Chapter11 Two-Sample Tests and One-Way ANOVA (Part I)s3.amazonaws.com/prealliance_oneclass_sample/0x9kObgW6M.pdf · Step1 Define the parameter(s) Step2 State the null and alternative

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 1σ and 2σ are known

- Use pooled-variance t test when 1σ and 2σ are unknown but equal

- Use separate-variance t test when 1σ and 2σ 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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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 1µ ≠ 2µ

z = -1.8137025

p = 0.06972353

1x = 5300

2x = 5760

1n = 45

2n = 50

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Step1

Let 1µ be the population mean repair cost for sport car R123

Let 2µ be the population mean repair cost for sport car S456

Step2

21: µµ =oH

21: µµ ≠AH

Step3

Level of significance = 0.05/2=0.025

Step 4

2-sample mean z test

Step5

statisticz = -1.8137 criticalz = p-value =

0.06972353

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 1µ 2µ

z =

p =

1x =

2x =

1n =

2n =

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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)

statisticz = criticalz = p-value =

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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 1µ 2µ

z =

p =

1x =

2x =

1n =

2n =

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Step2 State the null and alternative hypothesis

Step3Level of significance =




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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

1µ > 2µ t = 2.75012089

p = 3.7008E-03

fd. = 78

1x = 18

2x = 15.5

1sx = 4.2

2sx = 3.9

sp = 4.0335

1n = 35

2n = 45

Step1

Let 1µ be the population mean time for resolving software issues

Let 2µ be the population mean time for resolving hardware issues

Step2 21: µµ ≤oH

21: µµ >AH

Step3

Level of significance = 0.05

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Step 4

2-sample mean t test- pooled variance “on”

Step5

statistict = 2.7501 p-value = 0.0037

degree of freedom = criticalt =

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

1µ > 2µt = 4.28210973p = 2.5201E-04

fd. = 17

1x = 10.375

2x = 5.6363

1s = 2.2638

2s = 2.4605

sp = 2.3815

1n = 8

2n = 11

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Step1

Let 1µ be the population mean number of time using ATM machine for

younger adults (under 25)

Let 2µ be the population mean number of time using ATM machine for

senior (over 60)

Step2 21: µµ ≤oH

21: µµ >AH

Step3


Step 4

2-sample mean t test

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Step5



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 ($) 121 125 136 156 124 131

Audit ($) 120 93 118 133 139 110 129

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

1µ 2µt =p =

fd. =

1x =

2x =

1s =

2s =

sp =

1n =

2n =

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Step2 State the null and alternative hypothesis




statistict = degree of freedom = criticalt =

p-value =


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 AfterAlex 14 11

Brian 12 8

Clare 15 10

Diana 16 9

Ellen 9 6

Felix 17 12

Gary 11 14

Howard 10 12

Ivan 19 15

Johnny 20 16

Calculator Output

1-Sample t Test

1µ > 0

t = 3.0338

p = 7.0785E-03

x = 3

s = 3.1269n = 10

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Step1

d = before – after

Let dµ 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.

Step2 0: ≤doH µ

0: >dAH µ

Step3


Step 4

Paired observation mean t-test

Step5



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 nine-

item 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

4 liters

$3.94 $3.53

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 kg

$8.99 $9.49

Unicare vegetable oil, 3 liter $3.97 $4.97

Tropicare Pure Premium orange

juice, original or homestyle 3.78 L

$6.49 $6.49

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 Testµ

t =

p =

x = x s =

n =






statistict = degree of freedom = criticalt =

p-value =


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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 =

1p̂ = x 2p̂ =x p̂ =

1n = 1n 2n =

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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 =

1p̂ = x 2p̂ =x p̂ =

1n = 1n 2n =

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Chapter11 Two-Sample Tests and One-Way ANOVA (Part I)s3.amazonaws.com/prealliance_oneclass_sample/0x9kObgW6M.pdf · Step1 Define the parameter(s) Step2 State the null and alternative

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