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QUANTITATIVE METHODS Martin Huard Winter 2005

Review SOLUTIONS

1. Sarah wants to determine the mean number of hours per week a student at St. Lawrence

studies. To accomplish this, she passed a questionnaire to 10 students chosen at random from their registration number. The results were:

4 12 7 10 15 5 2 22 15 18 2 4 5 7 10 12 15 15 18 22

Find the following:

a) the mean 110

1110

xx

n= = =∑ hours

b) the median 10 12

112

Me+

= = hours

c) the mode 15Mo = hours d) the range 22 2 20= − = hours

e) e) the sample variance ( )2

22 1102 101596

42.891 9

x

nxs

n

∑− −= = =

−∑ hours2

f) f) the standard deviation 2 42.89 6.54s s= = = hours g) Q1 = 5 hours h) Q3 = 15 hours

i) Coefficient of variation: 6.54

0.595 59.5%11

sCV

x= = = =

j) Draw a box and whisker graph.

0

4

8

12

16

20

# hours per week

Box-and-whisker plot for the number of hours SLC student study per week

24

QM Review - Solutions

Winter 2005 Martin Huard 2

2. Here is a frequency distribution showing the ages of 121 randomly selected people who have a bachelor’s degree or higher.

Ages Frequency x xf x2f 18-24 8 21 168 3528 25-34 32 29.5 944 27848 35-44 35 39.5 1382.5 54608.75 45-54 23 49.5 1138.5 56355.75 55-64 11 59.5 654.5 38942.75 65-84 9 74.5 670.5 49952.25 85-104 3 94.5 283.5 26790.75

Total 121 368 5241.5 258026.3 Use this frequency distribution to estimate

a) the mean 5241.5

43.3121

xfx

n= = =∑ years

b) the variance ( )2

22 5241.52 121258026.3

258.11 121 1

xf

nx fs

n

∑− −= = =

− −∑ years2

c) the standard deviation 2 258.1 16.1s s= = = years



3. The following data on the stress scores before a QM test and the QM test scores for seven students are:

Stress Score (x)

Test score (y) x2 xy y2

6.5 81 42.25 526.5 6561 4 96 16 384 9216

2.5 93 6.25 232.5 8649 7.2 68 51.84 489.6 4624 8.1 63 65.61 510.3 3969 3.4 84 11.56 285.6 7056 5.5 71 30.25 390.5 5041

37.2x =∑ 556y =∑ 2 223.76x =∑ 2819xy =∑ 2 45116y =∑

( )2

22 37.2

223.76 26.077x

xSS x

n= − = − =∑∑

37.25.314

7x

xn

= = =∑

( )2

22 556

45116 953.77y

ySS y

n= − = − =∑∑

55679.43

7y

yn

= = =∑

37.2 556

2819 135.77xy

x yS xy

n⋅

= − = − = −∑ ∑∑

a) Find the slope of the least-squares line.

Slope : 135.7

5.20726.07

xy

x

SSb

SS−

= = = −

b) Find the intercept of the least-squares line. y-intercept : 79.43 5.207 5.314 107.1a y bx= − = + ⋅ = c) Find the equation of the least squares line.

5.207 107.1y x= − + d) Find the coefficient of correlation r.

135.7

0.86126.07 953.7

xy

x y

SSr

SS SS−

= = = −⋅

e) Find the coefficient of determination. 2 20.861 0.741 74.1%r = = = f) If a student obtained a stress score of 7.0, what is the expected test score? ( )ˆ 5.207 7.0 107.1 70.7y = − + = g) Draw a scatter diagram.

Scatter Diagram for the Stress and Tess scores for a

QM test

50

7090

110

2 3 4 5 6 7 8 9

Stress Score

Test

Sco

re



4. A box contains 3 marbles, a red, a blue and a green marble. If two marbles are picked at

random, what is the probability that a) one is red and the other is green?

