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İŞL 276
Fundamentals of Hypothesis
Testing: One-Sample Tests
What is a Hypothesis?
A
hypothesis is a claim
(assumption) about a
population parameter:

population mean
Example: The mean monthly cell phone bill of
this city is μ = $42

population proportion
Example: The proportion of adults in this city
with cell phones is π = 0.68
The Null Hypothesis, H0
 States
the claim or assertion to be tested
Example: Average number of mobile phone in a
Turkish family is equal to 3 ( H0 : μ  3 )
 Is
always about a population parameter,
not about a sample statistic
H0 : μ  3
H0 : X  3
The Null Hypothesis, H0
(continued)
 Begin
with the assumption that the null
hypothesis is true
 Similar to the notion of innocent until
proven guilty
 Refers to the status quo
 Always contains “=” , “≤” or “” sign
 May or may not be rejected
The Alternative Hypothesis, H1
Is the opposite of the null hypothesis
 e.g., The average number of TV sets in U.S.
homes is not equal to 3 ( H1: μ ≠ 3 )
 Challenges the status quo

Never contains the “=” , “≤” or “” sign
 May or may not be proven
 Is generally the hypothesis that the researcher is
trying to prove

Hypothesis Testing Process
Claim: the
population
mean age is 50.
(Null Hypothesis:
H0: μ = 50 )
Population
Is X 20 likely if μ = 50?
If not likely,
REJECT
Null Hypothesis
Suppose
the sample
mean age
is 20: X = 20
Now select a
random sample
Sample
Reason for Rejecting H0
Sampling Distribution of X
20
μ = 50
If H0 is true
If it is unlikely that
we would get a
sample mean of
this value ...
... if in fact this were
the population mean…
X
... then we reject
the null
hypothesis that μ
= 50.
Level of Significance, 

Defines the unlikely values of the sample statistic
if the null hypothesis is true


Defines rejection region of the sampling
distribution
Is designated by  , (level of significance)

Typical values are 0.01, 0.05, or 0.10

Is selected by the researcher at the beginning

Provides the critical value(s) of the test
Level of Significance and the Rejection Region
Level of significance =
H0: μ = 3
H1: μ ≠ 3

 /2
Two-tail test
 /2

0
Upper-tail test
H0: μ ≥ 3
H1: μ < 3

Lower-tail test
Rejection
region is
shaded
0
H0: μ ≤ 3 H1:
μ>3
0
Represents
critical value
Hypothesis Tests for the Mean
Hypothesis
Tests for 
 Known
(Z test)
 Unknown
(t test)
Z Test of Hypothesis for the Mean (σ Known)

Convert sample statistic ( X ) to a Z test statistic
Hypothesis
Tests for 
Known
σKnown
(Z test)
The test statistic is:
X μ
Z 
σ
n
Unknown
σUnknown
(t test)
Critical Value Approach to Testing

For a two-tail test for the mean, σ known:

Convert sample statistic (
statistic )

Determine the critical Z values for a specified
level of significance  from a table or computer

