Fundamental
Hypothesis
Testing
MB 1201 – Business Statistics
Tutorial Week 9 (March, 12th 2020)
What is a Hypothesis?
The Null Hypothesis, H0
The Alternative Hypothesis, H1
6 Steps in Hypothesis Testing
1. State the null hypothesis, H0 and the alternative
hypothesis, H1
2. 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 non-rejection regions
5. Collect data and compute the value of the test
statistic
6. Make the statistical decision, and state the
managerial conclusion
Hypothesis Test for the Mean
Hypothesis Test for 
σ Known
(Z test)
σ Unknown
(t test)
Hypothesis Test for the
Proportion
Hypothesis Test for the
Proportion
Critical Values
Level of Significance, 
and the Rejection Region
p-Value Approach to Testing
• p-value : Probability of obtaining a test
statistic equal to or more extreme (  or  )
than the observed sample value given H0 is
true
• The p-value is also called the observed level
of significance
• It is the smallest value of  for which H0 can
be rejected
p-Value Approach to Testing:
Interpreting the p-value

Convert Sample Statistic 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
• Remember
• If the p-value is low then H0 must go
The 5 Step p-value approach to
Hypothesis Testing
1. State the null hypothesis, H0 and the alternative
hypothesis, H1
2. Choose the level of significance, , and the sample
size, n
3. Determine the appropriate test statistic and
sampling distribution
4. Collect data and compute the value of the test
statistic and the p-value
5. Make the statistical decision and state the
managerial conclusion. If the p-value is < α then
reject H0, otherwise do not reject H0. State the
managerial conclusion in the context of the
problem
Errors in Decision Making
• Type I Error
– Reject a true null hypothesis
– Considered a serious type of error
The probability of Type I Error is 
•
• Called level of significance of the test
• Set by the researcher in advance
Type II Error
Fail to reject a false null hypothesis
The probability of Type II Error is β
Outcomes and Probabilities
Possible Hypothesis Test Outcomes
Actual Situation
Key:
Outcome
(Probability)
Decision
H0 True
H0 False
Do Not
Reject
H0
No error
(1 -  )
Type II Error
(β)
Reject
H0
Type I Error
()
No Error
(1-β)
HOMEWORK
Problem 9.14
The quality control manager at a compact fluorescent light
bulb (CFL) factory needs to determine whether the mean life
of a large shipment of CFLs is equal to 7,500 hours. The
population standard deviation is 1,000 hours. A random
sample of 64 CFLs indicates a sample mean life of 7,250
hours.
a. At the 0.05 level of significance , is there evidence that
the mean life is different from 7,500 hours?
b. Compute p-value and interpret its meaning.
Given
H0: μ = 7,500 hours
H1: μ ≠ 7,500 hours
Xbar = 7,250 hours
ϭ= 1,000 hours
α = 0.05
α/2 = 0.025
n = 64
Decision rule:
Reject H0 if Zstat < -1.96 or
Zstat > 1.96
𝑥−𝜇
ϭ
𝑛
7,250 − 7,500
=
1,000
64
= −2
𝑍𝑠𝑡𝑎𝑡 =
Since Zstat < -1.96, then reject Ho
we can conclude that, there is enough evidence to
state that population mean life is different from
7,500 hours
b. Probability of Z2.0 is 0.0228
Since it is 2-tail, then the p-value would be: 0.0228 *2 = 0.0456
The probability of getting a sample mean of 7,250 hours or less if the population
mean is 7,500 hours is 0.0456
Excel Guide
Problem 9.24
You are the manager of a restaurant for a fast-food
franchise. Last month, the mean waiting time at the drivethrough window for branches in your geographical region,
as measured from the time a customer places an order
until the time the customer receives the order, was 3.7
minutes. You select a random sample of 64 orders. The
sample mean waiting time is 3.57 minutes, with a sample
standard deviation of 0.8 minute.
a.
At the 0.05 level of significance, is there evidence that
the population mean waiting time is different from 3.7
minutes?
b.
Because the sample size is 64, do you need to be
concerned about the shape of the population
distribution when conducting the t test in (a)? Explain
Given
H0: μ = 3.7 minutes
H1: μ ≠ 3.7 minutes
Xbar = 3.57 minutes
S = 0.8 minutes
α = 0.05
α/2 = 0.025
n = 64
df = 64 – 1= 63
Decision rule:
Reject H0 if tstat < -1.9983
or tstat > 1.9983
𝑥−𝜇
𝑡𝑠𝑡𝑎𝑡 = 𝑠
𝑛
3.57 − 3.7
=
0.8
64
= −1.3
Since tstat > -1.9983 and tstat < 1.9983, then do not
reject Ho
we can conclude that, there is not enough evidence
to state that population mean waiting time is
different from 3.7 minutes.
b. Because n = 64, the sampling distribution of the t test statistic is
approximately normal.