( ) ( )( ) ( ) ( ) ( )

1 2 1 2

1 2 1 1 2 1

one Red and one Green

| |

1 1 1 1 13 2 3 2 3

P P R G G R

P R P G R P G P R G

= +

= +

= + =

b) both are red? ( ) ( ) ( )1 2 1 2 1|

10 0

3

P R R P R P R R=

= =

5. Two dice are rolled, where one is black and the other is white. Find the following

probabilities.

a) P(white die is an odd number) 18 136 2

= =

b) P(sum is 6) 5

36=

c) P(both dice show odd numbers) 9 1

36 4= =

d) P(number on black die is larger than number on white die)15 536 12

= =



6. Two thousand randomly selected adults were asked if they think they are financially better off than their parents. The following table gives the two-way classification of the responses based on the education levels of the persons included in the survey and whether they are financially better off, the same, or worse off than their parents.

Education Level High school or less CEGEP More than CEGEP Better off 140 450 420 1010 Same 60 250 110 420 Worse off 200 300 70 570 400 1000 600 2000

Suppose one adult is selected at random from these 2000 adults. Find the following probabilities

a) P(better off) ( ) 1010 1012000 200

P B= = =

b) P(better off and CEGEP) ( ) 450 9 and

2000 40P B C= = =

c) P(better off or CEGEP) ( )( ) ( ) ( )

or

and

1010 1000 450 392000 2000 2000 50

P B C

P B P C P B C

=

= + −

= + − =

d) P(better off given CEGEP) ( ) 450 9|

1000 20P B C= = =

e) Are the events better off and CEGEP independent? No since P(better off) = 101

200 ≠ P(better off given CEGEP) = 920

f) Are the events better off and CEGEP mutually exclusive? No since P(better off and CEGEP) = 9

40 0≠ 7. A single card is drawn from a deck of 52 cards. What is the probability that the card is

a) a king of diamonds? ( ) 152

P K♦ =

b) a king or a diamond? ( ) ( ) ( ) ( ) 4 13 1 4 or

52 52 52 13P K P K P P K♦ = + ♦ − ♦ = + − =

c) a face card? ( ) 12 352 13

P F = =

d) a red face card? ( ) 6 352 36

P RF = =

e) not an ace? ( ) 48 12not

52 13P A = =



8. A consume agency surveyed all 2500 families living in a small town to collect data on the number of television sets owned by them. The following table lists the frequency distribution of the data collected by this agency.

Number of TV sets owned 0 1 2 3 4 Number of families 120 970 730 410 270

a) Construct a probability distribution table for the number of television sets owned by

these families. Probability Distribution for

the number of TV sets owned

x ( )p x ( )xp x ( )2x p x

0 0.048 0 0 1 0.388 0.388 0.388 2 0.292 0.584 1.168 3 0.164 0.492 1.476 4 0.108 0.432 1.728 ( ) 1p x =∑ ( ) 1.896xp x =∑ ( )2 4.76x p x =∑

b) Draw a histogram.

c) Find ( )2P x > ( ) ( )3 4 0.164 0.108 0.272P P= + = + =

d) Find ( )1P x ≤ ( ) ( )0 1 0.048 0.388 0.436P P= + = + =

e) Find ( )1 2P x≤ ≤ ( ) ( )1 2 0.388 0.292 0.680P P= + = + = f) How many TV sets do you expect a family chosen at random to have?

( ) 1.896xp xµ = =∑ TV Sets g) Find the standard deviation of this distribution.

( )2 2 2 2

2

4.76 1.896 1.165

1.165 1.08 TV sets

x p xσ µ

σ σ

= − = − =

= = =

∑

Probability Histogram

00.20.40.6

0 1 2 3 4

x : # TV sets owned

p(x

)



9. Suppose that the time taken to run a road race is normally distributed with a mean of 195 minutes and a standard deviation of 21 minutes. If a runner is selected at random, what is the probability that this runner will complete the this road race

a) in less than 150 minutes ( ) ( )150 2.14

0.0162

P x P z< = < −

=

150 1952.14

21x

zµ

σ− −

= = = −

b) in 205 to 245 minutes

( ) ( )205 245 0.48 2.38

0.9913 0.68440.3069

P x P z< < = < <

= −=

1

2

205 1950.48

21245 195

2.3821

xz

z

µσ− −

= = =

−= =

10. Express Courier Service has found that the delivery times for packages are normally

distributed with mean 14 hours and standard deviation 2 hours. a) For a package selected at random, what is the probability that it will be delivered in

18 hours or less? ( ) ( )18 2

0.9772

P x P z< = <

=

18 142

2x

zµ

σ− −

= = =

b) For a package selected at random, what is the probability that it will be delivered in more than 15 hours?