Decision Rule: If the test statistic falls in the
rejection region, reject H0 ; otherwise do not reject
H0
X ) to test statistic (Z
Two-Tail Tests
 There are two
cutoff values
(critical values),
defining the
regions of
rejection
H0: μ = 3
H1 : μ  3
/2
/2
X
3
Reject H0
-Z
Lower
critical
value
Do not reject H0
0
Reject H0
+Z
Upper
critical
value
Z
6 Steps in Hypothesis Testing
1.
2.
State the null hypothesis, H0 and the alternative
hypothesis, H1
Choose the level of significance, , and the sample
size, n
3.
Determine the appropriate test statistic and
sampling distribution
4.
Determine the critical values that divide the
rejection and nonrejection regions
6 Steps in Hypothesis Testing
(continued)
5.
Collect data and compute the value of the test
statistic
6.
Make the statistical decision and state the
managerial conclusion. If the test statistic falls into
the nonrejection region, do not reject the null
hypothesis H0. If the test statistic falls into the
rejection region, reject the null hypothesis.
Express the managerial conclusion in the context of
the problem
Hypothesis Testing Example
Test the claim that the true mean of mobile phone in a Turkish
family is equal to 3. Assume σ = 0.8 and sample results are n =
100, X = 2.84
1. State the appropriate null and alternative
hypotheses
 H0: μ = 3
H1: μ ≠ 3 (This is a two-tail test)
2. Specify the desired level of significance and the sample
size
 Suppose that  = 0.05 and n = 100 are chosen for this
test
Hypothesis Testing Example
(continued)
3. Determine the appropriate technique
 σ is known so this is a Z test.
4. Determine the critical values
 For  = 0.05 the critical Z values are ±1.96
5. Collect the data and compute the test statistic
 Suppose the sample results are
n = 100,X = 2.84 (σ = 0.8 is assumed known)
So the test statistic is:
Z 
X μ
2.84  3
 .16


 2.0
σ
0.8
.08
n
100
Hypothesis Testing Example
(continued)

6.
Is the test statistic in the rejection region?
 = 0.05/2
Reject H0 if
Z < -1.96 or
Z > 1.96;
otherwise do
not reject H0
Reject H0
-Z= -1.96
 = 0.05/2
Do not reject H0
0
Reject H0
+Z= +1.96
Here, Z = -2.0 < -1.96, so the test
statistic is in the rejection region
Hypothesis Testing Example
(continued)
6(continued). Reach a decision and interpret the result
 = 0.05/2
Reject H0
-Z= -1.96
 = 0.05/2
Do not reject H0
0
Reject H0
+Z= +1.96
-2.0
Since Z = -2.0 < -1.96, we reject the null hypothesis and
conclude that there is sufficient evidence that the mean
number of TVs in US homes is not equal to 3
p-Value Approach to Testing
 p-value:
Probability of obtaining a test
statistic more extreme ( ≤ or  ) than the
observed sample value given H0 is true

Also called observed level of significance

Smallest value of  for which H0 can be
rejected
p-Value Approach to Testing
(continued)

Convert Sample Statistic (e.g.,
(e.g., Z statistic )
)X
to Test Statistic

Obtain the p-value from a table or computer

Compare the p-value with 

If p-value <  , reject H0

If p-value   , do not reject H0
p-Value Example

Example: How likely is it to see a sample mean of 2.84 (or
something further from the mean, in either direction) if the
true mean is  = 3.0?
X = 2.84 is translated to a
Z score of Z = -2.0
P(Z  2.0)  0.0228
P(Z  2.0)  0.0228
/2 = 0.025
/2 = 0.025
0.0228
0.0228
p-value
= 0.0228 + 0.0228 = 0.0456
-1.96
-2.0
0
1.96
2.0
Z
p-Value Example

(continued)
Compare the p-value with 

If p-value <  , reject H0

If p-value   , do not reject H0
Here: p-value = 0.0456
 = 0.05
Since 0.0456 < 0.05, we
reject the null
hypothesis
/2 = 0.025
/2 = 0.025
0.0228
0.0228
-1.96
-2.0
0
1.96
2.0
Z
Connection to Confidence Intervals
• For sample mean is 2.84 and σ = 0.8 and n = 100, the
95% confidence interval is:
0.8
2.84 - (1.96)
to
100
0.8
2.84  (1.96)
100
2.6832 ≤ μ ≤ 2.9968
 Since this interval does not contain the hypothesized mean
(3.0), we reject the null hypothesis at  = 0.05
One-Tail Tests

