unit 5 statistical reasoning - wordpress.com€¦ · · 2017-03-134 a) calculate the mean,...

Unit 5 Statistical Reasoning 1

5.1 Exploring Data

Data: A set of values.

A set of data can be represented as { x1 , x2 , x3 , ... , xn }.

(I) Measures Of Central Tendency

Mean (Average): The sum of a set of values divided by the total number of

values.

The mean can be represented as xor .

Mean = ValuesOfNumber

ValuesOfSum

Median: The middle value in a set of ordered numbers.

Mode: The most frequent number that occurs in a set of data.

(II) Measures Of Dispersion

Range: In a set of data, Range = largest number – smallest number.

Standard Deviation: A measure indicating how data is spread out or dispersed

within a set of data. This measure identifies the distance

that each data value is from the mean.

Goals:

Exploring Measures of Central Tendency (mean, median and mode)

Exploring range as a measure of dispersion


Example 1

Tim and Luke are both enrolled in Mathematics 2201 and scored the following

marks on their last 5 unit tests.

a) Determine the measures of central tendency (mean, median, and mode)

for each student's test marks.

Tim Luke

Mean =

Mean =

Median:

Median:

Mode:

Mode:

b) Calculate the range for each student's test marks.

Tim Luke

Range =

Range =

c) Describe the similarities and differences in the student’s marks.

Unit 1 Unit 2 Unit 3 Unit 4 Unit 5

Tim 60 65 70 75 80

Luke 68 69 70 71 72


Example 2

In a science experiment, students tested whether compost helped plants

grow faster by counting the number of leaves on each plant. The following

results were obtained:

a) Calculate the mean, median, and mode for each group.

Plant Growth with Compost Plant Growth without Compost

Mean =

Mean =

Median:

Median:

Mode:

Mode:

Plant Growth with

Compost

(# of leaves per plant)

Plant Growth without

Compost

(# of leaves per plant)

7 5

4 9

6 1

4 3

8 2

7 4

Points To Consider

To compare two sets of data you need only the mean, the median, and the mode.

False. Different sets of data can give equal measures of central tendency but it does

not tell us anything about the spread of data.

Most sets of data are evenly distributed about their mean.

False. The range of scores can vary significantly between sets of data.


b) Calculate the range for each group.

Plant Growth with Compost Plant Growth without Compost

Range =

Range =

c) Describe the dispersion of the data for each group.

d) Which group of plants grew better? Justify your decision.

Example 3 10 numbers have a mean of 37. If one number is removed the mean is 38.

What was the number removed?


Example 4

Calculate the range of each group. Explain why the range, by itself, can be a

misleading measure of dispersion.

Group A: 8, 13, 13, 14, 14, 14, 15, 15, 20 Group B: 7, 7, 8, 9, 11, 13, 15, 15, 17, 18

Group A Group B

Range =

Range =


Example 5

Paulo is trying to decide between two different brands of batteries for his car. Both brands

are the same price. He obtains data for the lifespan, in years, for 30 batteries of each brand.

http://easycalculation.com/statistics/mean-median-mode.php

a) Determine the measures of central tendency and the range. Describe any

similarities and differences between the two sets of data.

Brand X Brand Y

Mean = Mean =

Median:

Median:

Mode:

Mode:

Range:

Range:

Measured Lifespans of 30 Car Batteries (years)

Brand X Brand Y

5.1 7.3 6.9 4.7 5.0

6.2 6.4 5.5 5.7 6.8

6.0 4.8 4.1 5.2 8.1

6.3 7.5 5.0 5.7 8.2

3.3 3.1 4.3 5.9 6.6

5.8 6.4 6.1 4.6 5.7

5.4 6.3 4.8 5.9 5.5

4.7 6.0 4.5 6.6 6.0

5.0 6.5 5.8 5.4 5.1

5.7 6.8 5.6 4.9 6.1

4.9 5.7 6.2 7.0 5.8

6.8 5.9 5.3 5.6 5.9


b) Is the mode useful to compare in this situation?

c) How can we compare the data to help Paulo decide which brand of battery to buy?

Questions: Pages 239 – 240, # 1a, b, d, 2, 3


5.2 Frequency Tables, Histograms, and Frequency Polygons

(I) Frequency Distributions:

(II) Histograms:

(III) Frequency Polygons:

Goals:

Creating Frequency Tables, Histograms and Frequency Polygons


Example 1

The marks earned on a 10 question Math quiz for a class of 20 students are given

below.