( ) ( )15 0.5

1 0.69150.3085

P x P z> = >

= −=

15 14

0.52

xz

µσ− −

= = =



c) For a package selected at random, what is the probability that it will be delivered in between 10 and 20 hours?

( ) ( )10 20 2 3

0.9987 0.02280.9759

P x P z< < = − < <

= −=

1

2

10 142

220 14

32

xz

xz

µσ

µσ

− −= = = −

− −= = =

d) What should the guaranteed delivery time on all packages be in order to be 95% sure that a given package will be delivered within this time?

Area to the left = 0.95 1.645z =

14 1.645 2 17.3x zµ σ= + = + ⋅ = Thus the guarantee delivery should be 17.3 hours. 11. Quick Start Company makes 12-volt car batteries. After many years of product testing, the

company knows that the average life of a Quick Start battery is normally distributed with a mean of 45 months and a standard deviation of 8 months.

a) If Quick Start guarantees a full refund on any battery that fails within the 36-month period after purchase, what percentage of its batteries will the company expect to replace?

( ) ( )36 1.12

0.1314

P x P z< = < −

=

36 451.12

8x

zµ

σ− −

= = = −

Therefore the company can expect to replace 13.1% of its batteries.

b) If Quick Start does not want to replace more than 10% of its batteries under the full-refund guarantee policy, for how long should the company guarantee the batteries (to the nearest month)?

Area to the left = 0.10 1.28z = −

45 1.28 8 34.76x zµ σ= + = − ⋅ = Thus the company should guarantee the batteries for 35 months.



12. The ages of all university students follow a distribution with a mean of 23 years and a standard deviation of 4 years. Find the probability that the mean age for a random sample of 36 students would be

a) between 22 and 24 years

( ) ( )22 24 1.5 1.5

0.9332 0.06680.8664

P x P z< < = − < <

= −=

1 436

2 436

22 231.5

24 231.5

n

xz

z

σ

µ− −= = = −

−= =

b) more than 22 years ( ) ( )22 1.5

1 0.06680.9332

P x P z> = > −

= −=

436

22 231.5

n

xz

σ

µ− −= = = −

13. A new muscle relaxant is available. Researchers at the firm developing the relaxant have

done studies indicating that the time lapse between administration of the drug and beginning effects of the drug are normally distributed with mean 38 minutes and standard deviation 5 minutes.

a) The drug is administered to one patient selected at random. What is the probability that the time it takes to go into effect is 35 minutes or less?

( ) ( )35 0.6

0.2743

P x P z< = < −

=

35 380.6

5x

zµ

σ− −

= = = −

b) The drug is administered to a random sample of 10 patients. What is the probability that the average time before it is effective for all 10 patients is 35 minutes or less?

( ) ( )35 1.90

0.0287

P x P z< = < −

=

510

35 381.90

n

xz

σ

µ− −= = = −



14. Thirty randomly selected college students were asked how many cavities they had. These students had a mean of 3.2 cavities with a standard deviation of 1.65 cavities. Construct a 99% confidence interval for the mean number of cavities for all college students.

Step 1 Assumptions: 30 30n = ≥ Step 2 Area to the left = 0.995

0.99 2.58cz z= = Step 3 1.65

2.58 0.7830

sc n

E z= = =

Step 4 3.2 0.78 3.2 0.78

2.42 3.98

x E x Eµµµ

− < < +− < < +

< <

Step 5 The 99% confidence interval for the mean number of cavities for all college students is 2.42 to 3.98 cavities.

15. The administration at a college wishes to answer the question “How far (one way) does the

average community student commute to college each day?” A random sample of 100 commuting students was identified, and the one-way distance each commuted was obtained. The resulting sample mean distance was 18.5 kilometers with a standard deviation of 9.1 kilometers. Make a 97% confidence interval for the mean one-way commuting distance of the college students.

Step 1 Assumptions: 100 30n = ≥ Step 2 Area to the left = 0.985

2.17cz = Step 3 9.1

2.17 1.97100

sc n

E z= = =

Step 4 18.5 1.97 18.5 1.97

16.53 20.47

x E x Eµµµ

− < < +− < < +

< <

Step 5 The 97% confidence interval for the mean one-way commuting distance of college students is 16.53 to 20.47 kilometers.

