesis.
Example: z, t, F, 2
CRITICAL VALUE The dividing point between the region where the null
hypothesis is rejected and the region where it is not rejected.
10-7
One-tail vs. Two-tail Test—
H :<A
depending on H1
1
H1:  ≠ A
H1:  > A
10-8
Hypothesis Setups for Testing a
Population Mean ()
Required condition: the data is approximately normally distributed.
Two-tail test
Left-tail test
=
Right-tail test
=
Z > Zα/2 or Z < Zα/2
t > tα/2, n-1 or t < tα/2, n-1
With σ unknown:
With σ known:
10-9
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.
In other words, is the mean number of
desks produced at the Fedonia Plant
different from 200 at the .01
significance level?
10-10
Testing for a Population Mean with a
Known Population Standard Deviation- Example
To investigate this issue, the weekly production of 50
weeks from last year was obtained (the plant was shut
down 2 weeks for vacation). The sample mean is 203.5.
What we know from the description:
  16,
X  203.5, n  50 ,   0.01
10-11
Testing for a Population Mean with a
Known Population Standard Deviation- Example
Step 1: State the null hypothesis and the alternate
hypothesis.
H0:  = 200
H1:  ≠ 200
(note: keyword in the problem “has changed”)
Step 2: Select the level of significance.
α = 0.01 as stated in the problem
Step 3: Select the test statistic.
Use Z-distribution since σ is known
10-12
Testing for a Population Mean with a
Known Population Standard Deviation- Example
Step 3: Select the test statistic.
Use Z-distribution since σ is known
Z
203.5  200
 1.55
16 / 50
10-13
Testing for a Population Mean with a
Known Population Standard Deviation- Example
Step 4: Formulate the decision rule.
Rejection Region: Reject H0 if |Z| > Z/2
Z  Z / 2 or Z   Z / 2
 Z / 2   Z.01/ 2  2.58
Rejection H0 if Z  2.58 or Z  -2.58
=NORMSINV(1-0.005)
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.
10-14
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 is more than 200?
Recall: σ=16, n=50, α=.01
10-15
Testing for a Population Mean with a Known Population
Standard Deviation- Example
Step 1: State the null hypothesis and the alternate
hypothesis.
H0:  = 200
H1:  > 200
(note: keyword in the problem “an increase”)
Step 2: Select the level of significance.
α = 0.01 as stated in the problem
Step 3: Select the test statistic.
Use Z-distribution since σ is known
Z
203 .5  200
 1.55
16 / 50
10-16
One-Tailed Test versus Two-Tailed Test
=
10-17
Testing for a Population Mean with a Known Population
Standard Deviation- Example
Step 4: Formulate the decision rule.
Reject H0 if Z > Z
Z  Z
Z 0.01  2.33
Rejection H 0 if Z  2.33
1.55
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
10-18
Testing for the Population Mean:
Population Standard Deviation Unknown


When the population standard deviation (σ) 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:
10-19
Characteristics of the t-distribution
1. It is a continuous distribution, bell-shaped and symmetrical.
2. There is a family of t distributions. The shape of the t distribution is
controlled by degrees of freedom, denoted by ν, where ν = n – 1.
3. All t distributions have a mean of 0, but their standard deviations
differ according to the sample size, n.
4. The t distribution is more spread out and flatter at the center than
the standard normal distribution.
5. As the sample size increases, the t distribution approaches the
standard normal distribution; when the degrees of freedom is ∞, t
distribution becomes standard normal distribution.
9-20
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 costcutting 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 .01 significance level is it reasonable a claim is now
less than \$60? Data McFarland Insurance Claim
10-21
Testing for a Population Mean with a
Known Population Standard Deviation- Example
Step 1: State the null hypothesis and the alternate
hypothesis.
H0:  = \$60
H1:  < \$60
(note: keyword in the problem “now less than”)
Step 2: Select the level of significance.
α = 0.01 as stated in the problem
Step 3: Select the test statistic.
Use t-distribution since σ is unknown
t
56.42  60
 1.818
10.04 / 26
10-22
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
t  t ,n 1
 t0.01, 261  2.485
Rejection H 0 if t  -2.485
=TINV(2α, d.f.)=TINV(0.02, 25)
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.
10-23
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?
10-24
Testing for a Population Mean with an Unknown Population
Standard Deviation- Example
Step 1: State the null and the alternate hypothesis.
H0: µ = 250
H1: µ > 250
Step 2: Select the level of significance.
It is .05.
Step 3: Find a test statistic.
X   256  250
t

 3.162
s/ n
6 / 10
10-25
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  t ,n 1
t0.05,101  1.833
Step 5: Make a decision and interpret the results.
The null hypothesis is rejected. There is enough evidence
in support of the alternative hypothesis. The mean number
produced is more than 250 per hour.
10-26
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.