Excel Guide
Problem 9.48
A quality improvement project was conducted with the objective of
improving the wait time in county health department (CHD) adult
primary care unit (APCU). The evaluation plan included waiting room
as one key waiting time process measure. Waiting time was defined
as the time elapsed between requesting that the patient be seated in
the waiting room and the time he was called to be placed in an exam
room. Suppose that initially a targeted wait time goal of 25 minutes
was set. After implementing an improvement, the quality
improvement team collected data on 355 patients. In this sample,
the mean wait time was 23.05 minutes, with standard deviation of
16.83 minutes.
a.
If you test null hypothesis at 0.01 level of significance, is there
evidence that the population mean wait time is less than 25
minutes?
b.
Interpret the meaning of the p value in this problem
Given
H0: μ ≥ 25 minutes
H1: μ < 25 minutes
Xbar = 23.05 minutes
S = 16.83 minutes
α = 0.01
n = 355
df = 355 – 1= 354
Decision rule:
Reject H0 if tstat < -2.3369
𝑥−𝜇
𝑠
𝑛
23.05 − 25
=
16.83
355
= −2.1831
𝑡𝑠𝑡𝑎𝑡 =
Since tstat > -2.3369, then do not reject Ho
P-value = 0.0148 > 0.01, then do not reject Ho
we can conclude that, there is not enough evidence
to state that population mean waiting time is less
than 25 minutes.
b. The probability of getting a sample mean of 23.05 minutes or less if the
population mean is 25 minutes is 0.0148
Excel Guide
Problem 9.58
How do professionals stay on top of their careers? Of
935 surveyed U.S. linkedIn members, 543 reported that
they engaged in professional networking within the last
month. At the 0.05 level of significance, is there
evidence that the proportion of all LinkedIn members
who engaged in professional networking within the last
month is different from 52%?
Given:
H0: π = 0.52
H1: π ≠ 0.52
P = X/n = 543/935 = 0.5807
α = 0.05
α/2 = 0.025
n = 935
Rules:
X≥5
n- X ≥ 5
Decision rule:
Reject H0 if Zstat < -1.96 or Zstat > 1.96
Or
Reject H0 if P-value < 0.05
𝑡𝑠𝑡𝑎𝑡 =
=
𝑝−𝜋
𝜋(1 − 𝜋)
𝑛
0.5807 − 0.52
0.52(0.48)
935
=3.7181
Since tstat > 1.96, then reject H0
Probability of Z3.7181 is 0.99990
Since it is 2-tail, then the probability would
be: 1-0.99990 = 0.0001 *2 = 0.0002
Since p value < 0.05, then reject H0
we can conclude that there is enough evidence to say that the proportion of
LinkedIn members who engaged in professional networking within the last
month is different from 52%.
Excel Guide
EXERCISE
Problem 9.31
One of the major measures of the quality of service
provided by any organization is the speed with which it
responds to customer complaints. A large family-held
department store selling furniture and flooring, including
carpet, had undergone a major expansion in the past
several years. In particular, the flooring department had
expanded from 2 installation crews to an installation
supervisor, a measurer, and 15 installation crews. The
store had the business objective of improving its
response to complaints. The variable of interest was
defined as the number of days between when the
complaint was made and when it was resolved. Data were
collected from 50 complaints that were made in the past
year. These data are as below:
5
54
35
137
31
27
152
2
123
81
74
27
11
19
126
110
110
29
61
35
94
31
26
5
12
4
165
32
29
28
29
26
25
1
14
13
13
10
5
27
4
52
30
22
36
26
20
23
33
68
a. What test should you conduct if you face this
problem?
b. The installation supervisor claims that the
mean number of days between the receipt of
a complaint and the resolution of the
complaint is 20 days. At 0.05 level of
significance, is there evidence that the claim
is not true?
a.
According to the data, since the standard deviation is not known, we
should run t test.