6 7 5 7 7 8 7 6 9 7

4 10 6 8 8 9 5 6 4 8

a) Construct a frequency table. b) Display the data on a histogram

and construct the frequency polygon.

Mark Tally Frequency


Example 2

a) How does the above histogram differ from a bar graph?

b) How many residents are between the ages of 60 and 70?

c) How many residents are between the ages of 70 and 75?

d) How many residents, in total, reside at SunnyView Seniors Home?

e) Which interval would include someone who is exactly 80 years of age?

Example 3 Which of the distribution of scores has the larger dispersion? Justify your answer.


Example 4

Jason does a survey at the local dog park. He asks 36 owners how long it takes

them to walk their dog to the park. Their responses (in minutes) are given below.

7 32 13 27 9 27 17 12 12 26 23 16 25 18 19 16 17 19

8 20 24 20 2 28 18 19 18 14 19 12 13 19 22 21 22 20

a) Construct a frequency table for the data above.

Time (Min) Tally Frequency

0–4

4–8

8–12

12–16

16–20

20–24

24–28

28–32

32–36


b) Construct a frequency polygon to display the data.

Time Midpoint Frequency

0–4

4–8

8–12

12–16

16–20

20–24

24–28

28–32

32–36

Questions: Pages 249–251, # 2, 3, 4, 5, 6


5.3 Standard Deviation

Deviation: Standard Deviation:

Goals: Determining Standard Deviation and using it to solve problems and

make decisions

Population Standard Deviation Sample Standard Deviation

n

x

DataOfNumber

SumTotalx

2

11

2

n

xx

DataOfNumber

SumTotalS x

x population standard deviation xS sample standard deviation

calculate the sum calculate the sum

x each individual data value x each individual data value

the population mean x the sample mean

n number of values in data set n number of values in data set

Note: A set of data will have a standard deviation of 0 when all the data values are the same.

Example: Given the set of data { 20 , 20 , 20 , 20 , 20 }. Standard Deviation = 0


Example 1

Remember the sample of Tim's and Luke's marks in Math 2201. Both scored the

following marks on their last five unit tests.

We previously calculated the mean of Tim's marks to be 70 and the median to be 70.

Also, we found the mean of Luke's marks to be 70 and the median to be 70.

a) Whose marks are more dispersed?

b) What does this mean in terms of a high or low standard deviation?

c) If the data is clustered around the mean, what does this tell us about the

value of the standard deviation?

d) Who was more consistent over the five unit tests?


e) Calculate the sample standard deviation for each student. What does the standard

deviation tell you about the dispersion of Tim's and Luke's test scores?


Example 2

The following data represents the time, in minutes, it takes each member of the

entire track team to complete the assigned circuit. Determine the population

standard deviation.

9 , 10 , 12 , 14, 8 , 10 , 14

Data

( x )

Mean

( )

( Data – Mean )

( x – )

( Data – Mean )2

( x – )2

9

10

12

14

8

10

14

Example 3 For the following would we determine the sample standard deviation, or the

population standard deviation, based on the data collected.

a) We want to determine how many gold coins pirates have on a pirate ship but

we only have time to interview 5 pirates.

b) A teacher recorded the grades, on a Math test, for all the students in his class.

c) A coach records all the goals scored by his team for a 20 game season.

d) The owner of a restaurant is interested in how much money each customer

spends at the restaurant. He examines 10 randomly selected receipts and

records the data.


Example 4

Two companies produce cinnamon heart candies. Quality control randomly

chooses and weighs 25 boxes from each company. The following masses are

recorded (in grams).

http://easycalculation.com/statistics/standard-deviation.php

a) Use technology to calculate the standard deviation of Company A and Company B.

b) If the standard deviation must be less than 0.5 for production to continue,

what decision must be made by Company A and Company B?

Example 5

The principals, of two different high schools, kept a record of the number of

students given detention over a 5 day period. The following results were obtained:

School A: 4 8 13 2 5 School B: 9 6 11 10 8

Standard Deviation = 3.83 Standard Deviation = 1.72

Which school has the greatest variation? Why?