16. A high-tech company wants to estimate the mean number of years of college education its

employees have completed. A good estimate of the standard deviation for the mean number of years of college is 1.2. How large a sample needs to be taken to estimate µ to within 0.5 of a year with 99% confidence?

Area to the left = 0.995 0.99 2.58cz z= =

2 22.58 1.238.3

0.5cz s

nE

⋅ = = =

Thus 39 employees should be chosen.



17. A department store manager wants to estimate at a 90% confidence level the mean amount spent by all customers at this store. From an earlier study, the manager knows that the standard deviation of amounts spent by customers at this store is $27. What sample size should he choose so that the estimate is within 3$ of the population mean?

Area to the left = 0.90 1.645cz =

2 21.645 27219.1

3cz s

nE

⋅ = = =

Thus 220 customers should be chosen. 18. To determine the views of students at SLC on whether an extremist hate group should be

given a permit to demonstrate, a seven-point attitude scale (1 = strongly opposed through 7 = strongly favor) was administered to a random sample of 15 students. This survey yielded a sample mean of 2.1 and a standard deviation of 1.5. Assuming that the scores are approximately normally distributed, construct a 95% confidence interval for the mean population score.

Step 1 Assumptions: Population is normally distributed Step 2 0.95c = 2.145ct = 1 14df n= − = Step 3 1.5

2.145 0.8315

sc n

E t= = =

Step 4 2.1 0.83 2.1 0.83

1.27 2.93

x E x Eµµµ

− < < +− < < +

< <

Step 5 The 95% confidence interval for the mean population score is 1.27 to 2.93.



19. Suppose a researcher wanted to examine the extent of cooperation in kindergarten children. To do so, she unobtrusively observes a group of children at play for 30 minutes and notes the number of cooperative acts engaged in by each child. Here are the number of cooperative acts exhibited by each child:

1 5 2 3 4 1 2 2 4 3

Construct a 90% confidence interval for the mean number of cooperative acts exhibited by children, assuming that the number of cooperative acts exhibited by children is normally distributed.

Step 1 Assumptions: Population is normally distributed Step 2 0.90c = 1.833ct = 1 9df n= − = Step 3 27

2.710

xx

n= = =∑

( )222 27

2 10891.79

1 9

x

nxs

n

∑− −= = =

−∑

2 1.79 1.34s s= = =

1.341.833 0.78

10s

c nE t= = =

Step 4 2.7 0.78 2.7 0.78

1.92 3.48

x E x Eµµµ

− < < +− < < +

< <

Step 5 The 90% confidence interval for the mean number of cooperative acts exhibited by children is 1.92 to 3.48.

20. A random sample of 250 registered voters revealed that 63 of them feel that education is the

most important issue when deciding on a candidate. Construct a 90% confidence interval for the proportion of all registered voters who feel that education is the most important issue when deciding on a candidate.

Step 1 Assumptions: ( )ˆ ˆ63 5 1 187 5np n p= > − = > Step 2 Area to the left = 0.95

1.645cz = Step 3 ( )63 63

250 2501ˆ ˆ2501.645 0.0452pq

c nE z−

= = = Step 4

63 63250 250

ˆ ˆ

0.0452 0.0452

0.2068 0.2872

p E p p Ep

p

− < < +− < < +

< <

Step 5 The 90% confidence interval for the proportion of all registered voters who feel that education is the most important issue when deciding on a candidate is 20.68% to 28.72%.



21. A mail-order company promises its customers that the products ordered will be mailed within 72 hours after an order is placed. The quality control department at the company checks from time to time to see if this promise is fulfilled. Recently the quality department took a sample of 50 orders and found that 42 of them were mailed within 72 hours of the placement of the orders. Construct a 98% confidence interval for the percentage of all orders that are mailed within 72 hours of their placement.

Step 1 Assumptions: ( )ˆ ˆ42 5 1 8 5np n p= > − = > Step 2 Area to the left = 0.98

2.33cz = Step 3 ( )42 42

50 501ˆ ˆ502.33 0.1208pq

c nE z−

= = = Step 4

42 4250 50

ˆ ˆ

0.1208 0.1208

0.7192 0.9608

p E p p Ep

p

− < < +− < < +

< <

Step 5 The 98% confidence interval for the percentage of all orders that are mailed within 72 hours of their placement is 71.9% to 96.1%.