In many cases, the alternative hypothesis focuses
on a particular direction
H0: μ ≥ 3
H1: μ < 3
H0: μ ≤ 3
H1 : μ > 3
This is a lower-tail test since the
alternative hypothesis is focused on the
lower tail below the mean of 3
This is an upper-tail test since the
alternative hypothesis is focused on the
upper tail above the mean of 3
Lower-Tail Tests
H0: μ ≥ 3
 There is only one
critical value, since the
rejection area is in only
one tail
H1 : μ < 3

Reject H0
-Z
Do not reject H0
0
μ
Critical value
Z
X
Upper-Tail Tests
 There is only one
critical value, since the
rejection area is in only
one tail
Z
_
X
H0: μ ≤ 3
H1: μ > 3

Do not reject H0
0
Zα
Reject H0
μ
Critical value
Ex:Upper-Tail Z Test for Mean ( Known)
A phone industry manager thinks that customer
monthly cell phone bills have increased, and now
average over $52 per month. The company
wishes to test this claim. (Assume  = 10 is
known)
Form hypothesis test:
H0: μ ≤ 52 the average is not over $52 per month
H1: μ > 52
the average is greater than $52 per month
(i.e., sufficient evidence exists to support the
manager’s claim)
Example: Find Rejection Region
(continued)

Suppose that  = 0.10 is chosen for this test
Find the rejection region:
Reject H0
 = 0.10
Do not reject H0
0
1.28
Reject H0
Reject H0 if Z > 1.28
Review:One-Tail Critical Value
What is Z given  = 0.10?
0.90
Standardized Normal
Distribution Table (Portion)
0.10
 = 0.10
Z
.07
.08
.09
1.1 .8790 .8810 .8830
0.90
1.2 .8980 .8997 .9015
z
0
1.28
Critical Value
= 1.28
1.3 .9147 .9162 .9177
Example: Test Statistic
(continued)
Obtain sample and compute the test statistic
Suppose a sample is taken with the following results:
n = 64, X = 53.1 (=10 was assumed known)

Then the test statistic is:
Xμ
53.1  52
Z 

 0.88
σ
10
n
64
Example: Decision
(continued)
Reach a decision and interpret the result:
Reject H0
 = 0.10
Do not reject H0
1.28
0
Z = 0.88
Reject H0
Do not reject H0 since Z = 0.88 ≤ 1.28
i.e.: there is not sufficient evidence that the
mean bill is over $52
p -Value Solution
Calculate the p-value and compare to 
(continued)
(assuming that μ = 52.0)
p-value = 0.1894
Reject H0
 = 0.10
0
Do not reject H0
1.28
Z = 0.88
Reject H0
P( X  53.1)
53.1  52.0 

 P Z 

10/ 64 

 P(Z  0.88)  1 0.8106
 0.1894
Do not reject H0 since p-value = 0.1894 >  = 0.10
t Test of Hypothesis for Mean (σ Unknown)

Convert sample statistic ( X) to a t test statistic
Hypothesis
Tests for 
Known
σKnown
(Z test)
Unknown
σUnknown
(t test)
The test statistic is:
t n-1
X μ

S
n
Example: Two-Tail Test( Unknown)
The average cost of a hotel
room in New York is said
to be $168 per night. A
random sample of 25
hotels resulted in X =
$172.50 and
S = $15.40. Test at the
 = 0.05 level.
(Assume the population distribution is normal)
H0: μ = 168
H1: μ  168
Example Solution: Two-Tail Test
H0: μ = 168
H1: μ  168

 = 0.05

n = 25

 is unknown, so
use a t statistic

Critical Value:
t24 = ± 2.0639
/2=.025
/2=.025
Reject H0
-t n-1,α/2
-2.0639
t n1 
Do not reject H0
Reject H0
0
1.46
t n-1,α/2
2.0639
X μ
172.50  168

 1.46
S
15.40
n
25
Do not reject H0: not sufficient evidence that true
mean cost is different than $168
Hypothesis Tests for Proportions

Involves categorical variables

Two possible outcomes


“Success” (possesses a certain characteristic)

