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QMS202-Business Statistics II
Chapter11
Chapter11 Two-Sample Tests and
One-Way ANOVA (Part I)
Outcomes:
1. Conduct a test of hypothesis for two independent population means
- Use z-test when σ1 and σ2 are known
- Use pooled-variance t test when σ1 and σ2 are unknown but equal
- Use separate-variance t test when σ1 and σ2 are unknown and
unequal
2. Conduct a test of hypothesis for paired or dependent observations,
using the paired t-test
3. Conduct a test of hypothesis for two population proportions
using the z test
4. List the characteristics of the F distributions
5. Conduct a test of hypothesis to determine whether the variances
of two populations are equal, using the F Test
6. Discuss the general idea of Analysis of Variance (ANOVA) and its
assumptions
7. Conduct the F Test when there are more than two means
8. Discuss multiple comparisons: The Tukey-Kramer procedure
9. Conduct Levene’s Test for Homogeneity of Variance
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QMS202-Business Statistics II
Chapter11
Example1
Ryerson Car magazine is comparing the total repair costs incurred during
the first three years on two sport cars, The R123 and the S456. Random
samples of 45 R123 cars are $5300 for the first three years. For the 50
S456 cars, the mean is $5760. Assume that the standard deviations for the
two populations are $1120 and $1350, respectively. Using the 5%
significance level, can we conclude that such mean repair costs are
different for these two types of cars?
Calculator Output
2-Sample z Test
µ1 ≠ µ2
z
= -1.8137025
p = 0.06972353
x1 = 5300
x 2 = 5760
n1
= 45
n 2 = 50
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Step1
Let µ1 be the population mean repair cost for sport car R123
Let µ2 be the population mean repair cost for sport car S456
Step2
H o : µ1 = µ2
H A : µ1 ≠ µ2
Step3
Level of significance = 0.05/2=0.025
Step 4
2-sample mean z test
Step5
= -1.8137
0.06972353
z statistic
z critical
=
p-value =
Step6
Since the p-value > 0.05, do not reject the null hypothesis.
There is not enough evidence to conclude that such mean repair costs are
different for these two types of cars
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Example2
Mark is the owner of the Appliance Patch. Mark observed a difference in
the dollars value of sales between the men and women he employed as
sales associates. A sample of 50 days revealed the men sold a mean of
$1560 worth of appliances per day. For a sample of 60 days, the women
sold a mean of $1650 worth appliances per day. Assume the population
standard deviation for men is $204 and for women $259. At the 0.05
significance level, can Mark conclude that the mean amount sold per day is
larger for women?
Calculator Output
2-Sample z Test
µ1
µ2
z =
p =
x1 =
x2 =
n1 =
n2 =
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Chapter11
Step1 Define the parameter(s)
Step 2 State the null and alternative hypothesis
Step3
Level of significance =
Step 4 Test statistic
Step5 (Determine the test statistic, the p-value, and the critical value)
z statistic
=
z critical
=
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p-value =
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Chapter11
Step6 (Statistical Decision and Business Conclusion)
Example3
The operations manager of Ryerson Light Bulb factory wants to determine
if there is any difference in the average life expectancy of bulbs
manufactured on two different types of machine. The process population
standard deviation of machine A is 112 hours and of machine B is 127
hours. A random sample of 35 light bulbs obtained from machine A
indicates a sample mean of 379 hours, and a similar sample of 35 from
machine B indicates a sample mean of 368. Using the 1% significance
level.
Calculator Output
2-Sample z Test
µ1
µ2
z =
p =
x1 =
x2 =
n1 =
n2 =
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Chapter11
Step1 Define the parameter(s)
Step2 State the null and alternative hypothesis
Step3
Level of significance =
Step 4 Test statistic
Step5 (Determine the test statistic, the p-value, and the critical value)
z statistic
=
z critical
=
p-value =
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Chapter11
Step6 (Statistical Decision and Business Conclusion)
Example4
Ryerson Computer Manufacturer offers a help line that purchasers call for
help 24 hours per day, 7 days a week. Clearing these calls for help in a
timely fashion is important to the company’s image. After telling the caller
that resolution of the problem is important the caller is asked whether the
issue is “software” or “hardware” related. A mean time takes a technician
to resolve a software issue is 18 minutes with a standard deviation of 4.2
minutes. This information was obtained from a sample of 35 monitored
calls. For a study of 45 hardware issues, the mean time for a technician to
resolve a problem was 15.5 minutes with a standard deviation of 3.9
minutes. This information was also obtained from monitored calls. Assume
both population standard deviations are the same. At the 0.05
significance level is it reasonable to conclude that it takes longer to resolve
software issue?