22. A consumer agency wants to estimate the proportion of all drivers who wear seat belts while

driving. Assume that a preliminary study has shown that 76% of drivers wear seat belts while driving. How large should the sample size be so that the 99% confidence interval for the population proportion has a maximum error of 3%?


( ) ( )2 * * 2

2 2

1 2.58 0.76 1 0.761349.03

0.03cz p p

nE

− ⋅ ⋅ −= = =

Thus the sample should have 1350 drivers. 23. A researcher wants to determine what proportion of all high school students have Internet

access at home. He has no idea what the sample proportion will be. How large a sample is required to be 95% sure that the sample proportion is off by no more than 5%?

Area to the left = 0.975 0.95 1.96cz z= =

( )

2 2

22

1.96384.16

4 4 0.05cz

nE

= = =

Thus the sample should have 385 high school students. 24. A student wants to determine what percentage of college students smoke. How large a

sample should she take to be 90% confident that her sample proportion is off by no more than 4.5%?


( )

2 2

22

1.645334.1

4 4 0.045cz

nE

= = =

Thus the sample should have 335 college students.



25. In order to investigate whether conformity changes from one generation to the next, a test that measures and individual’s nonconformity rating was designed by sociologist. The larger the score on the test, the more nonconforming the individual. In 1970, the average score was 153. To see if a change occurred, the test was given to 45 randomly selected individuals in 2000 and the mean score was found to be 161 with a standard deviation of 17. Test at the 3% level of significance whether nonconformity has increased between 1970 and 2000. Try with both approaches, the classical and the p-value. Classical Approach

Step 1 45 30n = ≥ Step 2 : 153oH µ =

: 153AH µ > Step 3 Right-tailed test with a = 0.03


Step 4 1745

161 1533.16

sn

xz

µ− −= = =

Step 5 z is in the critical region Reject Ho.

∴ There is sufficient evidence, at the 3% level of significance, to conclude that nonconformity has increased between 1970 and 2000.

p-value approach

Step 1 45 30n = ≥ Step 2 : 153oH µ =

: 153AH µ > Step 3 Right-tailed test with a = 0.03 Step 4

1745

161 1533.16

sn

xz

µ− −= = =

p-value = ( )3.16 1 0.9992 0.0008P z > = − = Step 5 p-value = 0.0008 < a = 0.03

Reject Ho. ∴ There is sufficient evidence, at the 3% level of significance, to conclude that nonconformity has increased between 1970 and 2000.



26. A sample of 225 parents were asked how much time they spent per week on school work or school-related activities. This sample produced a mean of 5.6 hours per week, with a standard deviation of 4.4 hours. At the 1% level of significance, test the claim that the mean number of hours spent by parents on school work or school-related activities is 5 hours per week. Try with both approaches, the classical and the p-value. Classical Approach

Step 1 225 30n = ≥ Step 2 : 5oH µ = hours

: 5AH µ ≠ hours Step 3 Two-tailed test with a = 0.01


Step 4 4.4225

5.6 52.05

sn

xz

µ− −= = =

Step 5 z is not in the critical region Fail to reject Ho.

∴ There is not sufficient evidence, at the 1% level of significance, to conclude that the mean number of hours spent by parents on school work or school-related activities is not 5 hours per week.

p-value approach

Step 1 225 30n = ≥ Step 2 : 5oH µ = hours

: 5AH µ ≠ hours Step 3 Two-tailed test with a = 0.01 Step 4

4.4225

5.6 52.05

sn

xz

µ− −= = =

p-value = ( )2 2.05 2 0.0202 0.0404P z < − = ⋅ = Step 5 p-value = 0.0404 > a = 0.01

Fail to reject Ho. ∴ There is not sufficient evidence, at the 1% level of significance, to conclude that the mean number of hours spent by parents on school work or school-related activities is not 5 hours per week.



27. A telephone company claims that the mean duration of all long-distance phone calls made by its residential customers is 10 minutes. A random of 100 long-distance calls made by its residential customers taken from the records of this company showed that the mean duration of calls for this sample is 9.0 minutes with a standard deviation of 5.2 minutes. Test, at the 2% level of significance whether the mean duration of all long-distance calls is less than 10 minutes. Try with both approaches, the classical and the p-value. Classical Approach

Step 1 100 30n = ≥ Step 2 : 10oH µ = minutes

: 10AH µ < minutes Step 3 Left-tailed test with a = 0.02

Area to the left = 0.02 2.05cz = −

Step 4 5.2100

9.0 101.92

sn

xz

µ− −= = = −

Step 5 z is not in the critical region Fail to reject Ho.