“Failure” (does not possesses that characteristic)
Fraction or proportion of the population in the
“success” category is denoted by π
Proportions
(continued)

Sample proportion in the success category is denoted
by p
X number of successes in sample
 p n 
sample size

When both nπ and n(1-π) are at least 5, p can be
approximated by a normal distribution with mean
and standard deviation

 (1  )
μp  
σp 
n
Hypothesis Tests for Proportions

The sampling
distribution of p is
approximately
normal, so the test
statistic is a Z value:
Z
pπ
π(1  π )
n
Hypothesis
Tests for p
nπ  5
and
n(1-π)  5
nπ < 5
or
n(1-π) < 5
Not discussed
in this chapter
Z Test for Proportion

An equivalent form
to the last slide, but
in terms of the
number of
successes, X:
X  n
Z
n (1  )
Hypothesis
Tests for X
X5
and
n-X  5
X<5
or
n-X < 5
Not discussed
in this chapter
Example: Z Test for Proportion
A marketing company
claims that it receives 8%
responses from its mailing.
To test this claim, a
random sample of 500
were surveyed with 25
responses. Test at the  =
0.05 significance level.
Check:
n π = (500)(.08) = 40
n(1-π) = (500)(.92) = 460

Z Test for Proportion: Solution
Test Statistic:
H0: π = 0.08
H1: π  0.08
Z
 = 0.05
n = 500, p = 0.05
Decision:
Critical Values: ± 1.96
Reject
p 
.05  .08

 2.47
 (1   )
.08(1  .08)
n
500
Reject
Reject H0 at  = 0.05
Conclusion:
.025
.025
-1.96
-2.47
0
1.96
z
There is sufficient
evidence to reject the
company’s claim of 8%
response rate.
p-Value Solution
(continued)
Calculate the p-value and compare to 
(For a two-tail test the p-value is always two-tail)
Do not reject H0
Reject H0
Reject H0
/2 = .025
p-value = 0.0136:
/2 = .025
P(Z  2.47)  P(Z  2.47)
0.0068
0.0068
-1.96
Z = -2.47
0
 2(0.0068)  0.0136
1.96
Z = 2.47
Reject H0 since p-value = 0.0136 <  = 0.05
Example 1
Price/earnings ratios for stocks.
 Theory: stable rate of P/E in market = 13.
 If P/E (market) < 13, you should invest in the
stock market.
 If P/E (market) > 13, you should take your
money out.
 We have a sample of 50
 = 12.1.
 Historical σ = 3.0456

44
Estimate Steps
Can we estimate if the population P/E is 13 or not?
 Common steps:
 Set hypothesis:
H0: μ = 13
HA: μ ≠ 13
 Select  = .05.

45
Calculating Test Statistic

3.0456
Standard Error  x 

 0.4307
n
50
z  value 
46
x - 0
x
12.1  13

 2.0896
0.4307
p-value Approach

47
Calculate the p-value.
 We will calculate for the lower tail
→ then make an adjustment for the upper tail.
p-value, Two-Tailed Test
p(z > 2.09) = ??
p(z < –2.09) = ??
Z=–2.09
0
Z=2.09
We can just calculate one value, and double it.
48
z
Calculating the p-value cont’d

Find 2.090 on the Standard Normal Distribution
tables:
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
…
1.9
2.0 .4772 .4778 .4783 .4788 .4793 .4798 .4803 .4808 .4812 .4817
2.1
…
49
p-value, Two-Tailed Test
0.4817
p(z > 2.09) = 0.5 -.4817
= 0.0183
p(z < –2.09) = ??
Z=–2.09
0
Z=2.09
Doubling the value, we find the p-value = 0.0366
50
z
Should We Reject the Null Hypothesis?
Yes!
 p-value = 0.0366 <  = 0.05.
 There is only a 3.66% chance that the measured
price/earnings ratio sample mean of 12.1 is not equal to
the stable rate of 13 by random chance.