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Calculator Output
2-Sample t Test
µ1 > µ2
t = 2.75012089
p = 3.7008E-03
d. f
= 78
= 18
x1
x 2 = 15.5
sx 1 = 4.2
sx 2 = 3.9
sp = 4.0335
n1 = 35
n 2 = 45
Step1
Let µ1 be the population mean time for resolving software issues
Let µ2 be the population mean time for resolving hardware issues
Step2
H o : µ1 ≤ µ2
H A : µ1 > µ2
Step3
Level of significance = 0.05
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Step 4
2-sample mean t test- pooled variance “on”
Step5
t statistic
= 2.7501
degree of freedom =
p-value = 0.0037
t critical
=
Step6
Since the p-value <0.05, reject the null hypothesis.
There is enough evidence to conclude that it takes longer to resolve
software issue
Example5
The Commercial Bank and the Trust Company are studying the use of it
automatic teller machines (ATM). Of particular interest is whether young
adults (under 25 years) use the machines more than senior citizens. To
investigate further, samples of customers under 25 years of age and
customers over 60 years of age were selected. The number of ATM
transactions last month was determined for each selected individual, and
the results are shown below. Assume the number of times of using ATM
machine for both groups are to be normally distributed and both population
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standard deviations are equal. At the .05 significance level, can bank
management conclude that younger customers use the ATM more?
Under25 10 10 11 15
7 11 10 9
Over60
4
8
7
7
4
5
1 7 4 10 5
Calculator Output
2-Sample t Test
µ1 > µ2
t = 4.28210973
p = 2.5201E-04
d . f = 17
x1 = 10.375
x 2 = 5.6363
s1 = 2.2638
s 2 = 2.4605
sp = 2.3815
n1 = 8
n 2 = 11
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Step1
Let µ1 be the population mean number of time using ATM machine for
younger adults (under 25)
Let µ2 be the population mean number of time using ATM machine for
senior (over 60)
Step2
H o : µ1 ≤ µ2
H A : µ1 > µ2
Step3
Level of significance = 0.05
Step 4
2-sample mean t test
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Step5
t statistic
= 4.2821
p-value = 0.000252
degree of freedom =
t critical
=
Step6
Since the p-value <0.05, reject the null hypothesis.
There is enough evidence to conclude that that younger customers use the
ATM more.
Example6
Mark is the budget director for Aplus Proces Company. He would like to
compare the daily travel expenses for the sales staff and the audit staff. He
collected the following sample information.
Sales ($)
Audit ($)
121
120
125
93
136
118
156
133
124
139
131
110
129
At the 0.10 significance level, can she conclude that the mean daily
expenses are greater for the sales staff than the audit staff?
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Calculator Output
2-Sample t Test
µ1
t
µ2
=
=
d. f =
p
x1
x2
s1
s2
sp
=
=
=
=
=
n1 =
n2 =
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Step1 Define the parameter(s)
Step2 State the null and alternative hypothesis
Step3
Level of significance =
Step 4 Test statistic
Step5 (Determine the test statistic, the p-value, and the critical value)
t statistic =
t critical
degree of freedom =
=
p-value =
Step6 (Statistical Decision and Business Conclusion)
Example7
Bradley is the Vice president of Human Resources for Ryerson
Manufacturing Company. In recent years he has noticed an increase in
absenteeism that he thinks is related to the general health of employees.
Three years ago, in an attempt to improve the situation, he began a yoga
program in which employees practice yoga during their lunch hour. To
evaluate this program, he selected a random sample of 10 participants and
found the number of days each was absent in the last twelve months
before the yoga program began and in the last twelve months. Below are
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the results. At the 5% significance level, can he conclude that the number
of absences has declined?