∴ There is not sufficient evidence, at the 2% level of significance, to conclude that the mean duration of all long-distance calls is less than 10 minutes.

p-value approach

Step 1 100 30n = ≥ Step 2 : 10oH µ = minutes

: 10AH µ < minutes Step 3 Left-tailed test with a = 0.02 Step 4

5.2100

9.0 101.92

sn

xz

µ− −= = = −

p-value = ( )1.92 0.0274P z < − = Step 5 p-value = 0.0274 > a = 0.02

Fail to reject Ho. ∴ There is not sufficient evidence, at the 2% level of significance, to conclude that the mean duration of all long-distance calls is less than 10 minutes.



28. At the 1% level of significance, test the claim that the hours worked by college students is greater than 15 hours per week. A random sample of 25 students produced a sample mean of 20.83 hours per week with a standard deviation of 14.20 hours per week. Assume that the number of hours worked by college students is normally distributed.

Step 1 Assumptions: Population is normally distributed Step 2 : 15oH µ = hours per week

: 15AH µ > hours per week Step 3 Right-tailed test with a = 0.01

2.492ct = 1 24df n= − = Step 4

14.2025

20.83 152.05

sn

xt

µ− −= = =

Step 5 t is not in the critical region Fail to reject Ho.

∴ There is not sufficient evidence, at the 1% level of significance, to conclude that the hours worked by college students is greater than 15 hours per week.

29. How many pair of shoes do female college students own? A random sample of 15 female

college students produced a sample mean of 8.7 pairs of shoes, with a standard deviation of 0.85 pairs. Use these data to test the claim that the mean number of pairs of shoes owned by female college students is less than 10 at the 5% level of significance. Assume that the number of pair of shoes owned by female students is normally distributed.

Step 1 Assumptions: Population is normally distributed Step 2 : 10oH µ = shoes

: 10AH µ < shoes Step 3 Left-tailed test with a = 0.05

1.761ct = − 1 14df n= − = Step 4

0.8515

8.7 105.92

sn

xt

µ− −= = = −

Step 5 t is in the critical region Reject Ho.

∴ There is sufficient evidence, at the 5% level of significance, to conclude that the mean number of pairs of shoes owned by female college students is less than 10.



30. A past study claims that adult Canadians spend an average of 18 hours a week on leisure activities. A researcher wanted to test this claim. She took a random sample of 10 adults and asked them about the time they spend per week on leisure activities. A mean of 16.3 hours per week with a standard deviation of 3.8 hours per week was obtained. Is this sufficient evidence, at the 10% level of significance, to conclude that the claim is no longer valid? Assume that the time spent on leisure activities by all adults is normally distributed.

Step 1 Assumptions: Population is normally distributed Step 2 : 18oH µ = hours per week

: 18AH µ ≠ hours per week Step 3 Two-tailed test with a = 0.10

1.833ct = ± 1 9df n= − = Step 4

3.810

16.3 181.41

sn

xt

µ− −= = = −

Step 5 t is not in the critical region Fail to reject Ho.

∴ There is not sufficient evidence, at the 10% level of significance, to conclude that the claim is no longer valid.



31. A mail-order company claims that 60% of all orders are mailed within 48 hours. From time to time the quality control department at the company checks to see if this promise is fulfilled. Recently, the quality control department at this company took a sample of 400 orders and found that 212 of them were mailed within 48 hours of the placement of the orders. Testing at the 3% significance level, can you conclude that less then 60% of all orders are mailed within 48 hours? Try with both approaches, the classical and the p-value.