Company A Company B

24.8 25.3 24.8 25.2 24.5 24.7 25.8

25.2 25.1 24.9 24.5 24.6 25.0 24.7

24.3 24.7 25.4 24.9 25.1 24.0 24.6

25.1 24.4 24.8 25.0

72.8 71.9 72.0 72.3 72.9 71.0 71.3

72.1 72.8 71.7 71.8 71.6 72.5 72.9

71.0 70.9 71.1 72.6 72.5 72.0 70.0

71.4 72.1 71.8 72.1

http://easycalculation.com/statistics/standard-deviation.php


Example 6

Mr. Jones organized the results of the last math test he gave his Math 2201 class.

Mark Number Of Students

40 – 50 1

50 – 60 2

60 – 70 8

70 – 80 10

80 – 90 9

90 – 100 2

a) How many students wrote the test? b) What is the mode of the data?

c) What is the range of the data? d) Determine the standard deviation.

Data

( x )

Mean

( )

( Data – Mean )

( x – )

( Data – Mean )2

( x – )2

45

55

65

75

85

95

Questions: Pages 261 – 263, # 1a, c, 2, 5(use technology), 6, 7a, b, 9, 11


5.4 Normal Distribution

Analyzing the Distribution of Data

Sometimes the distribution of data has a special shape.

Example: The graph to the right has one peak

and the shape has one mode.

Describe the shape of each distribution below:

(a) (b) (c)

Goals: Determining Properties of a Normal Distribution, and compare

normally distributed data.


Properties Of The Normal Distribution Curve

• The graph is symmetrical.

• The mean, median, and mode are equal (or close) and fall at the line of symmetry.

• The normal curve is shaped like a bell peaking in the middle, sloping down toward

the sides, and approaching zero at the extremes.

• 50% of the data is to the right of the mean.

• 50% of the data is to the left of the mean.

• The area under the curve can be considered as 1 unit since it represents 100% of the data.

The 68 – 95 – 99.7% Rule

• About 68% of the data lie within one standard deviation of the mean.

• About 95% of the data lie within two standard deviations of the mean.

• About 99.7% of the data lie within three standard deviations of the mean.

Total area under the curve is 100% or a unit of 1

3223

0.15% 0.15%


Example 1

The following sample data represents the approximate number of hours of playing

video games in a week, for 15 year olds.

a) Determine the mean, median, mode, and standard deviation (use technology).

What do you notice about the values?

b) Construct a frequency table and generate a histogram. Use an interval width

equal to the standard deviation.

Hours Tally Frequency

c) Discuss the symmetry of the histogram.

21 21 17 23 14 20 19 17 24 22

18 19 18 17 20 15 21 16 13 20

14 19 24 22 17 24 21 19 21 21

16 23 19 19 15 20 27 17 19 18

Number of hours playing video games in a week. (15 year olds)


d) Draw a frequency polygon and explain its shape.

e) Where do the mean, median and mode lie?

f) Approximate the percentage of data that is within 1 standard deviation of the mean.

g) Approximate the percentage of data that is within 2 standard deviations of the mean.

h) Approximate the percentage of data that is within 3 standard deviations of the mean.

Example 2

Which normal distribution curve has the largest standard deviation? Explain your

reasoning.

A) B)

C) D)


Example 3

A data set of 50 items is given below with a standard deviation of 1.8. Answer the

following questions:

a) What is the value of the mean, median and mode?

b) Is the data normally distributed? Explain your reasoning.

6 9 6 7 6 7 6 7 6 6

5 8 5 8 6 7 7 4 8 6

5 4 6 8 6 6 7 6 6 8

8 7 6 6 0 4 7 8 5 3

6 7 6 7 1 7 7 4 4 3


Example 4

The weight of adult female Siberian huskies

is normally distributed with a population

mean ( ) of 48.5 lb and a standard

deviation ( ) of 2.1 lb.

a) Sketch a normal distribution curve

that represents the given data.

b) What percentage of adult female dogs would you expect to weigh between:

i) 46.4 lb and 50.6 lb ii) 44.3 lb and 52.7 lb iii) 42.2 lb and 54.8 lb

iv) 48.5 lb and 52.7 lb v) 44.3 lb and 50.6 lb vi) More than 50.6 lb

c) If there are 820 adult female Siberian huskies registered in Alberta how

many would weigh between 44.3 lb and 50.6 lb?