Employee
Before
After
Alex
Brian
Clare
Diana
Ellen
Felix
Gary
Howard
Ivan
Johnny
14
12
15
16
9
17
11
10
19
20
11
8
10
9
6
12
14
12
15
16
Calculator Output
1-Sample t Test
µ1 > 0
t = 3.0338
p = 7.0785E-03
x =3
s = 3.1269
n = 10
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Step1
d = before – after
Let µd be the population mean difference in the number of days absent in
the last twelve months before the yoga program had begun and in the next
twelve months after the yoga program had begun.
H o : µd ≤ 0
Step2
H A : µd > 0
Step3
Level of significance = 0.05
Step 4
Paired observation mean t-test
Step5
t statistic
= 3.0338
degree of freedom =
p-value = 0.0070785
t critical
=
Step6
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Since the p-value <0.05, reject the null hypothesis. There is enough
evidence to conclude that the number of absences has declined
Example8
On-line grocery stores provide customers with the convenience of shopping
at home, but this convenience often comes with a lofty price tag due to the
high costs on-line grocers incur related to handling and delivery. In 2008
the largest on-line grocer was Ryerson Inc. with over $58 million in
revenue. A new company, Yes Frills Inc. began operations in 2009, a nineitem shopping list from the two on-line grocers in downtown Toronto for
January 8,2010 is given in the following.
ITEM
Ryerson
Yes Frills
ABC 2% or 1% or skim milk
$3.94
$3.53
4 liters
ABC Ultra, liquid detergent
$7.58
$8.79
Fuji apples, 3lbs
$2.41
$2.97
APlus Peanut butter Cookies, 350g $2.60
$2.50
Old Treasure long grain white rice, $8.99
$9.49
8 kg
Unicare vegetable oil, 3 liter
$3.97
$4.97
Tropicare Pure Premium orange $6.49
$6.49
juice, original or homestyle 3.78 L
No Name liquid dish detergent
$1.98
$1.79
Colgate, regular toothpaste, 85 ml
$1.19
$0.99
At the 0.05 level of significance, is there evidence of a difference in the
average price for items purchased from Ryerson and Yes Frills?
SPSS Output
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Calculator Output
1-Sample t Test
µ
t =
p =
x
=
s =
x
n =
Step1 Define the parameter(s)
Step 2 State the null and alternative hypothesis
Step3
Level of significance =
Step 4 Test statistic
Step5 (Determine the test statistic, the p-value, and the critical value)
t statistic =
t critical =
degree of freedom =
p-value =
Step6 (Statistical Decision and Business Conclusion)
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Chapter11
Questino#9
Of 150 adults who tried a new organic chocolate, 87 rated it excellent. Of
200 children sampled, 123 rated it excellent. Using the 0.10 level of
significance, can we conclude that there is a significant difference in the
proportion of adults and the proportion of children who rate the new organic
chocolate excellent?
2-Prop Z Test
p1 p2
z =
p =
ˆ1 =
p
ˆ2 =
p
x
ˆ =
p
x
n1 =
n1
n2 =
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QMS202-Business Statistics II
Chapter11
Step1 Define the parameter(s)
Step 2 State the null and alternative hypothesis
Step3
Level of significance =
Step 4 Test statistic
Step5 (Determine the test statistic, the p-value, and the critical value)
z statistic
=
z critical
=
p-value =
Step6 (Statistical Decision and Business Conclusion)
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Chapter11
Question#10
The research department at the head office of Ryerson Insurance
Company conducts ongoing research on the causes of automobile
accidents: the characteristics of the drivers, and so on. A random sample of
500 policies written on single persons revealed 120 had at least one
accident in the previous three-year period. Similarly, a sample of 700
policies written on married persons revealed that 150 had been in at least
one accident. At the 0.05 significance level, is there a significant difference
in the proportions of single and married persons having an accident during
a three-year period?
2-Prop Z Test
p1 p2
z =
p =
ˆ1 =
p
ˆ2 =
p
x
ˆ =
p
x
n1 =
n1
n2 =
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Chapter11
Step1 Define the parameter(s)
Step 2 State the null and alternative hypothesis
Step3
Level of significance =
Step 4 Test statistic
Step5 (Determine the test statistic, the p-value, and the critical value)
z statistic
=
z critical
=
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p-value =
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Chapter11
Step6 (Statistical Decision and Business Conclusion)
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