51
Example #2:

You want to test the hypothesis that a treatment conducted at
the ward you are working on in a hospital is more beneficial
than the average treatment used in other wards of the
hospital.Given that the mean wellness score of the people on
your ward is 87, and that the mean for the entire hospital (N
= 20) is 76 and the standard deviation of population is 15, is
your ward significantly better?
 State Ho and H1
 State if you’re using a one- or two-tailed test and why.
 State the p-value that supports your claim.
Hypothesis Testing w/ One Sample
87  76
11
z

 3.27
15
3.36
20
Smaller Portion = p < .0006
Example

A family therapist states that parents talk to their
teenagers an average of 27 minutes per week.
Surprised by that claim, a psychologist decided to
collect some data on the amount of time parents
spend in conversation with their teenage children.
For the n = 12 parents, the study revealed that
following times (in minutes) devoted to
conversation in a week:
Example
Do the psychologists findings differ significantly
from the therapist’s claim? If so, is the family
expert’s claim an overestimate or underestimate of
actual time spent talking to children? Use the 0.05
level of significance with two tails.
 Mean = 24.58, s2 = 12.24, (s/√n) = 1.00

29
22
19
25
27
28
21
22
24
26
30
22
Hypothesis Testing w/ One Sample

Example #2:
 H0 = μ = 27
 Critical t (df = 11) = ±2.201
 Reject H0, and conclude that the data are
significantly different from the therapist’s claim
J) Hypothesis Tests, 2 Means: ’s Unknown
Two datasets –> is the mean value of one larger than
the other?
 Is it larger by a specific amount?
 μ1 vs. μ2 –> μ1 – μ2 vs. D0.
 Often set D0 = 0 –> is μ1 = μ2?

57
Example: Female vs. Male Salaries
Saskatchewan 2001 Census data:
- only Bachelor’s degrees
- aged 21-64
- work full-time
- not in school
 Men: M = $46,452.48, sM = 36,260.1, nM = 557.
 Women: W = $35,121.94, sW = 20,571.3, nW = 534.
 M – W = $11,330.44 } our point estimate.
 Is this an artifact of the sample, or do men make
significantly more than women?

58
Hypothesis, Significance Level, Test Statistic


1.
2.
3.
We will now ONLY use the p-value approach, and
NOT the critical value approach.
Research hypothesis: men get paid more:
H0: μM – μW < 0
H1: μM – μW > 0
Select  = 0.05
Compute test t-statistic:
t
( xM  xW )  D0
sW2
sM2

nM nW
(46452.48  35121.94)  0

 6.375
1314794928 426178351

557
534
59
4. a. Compute the Degrees of Freedom

Can compute by hand, or get from Excel:
2
s s 
  
n1 n2 

=
888
df 
2 2
2 2
1  s1 
1  s2 
  
 
n1  1  n1  n2  1  n2 
2
1
2
2
60
4. b. Computing the p-value
Degrees of
Freedom
0.20
0.10
0.02
0.025
0.01
0.005
.845
.842
1.290
1.282
1.660
1.645
1.984
1.960
2.364
2.326
2.626
2.576
…
100

6.375 up here somewhere
The p-value <<< 0.005.
61
5. Check the Hypothesis
Since the p-value is <<< 0.05, we reject H0.
 We conclude that we can accept the alternative
hypothesis that men get paid more than women at a
very high level of confidence (greater than 99%).

62
Excel
t-Test: Two-Sample Assuming Unequal Variances
Male
Female
Mean
46452.47935 35121.94195
Variance
Observations
1314794928
557
423178351.3
534
Hypothesized Mean Diff.
df
t Stat
P(T<=t) one-tail
0
888
6.381034789
1.41441E-10 .00000000014
t Critical one-tail
P(T<=t) two-tail
1.646571945
2.82883E-10
t Critical two-tail
1.962639544
63