Example 5 The ages of the members of a seniors

curling club are normally distributed,

with a mean of 63 years and a standard

deviation of 4 years.

a) Construct a normal distribution curve.

b) What percentage of the curlers are in each of the following age groups?

i) between 59 and 67 years old ii) between 55 and 71 years old

iii) between 67 and 75 years old iv) older than 75 years old.

c) If there are 120 senior curlers in the club, how many of them are between 51

and 67 years old?

Questions: page 279–281, #2, 3, 4, 6, 8, 10, 11, 13


STATISTICS INCLASS ASSIGNMENT REVIEW SHEET

1. For the following data, determine the mean, the median, the mode, and the range

(a) 10 , 5 , 6 , 5 , 2 , 3 , 4 (b) 24, 10, 18, 11, 19, 12, 11, 7 (c) 21, 18, 25, 24, 22, 19, 26, 25, 23, 27

2. Given are the grades Mark and Mary received on 5 unit Math tests this year.

(a) Determine Mark’s and Mary’s average on their unit tests and the range value for their tests.

(b) Which student was more consistent in their grades?

3. (a) 15 numbers have a mean of 32. If one number is removed the mean is 30. What was the

number removed?

(b) 10 numbers have a mean of 24. If one number is added the mean is 28. What was the

number added?

4. Answer the following questions based on the given histogram.

(a) How many residents are between the ages of

60 and 85?

(b) How many residents are older than 75 years of age?

(c) How many residents reside at the Seniors Home?

5. The marks earned on a 10 question Science quiz for a class of 20 students are given below.

Construct a frequency table and construct a histogram.

5 7 5 7 7 9 6 8 9 7 4 10 7 6 8 4 5 6 3 8

Unit 1 Unit 2 Unit 3 Unit 4 Unit 5

Mark 50 68 77 59 86

Mary 45 65 55 85 75


6. The survey data of how long it takes 30 owners to walk their dog to the park, in minutes, is given below.

Complete the frequency table and the histogram and construct a frequency polygon.

6 12 26 8 26 16 11 22 15 24 17 18 15 16 18

7 21 23 1 17 18 17 13 18 11 12 21 20 21 19

7. A high school track team consists of 16 runners. The following data represents the time, in minutes,

it takes 8 members (chosen at random) of the entire track team to complete the assigned circuit.

9 10 12 15 8 11 10 13

(a) Determine the sample standard deviation.

(b) The track coach has decided to take his team to the national championships if the

standard deviation is less than 2 minutes. What decision should the coach make?

Time (Min) Tally Frequency

0–4

4–8

8–12

12–16

16–20

20–24

24–28

Data

( x )

Mean

( x )

( Data – Mean )

( x – x )

( Data – Mean )2

( x – x )2


8. Mr. Jones organized the results of the last test he gave his Math 2201 class.

Mark Number Of Students

40 – 50 1

50 – 60 3

60 – 70 4

70 – 80 7

80 – 90 3

90 – 100 2

(a) How many students wrote the test? (b) What is the mode of the data?

(c) Determine the population standard deviation.

Data

( x )

Mean

( )

( Data – Mean )

( x – )

( Data – Mean )2

( x – )2

9. At a bowling tournament, the bowling scores are normally distributed with a population mean ( )

of 171.5 and a standard deviation ( ) of 7.4.

a) Sketch a normal distribution curve that represents the given data.

b) What percentage of bowlers would you expect to score:

j) 164.1 and 178.9 ii) 156.7 and 186.3 iii) 149.3 and 186.3

iv) 156.7 and 178.9 v) 164.1 and 193.7 vi) More than 193.7

c) There are 120 bowlers at the tournament, approximately how many bowlers scored

between 178.9 lb and 193.7?

10. The ages of the members of a card club are normally distributed with a mean of 62 years

and a standard deviation of 7 years

(a) What percentage of the card players are in the following age groups?

i) between 41 and 83 years old ii) between 48 and 69 years old

iii) between 55 and 76 years old iv) younger than 48 years old

(b) If there are 40 card players in the club how many of them are between 48 and 76 years old?


SOLUTIONS

1(a) Mean = 5 Median = 5 Mode = 5 Range = 8 (b) Mean = 14 Median = 11.5 Mode = 11 Range = 17

(c) Mean = 23 Median = 23.5 Mode = 25 Range = 9

2(a) Mark: Mean 68 Range 36 Mary: Mean 65 Range 40 (b) Mark 3(a) 60 (b) 68 4(a) 29 (b) 17 (c) 31

7(a) sample standard deviation = 2.27 (b) Don’t take the track team to the national championships.