Classical approach

Step 1 Assumptions: 400 0.60 240 5np = ⋅ = >

( )1 160 5n p− = > Step 2 : 0.60

: 0.60o

a

H pH p

=<

Step 3 Left-tailed test with α = 0.03 Area to the left = 0.03 1.88cz = −

Step 4

( )212400ˆ 0.60

2.860.60 0.401

400

p pz

p pn

− −= = = −

⋅−


∴ There is sufficient evidence at the 3% level of significance to conclude that less than 60% of all orders are mailed within 48 hours.

p-value approach


( )1 160 5n p− = > Step 2 : 0.60

: 0.60o

a

H pH p

=<

Step 3 Left-tailed test with α = 0.03 Step 4

( )212400ˆ 0.60

2.860.60 0.401

400

p pz

p pn

− −= = = −

⋅−

p-value = ( )2.86 0.0021P z < − = Step 5 p-value = 0.0021 < α = 0.03

Reject Ho. ∴ There is sufficient evidence at the 3% level of significance to conclude that

less than 60% of all orders are mailed within 48 hours.



32. Thirty-five percent of the physicians in Canada were women in 1995. A recent sample of 300 Canadian physicians found that 147 of them were women. Using a 2% significance level, can you conclude that the current percentage of physicians in Canada is higher than it was in 1995? Try with both approaches, the classical and the p-value.

Classical approach


( )1 195 5n p− = > Step 2 : 0.35

: 0.35o

a

H pH p

=>

Step 3 Right-tailed test with α = 0.02 Area to the left = 0.98 2.05cz =

Step 4

( )147300ˆ 0.35

5.080.35 0.651

300

p pz

p pn

− −= = =

⋅−


∴ There is sufficient evidence at the 2% level of significance to conclude that the percentage of physicians in Canada who are women is higher now then it was in 1995.

p-value approach


( )1 195 5n p− = > Step 2 : 0.35

: 0.35o

a

H pH p

=>

Step 3 Right-tailed test with α = 0.02 Step 4

( )147300ˆ 0.35

5.080.35 0.651

300

p pz

p pn

− −= = =

⋅−

p-value = ( )5.08 1 1.000 0.000P z > = − = Step 5 p-value = 0.000 < α = 0.02


the percentage of physicians in Canada who are women is higher now then it was in 1995.



33. In 1995, 32% of Canadian households owned a personal computer. In a recent sample of 850 Canadian households, 305 own personal computers. Test at the 2% level of significance whether the current percentage of all Canadian households who own personal computers is different from 32%. Try with both approaches, the classical and the p-value.

Classical approach


( )1 578 5n p− = > Step 2 : 0.32

: 0.32o

a

H pH p

=≠

Step 3 Two-tailed test with α = 0.02 Area to the left = 0.99 2.33cz = ±

Step 4

( )305850ˆ 0.32

2.430.32 0.681

850

p pz

p pn

− −= = =

⋅−


∴ There is sufficient evidence at the 2% level of significance to conclude that the current percentage of all Canadian households who own personal computers is different from 32%.

p-value approach


( )1 578 5n p− = > Step 2 : 0.32

: 0.32o

a

H pH p

=≠

Step 3 Two-tailed test with α = 0.02 Step 4

( )305850ˆ 0.32

2.430.32 0.681

850

p pz

p pn

− −= = =

⋅−

p-value = ( ) ( )2 2.43 2 0.0075 0.0150P z < − = = Step 5 p-value = 0.0150 < α = 0.02


the current percentage of all Canadian households who own personal computers is different from 32%.



34. The manager of an assembly process wants to determine whether the number of defective articles manufactured depends on the day of the week the articles are produced. She collected the following information.

Monday Tuesdays Wednesday Thursday Friday Nondefective 85 (91) 90 (91) 95 (91) 95 (91) 90 (91) 455 Defective 15 (9) 10 (9) 5 (9) 5 (9) 10 (9) 45 100 100 100 100 100 500

Is there sufficient evidence to reject the hypothesis that the number of defective articles is independent of the day of the week on which they are produced? Use a 5% level of significance.

Step 1 Assumptions: The classes are all inclusive and mutually exclusive Step 2 Ho: the number of defective articles is independent of the day of the week on

which they are produced. HA: the number of defective articles is not independent of the day of the week

on which they are produced. Step 3 α = 0.05 df = (4)(1) = 4

2 9.49cχ = Step 4 ( )

( ) ( ) ( )

22

2 2 285 91 90 91 10 9

91 91 98.55

O EE

χ−

=

− − −= + + +

=

∑

L

Step 5 a) 2χ is not in the critical region b) Fail to reject Ho.

∴ There is not sufficient evidence at the 5% level of significance to conclude that the number of defective articles is not independent of the day of the week on which they are produced.

2χ

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