In testing a hypothesis, we can also compare the pvalue to the significance level ().

Decision rule using the p-value:
Reject H0 if p-value < significance level
10-27
p-Value in Hypothesis Testing - Example
Recall the last problem where the
hypothesis and decision rules were set up
as:
H0:  = 200
H1:  > 200
Reject H0 if Z > 2.33, where Z = 1.55
P-value: What is the probability of
observing a sample mean as high as
203.5 when the population mean is 200
and the population standard deviation is
16?
P-value=P(Z>1.55)=P(Z<-1.55)=.0606
=NORMDIST(-1.55)
Reject H0 if p-value < , where =0.01
Decision: Do not reject H0
10-28
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.
10-29
P-value—two-tail test example
Two-tail test
H0:  = 200
H1:  ≠ 200
Test statistic: Z  1.55
P-value:
2P(Z<neg test stat)=2P(Z<-1.55)=2(.0606)=.1212
P-value>0.01
Decision: do not reject the null hypothesis
Summary
Two-tail test
Left-tail test
=
p-value=2P(Z<neg test stat) p-value=P(Z<test stat)
p-value=2P(t<neg test stat) p-value=P(t<test stat)
Right-tail test
=
p-value=P(Z>test stat)
p-value=P(t>test stat)
P-value for t-test: =tdist(| t |, d.f., tail), where tail=1 for one-tail test
and tail=2 for two-tail test.
Assumptions in Testing a Population
Proportion using the z-Distribution






A random sample is chosen from the population.
the sample data collected are the result of counts.
the outcome of an experiment is classified into one of
two mutually exclusive categories—a “success” or a
“failure.”
“Success” refers to the category we want to study;
“failure” refers to any other categories.
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.
10-32
Hypothesis Setups for Testing
a Proportion ()
Z > Zα/2 or Z < Zα/2
10-33
Test Statistic for Testing a Single
Population Proportion
Hypothesized
population proportion
Sample proportion
z
p 
 (1   )
n
Sample size
10-34
Test Statistic for Testing a Single
Population Proportion - Example
The GO transportation system of
buses and commuter trains
operates on the honor system.
Train travelers are expected to buy
their tickets before boarding the
train. Only a small number of
people will be checked on the train
to see whether they bought a ticket.
Suppose that a random sample of
400 train travelers was sampled
ticket. Test to determine whether
the proportion of travelers who fail
to buy a ticket exceeds 15%. Use a
5% significance level.
10-35
Test Statistic for Testing a Single Population
Proportion - Example
Step 1: State the null hypothesis and the alternate
hypothesis.
H0:  = .15
H1:  > .15
(note: keyword in the problem “exceeds”)
Step 2: Select the level of significance.
α = 0.05 as stated in the problem
Step 3: Select the test statistic.
Use Z-distribution since the assumptions are met.
68
p 
.17  .15
p
 .17
Z

 1.12
400
 (1   )
.15(1  .15)
n
400
10-36
Testing for a Population Proportion Example
Step 4: Formulate the decision rule.
Reject H0 if Z < Z
Zα = Z.05 = 1.645
Rejection region: reject H0 if Z > 1.645
Z = 1.12
Step 5: Make a decision and interpret the result.
The test statistic (Z=1.12) is not in the rejection region, so the null hypothesis is
not rejected at the .05 level. This indicates that there is not enough evidence to
infer the prportion of travelers who fail to buy a ticket exceeds 15%.
10-37
Testing for a Population
Proportion - P-value
Hypotheses
H0:  = .15
H1:  > .15
Test statistic
z=1.12
P-value
P(Z > test stat)=P(Z > 1.12)=P(Z<-1.12)=0.1314
P-value>0.05
Decision
Do not reject the null hypothesis
```
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