8(a) 20 students (b) mode = 70 – 80 (c) population standard deviation = 9.42

9(b) i. 68% ii. 95% iii. 97.35% iv. 81.5% v. 83.85% vi. 0.15% (c) approximately 19 bowlers

10(a) i. 99.7% ii. 81.5% iii. 81.5% iv. 2.5% (b) 38 card players

TO KNOW

Mean = ValuesOfNumber

ValuesOfSum

Population Standard Deviation

n

x

DataOfNumber

SumTotalx

2

Sample Standard Deviation

11

2

n

xx

DataOfNumber

SumTotalS x


5.5 Z – Scores

How can we compare individual data scores within a set of data?

Example:

You play in a rock band. Band members practise an average of 16.5 h per week,

with a standard deviation of 4.2 h. You practise an average of 22 h per week. How

can we compare and estimate the percent of the band that, on average, practises

a greater number of hours than you.

Goals: Using Z – Scores to compare data, make predictions and solve

problems.

(I) Sketch the normal curve.

(II) Determine how many standard

deviations your practise time is above the

mean?

(III) Mark that standard deviation on the

normal curve.

(IV) Approximate the percentage of band

members that practise less hours than you.

(V) Determine the percentage of bamd

members that practise more than you.


The Z – Score:

The mean and standard deviation completely define a normal distribution.

Because populations have different means and standard deviations, their

curves will not be exactly the same but all normal distribution curves are bell

shaped.

Each normal distribution curve will have its own mean, 𝜇 and its own

standard deviation, 𝜎.

In order to compare different normal distribution curves, we must

standardize the normal distribution.

This means we rewrite the normal distribution with 𝜇 = 0 and 𝜎 = 1. This

produces the Standard Normal Distribution Curve.

To accomplish this we use a z – score formula:

z =x−μ

σ where x represents a data value

𝜇 represents the population mean

𝜎 represents the standard deviation


Standard Normal Distribution Curve

The Normal Curve is Standardized to:

Compare scores within a set of data.

Compare scores between different sets of data to see which is better.

To determine if individual data values are above or below average.

To determine how far away an individual data value is from the average

which is calculated by the z – score formula.

A positive z–score indicates a data value lies above the mean.

A negative z–score indicates a data value lies below the mean.

The area under the standard normal distribution curve is 1.


Example 1

IQ tests are normally distributed with a mean of 100 and a standard deviation of

15. What percentage of students achieved less than the 130 mark?

a) Draw the normal distribution curve, labeling the mean and standard

deviation. Use this diagram to determine what percentage of students

achieved less than a 130 mark.

b) Draw the standard normal distribution curve and indicate where the

130 mark is found.

c) Using the z–score formula and then the z–score table (P.35 & P.36) check

your answer. Was there any difference in your answers from (b) and (c)?

Explain.


d) Using your diagram from (a), estimate the percentage of students who achieved

a mark less than 120.

e) Using your diagram from (b), the z–score formula and the z–score table

determine the percentage of students who achieved less than 120.

f) Why is the z–score more reliable than estimating using standard deviation?

g) What percentage of students achieved more than 120?

Example 2

Two students competed in a nation–wide mathematics competition and received

the following scores. For the competition the mean (μ) was 66 and the standard

deviation (σ) was 10. Determine the z–score for Alma and Bruce.

Alma: 70 Bruce: 80


Example 3 On the math placement test at Memorial University of Newfoundland, the mean

score was 62 and the standard deviation was 11. If Mark’s z–score was 0.8 what

was his actual exam mark?

Example 4

On her first math test Susan scored 70%. The mean class score was 65% with a

standard deviation of 4%. On her second test she received 76%. The mean class

score was 73% with a standard deviation of 10%.

a) Without performing any calculations, on which test do you think she did better?

b) By calculating 2 separate z–scores, on which test did Susan perform better

with respect to the rest of her class?


Example 5

NHL hockey players sharpen their skates to ensure speed, balance and quick turns.

Sidney Crosby wants to sharpen his skates when only 25% of other NHL skaters

would sharpen their skates. If the population mean for skate sharpening is 20

hours with a standard deviation of 3 hours, when should Sidney Crosby sharpen his

skates?

z =x−μ

σ

Step I:

Indicate the area under the standardized normal curve.