Given
H0: μ = 20
H1: μ ≠ 20
Xbar = 43.04
S = 41.9261
α = 0.05
α/2 = 0.025
n = 50
df = 50 – 1= 49
Decision rule:
Reject H0 if tstat < -2.0096
or tstat >2.0096
𝑥−𝜇
𝑠
𝑛
43.04 − 20
=
41.9261
50
= 3.8841
𝑡𝑠𝑡𝑎𝑡 =
Since tstat > 2.0096, then Ho is rejected.
we can conclude that, there is enough evidence
that the claim about mean number of days between
the receipt of a complaint and the resolution is 20
days is not true.
Problem 9.55
According to a recent National Association of Colleges and
Employers (NACE) report, 48% of college student internships are
unpaid. A recent survey of 60 college interns at a local university
found that 30 had unpaid internships.
a. What test should they conduct if they face this problem?
b. Use the five-step p-value approach to hypothesis testing and a
0.05 level of significance to determine whether the proportion
of college interns that had unpaid internships is different from
NACE report.
c.
Assume that the study found that 37 of the 60 college interns had
unpaid internships and repeat (b). Does the conclusion remain the
same?
a. According to the problem, we should use Z test for proportion.
Given:
H0: π = 0.48
H1: π ≠ 0.48
α = 0.05
α/2 = 0.025
n = 60, X = 30
P = X/n = 30/60 = 0.5
Rules:
X≥5
n- X ≥ 5
𝑧𝑠𝑡𝑎𝑡 =
=
𝑝−𝜋
𝜋(1 − 𝜋)
𝑛
0.5 − 0.48
0.48(0.52)
60
= 0.3101
Probability of Z0.3101 is 0.6217
Since it is 2-tail, then the p-value would
be: 1-0.6217 = 0.3783 *2 = 0.7566
Decision rule:
Reject H0 if Zstat < -1.96 or Zstat > 1.96
Or
Reject H0 if P-value < 0.05
Since p value > 0.05, then do not reject H0
we can conclude that there is not enough evidence to say that the proportion
of college interns that had unpaid internships is different from NACE report.
Problem 9.60
Actuation consulting and Enterprise Agility recently conducted a
global survey of product teams with the goal of better
understanding the dynamics of product team performance and
uncovering the practices that make these teams successful. One
of the survey findings was that 37% of organizations have a
coherent business strategy that they stick to effectively
communicate. Suppose that another study is conducted to check
the validity of this result, with the goal of proving that the
percentage is less than 37%.
a. State the null and research hypothesis
b. A sample of 100 organizations is selected, the result indicate
that 34 organizations have a coherent business strategy that
they stick to effectively communicate. Use either critical value
hypothesis testing or p-value approach to determine the 0.05
level of significance whether there is evidence that the
percentage is less than 37%
Given:
H0: π ≥ 0.37
H1: π < 0.37
P = X/n = 34/60 = 0.34
α = 0.05
𝑧𝑠𝑡𝑎𝑡 =
=
Rules:
X≥5
n- X ≥ 5
Decision rule:
Reject H0 if Zstat < -1.645
Or
Reject H0 if P-value < 0.05
𝑝−𝜋
𝜋(1 − 𝜋)
𝑛
0.34 − 0.37
0.37(0.63)
100
= − 0.6214
Probability of Z-0.6214 is 0.2676
Since Zstat > -1.96, then do not reject Ho
p value > 0.05, then do not reject H0
we can conclude that there is insufficient evidence to say that the
proportion who respond that the organizations have a coherent business
strategy that they stick to effectively communicate is less than 37%
Problem 9.46
The Los Angeles County Metropolitan Transportation
Authority has set a bus mechanical reliability goal of
3,900 bus miles. Bus mechanical reliability is measured
specifically as the number of bus miles between
mechanical road calls. Suppose a sample of 100 buses
resulted in a sample mean of 3,975 bus miles and a
sample standard deviation of 275 bus miles.
a. What test should you utilise for this problem?
b. Is there evidence that the population mean bus miles is
greater than 3,900 bus miles? (Use a 0.05 level of
significance)
a.
According to the data, since the standard deviation is not known, we
should run t test.
Given
H0: μ ≤ 3,900
H1: μ > 3,900
Xbar = 3,975
S = 275
α = 0.05
df = 100 – 1= 99
𝑥−𝜇
𝑡𝑠𝑡𝑎𝑡 = 𝑠
𝑛
3,975 − 3,900
=
275
100
= 2.7273
Since tstat > 1.6604, then reject H0
Decision rule:
Reject H0 if tstat > 1.6604
we can conclude that there is enough evidence that
population mean bus miles is greater than 3,900
bus miles.
Thank You!
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