Step II:

Use the z – score table to determine a z – score

based on an area of 25% which is 0.25.

Step III:

Use the z – score from the table and the z – score formula to determine the individual

data score that represents when Sidney Crosby should sharpen his skates.


Example 6 Red candy hearts are packaged according to weight with a mean of 300 g and a

standard deviation of 8 g. Packages with weights less than 290 g and more than 312

g are rejected by quality control workers.

If 50 000 packages are produced each day, how many packages would quality

control expect to reject in a day? What advice would you give this company?

Step I:

Calculate TWO z – scores based on the given data.

Step II:

On the standardized normal curve indicate

the position of the z – scores from step I

and indicate the unacceptable and

acceptable regions.

Step III:

Use the z – score table to attain the area under the curve for each

z – score then determine the area of the unacceptable region. Use

that result to determine the number of packages to be rejected.

Questions: Pages 292–293, #1, 2, 3, 4, 9, 10, 11, 13, 14, 16, 17, 18


5.6 Confidence Intervals

What is the population?

Population: A group of all individuals persons, objects, or items from which

information is taken for statistical measurement.

Why not survey all of the eligible voting population?

Goals: Use the Normal Distribution to solve problems that involve confidence intervals.


Taking a portion or part of the population is a sample

Sample: A representative portion of the population

How can we be sure a sample represents the entire population?

How well a sample represents the larger population depends on two things:

(i) the confidence level

(ii) the margin of error

Consider the results of the survey posted above:

(a) What does “accurate to within 3.5% points” mean?

This is called the margin of error


(b) What does (c) What is a confidence interval?

“95 times out of 100” mean?

Example:

The confidence interval is calculated as:

42% ± 3.5%

Lower limit: 42% – 3.5% = 38.5%

Upper limit: 42% + 3.5% = 45.5%

This means that the committed NL voters are considered to accurately reflect the views of the NL voting

population so that if this survey was replicated 100 times, 95 times out of 100 percent of the time, voters

would support the liberals between 38.5% and 45.5%.

This is called the confidence level

This is called the confidence interval


Example 1

A Rent–A Car–company surveys customers and finds that 50 percent of the

respondents say its customer service is “very good.” The confidence level is cited

as 95 percent and the margin of error is ± 3 percent.

Determine the margin error, the confidence level, and the confidence interval.

Explain what they mean in this situation.

Example 2

A brand of battery has a mean life expectancy of 12.6 hours with a margin of error of

± 0.7 hours. Determine the confidence interval.


Example 3

A botanist collects a sample of 50 iris petals and measures the length of each. It is

found that the mean is 5.55 cm and the standard deviation is 0.57 cm. He then

reports that he is 95% confident that the average petal length is between 5.39 cm

and 5.71 cm.

a) Identify the margin of error, the confidence interval, and the confidence level.

b) Explain what information the confidence interval gives about the length of

iris petals.

c) If you did not know the margin of error but you knew that the confidence

interval is between 5.39 cm and 5.71 cm, how could you determine the

margin of error?

Example 4

A telephone survey of 600 randomly selected people was conducted in an urban

area. The survey determined that 76% of people, from 18 to 34, years of age, have

a social networking account. The results are accurate within ± 4 percentage points,

19 times out of 20.

a) Determine the margin of error, the confidence level, and the confidence interval.

b) What range of people, that were selected, have a social networking account?


Example 5

A recent study reports that 61% of students at Lewisporte Intermediate own a cell

phone. The results of the study are reported to be accurate, 19 times out of 20, with

a margin of errror of 3.6%.

a) What is the confidence level? b) What is the confidence interval?

c) According to the study, if there are 258 students at Lewisporte Intermediate

School what is the range of students who own a cell phone?

Example 6 In a national survey of 400 Canadians from the ages of 20 to 35, 37.5% of those

interviewed claimed they exercise for at least four hours a week. The results were

considered accurate within 4%, 9 times out of 10.

a) How many people in the survey claimed to exercise at least four hours a week?

b) What is the margin of error and the confidence level? What are the limitations

to this survey?

c) If there are approximately, 8.7 million people in Canada, what range of

Canadians exercise at least 4 hours a week?

Questions: Pages 302–303, # 1, 2a, 3, 4, 5, 6, 8a, b

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