STATISTICS FOR DECISION MAKING

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5.ExaminingDistributionofDataConfidenceIntervals.pdf

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5.1 Binomial distribution

Binomial distribution

A binomial distribution is a discrete random variable distribution with two possible values that have �xed probabilities that add up to . The probability for the value of a random variable with a binomial distribution is

where is the number of trials, is the number of successes, and is the probability of a success for each trial.

Example 5.1.1: Flipping a coin.

Determine the distribution of probabilities for the number of times a coin comes up heads for , , and coin �ips.

Solution

Flipping a coin is an example of a binomial distribution. Each �ip is independent, and has a probability of of coming up heads.

The sample space when a coin is �ipped twice is . Each outcome has a probability of . However, since two ways exist for heads to come up once, the total probability for that outcome is .

The sample space when a coin is �ipped three times is . Each outcome has a

probability of , but three ways exist for heads to come up once or twice, so the probability of those outcomes is . The distribution of probabilities for ,

, or coin �ips is summarized in the table below.

# of �ips heads head heads heads

P(k) = (1 − pCn,kp k )n−k

n k p

1 2 3

0.5

{HT ,TH,HH,TT} 0.25

0.25 + 0.25 = 0.5

{HHH,HHT ,HTH,THH,HTT ,THT ,TTH,TTT} 0.125

0.125 × 3 = 0.375 1 2 3

0 1 2 3

0.5 0.5

0.25 0.5 0.25

0.125 0.375 0.375 0.125

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The formula for the binomial distribution can be used to calculate the same probabilities. Ex: the probability of heads coming up twice in three �ips is

Example 5.1.2: Cumulative probabilities for male kittens.

A cat has seven kittens. Assuming that male and female kittens are equally likely, what is the probability that at least kittens are male?

Solution

To �nd the probability of getting at least males, the probabilities of getting , , and males must be added.

The probability of at least kittens being male is therefore

PARTICIPATION ACTIVITY 5.1.1: Finding binomial probability.

1) Assuming a fair coin, if the coin is �ipped times, what is the probability of getting tails times? Type as #.##

P(k) = (1 − pCn,kpk )n−k

P(2) = (0.5 (0.5 3!

2!(1!) )2 )1

= 0.375

5 5 6 7

P(k ≥ 5)

P(k = 5)

P(k = 6)

P(k = 7)

= P(k) = P(k = 5) + P(k = 6) + P(k = 7)∑ k=5

= (0.5 (0.5 = 0.164 7!

5!(2!) )5 )2

= (0.5 (0.5 = 0.055 7!

6!(1!) )5 )2

= (0.5 (0.5 = 0.008 7!

7!(0!) )5 )2

P(k ≥ 5) = P(k) = 0.164 + 0.055 + 0.008 = 0.227∑ k=5

5 4

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2) If a six-sided die is rolled times, what is the probability of rolling numbers less than ? Type as #.###

3) If the forecast for the next days is a chance of rain each day, what is

the probability that rain will fall least of the days? Type as #.###

4) After a class took a midterm that of the class passed, the professor randomly assigned students to study groups of . What is the probability that fewer than students in a particular study group passed the midterm? Type as #.###

Mean, variance, and standard deviation of the binomial distribution

Since the mean of a discrete random variable is , the mean of one trial in a binomial distribution is . The mean of the distribution over trials is the sum of the individual means, and is therefore

Similarly, the variance of one trial is

Check Show answer

5 4

Check Show answer

4 50%

Check Show answer

70%

10 5

Check Show answer

μ = Σ(x ⋅ p(x)) μ = Σ(x ⋅ p(x)) = 0(1 − p) + 1(p) = p n

μ = np

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and the variance of the distribution is

The standard deviation of the binomial distribution is the square root of the variance,

Example 5.1.3: Shape of the binomial distribution.

Find the shape, mean, variance, and standard deviation of the binomial distribution for and .

Solution

The shape of the binomial distribution for a given and can be seen by plotting the probabilities of each possible number of successes. The distribution for and

is shown below.

σ2 = Σ((x − μ p(x)))2

= (0 − p (p) + (1 − p (p))2 )2

= − + p − 2 +p3 p2 p3

= p(1 − p)

= np(1 − p)σ2

σ = np(1 − p) − −−−−−−−

√

n = 10 p = 0.5

n p n = 10

p = 0.5

P(0)

P(1)

P(2)

P(3)

P(4)

P(5)

= ( ) = 0.00098 10!

0!(10 − 0)! 0.50 0.510

= 0.0098 P(6) = 0.2051

= 0.0439 P(7) = 0.1172

= 0.1172 P(8) = 0.0439

= 0.2051 P(9) = 0.0098

= 0.2461 P(10) = 0.00098

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The mean of the distribution is

The variance and standard deviation are

PARTICIPATION ACTIVITY 5.1.2: Finding the mean and standard deviation of a binomial distribution.

1) What is the mean of a binomial distribution with trials and a probability of success for each trial of

2) What is the variance of a binomial distribution with trials and a probability of success for each trial of

μ = np = 10 ⋅ 0.5 = 5

σ2

= np(1 − p) = 10(0.5)(0.5) = 2.5

= ≈ 1.5812.5−−−√

0.3

Check Show answer

0.6

Check Show answer

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3) What is the standard deviation of a binomial distribution with trials and a probability of success for each trial of

? Type as #.###

Bernoulli distribution

The Bernoulli distribution is the special case of a binomial distribution where . The Bernoulli distribution is the probability distribution of a single trial with two possible outcomes. The mean of the distribution is

the variance of the distribution is

and the standard deviation is

Example 5.1.4: Bernoulli distribution.

A six-sided die is rolled once. What is the probability that the number will be greater than ? Graph the distribution and give the mean, variance, and standard deviation.

Solution

Since two numbers on the die are greater than , the probability of rolling a number greater than is or . Since only two possible outcomes exist for one trial, the graph of the distribution has only two bars, showing and .

Check Show answer

0.5

Check Show answer

n = 1

μ = p

= p(1 − p)σ2

σ = p(1 − p) − −−−−−−

√

4 4 26 0.333

p 1 − p

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The mean of the distribution is , the variance is , and the standard deviation is .

PARTICIPATION ACTIVITY 5.1.3: Determining when to use a Bernoulli distribution.

Which of the following scenarios can be modeled using a Bernoulli distribution?

1) The probability of a tra�c light being red, yellow, or green when a car approaches the light.

2) The probability of getting tails on a coin �ip.

3) The probability of a politician winning an election.

4) The probability of rolling three tails in four coin �ips.

μ = p = 0.333 = p(1 − p) ≈ 0.222σ2

σ = ≈ 0.4710.222 − −−−√

Bernoulli distribution

Not Bernoulli distribution

Bernoulli distribution

Not Bernoulli distribution

Bernoulli distribution

Not Bernoulli distribution

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PARTICIPATION ACTIVITY 5.1.4: Using a Bernoulli distribution.

Consider the Bernoulli distribution of the probabilities for a student guessing the answer on a choice multiple choice question.

1) What is the probability that the student will guess the wrong answer? Type as #.##

2) What is the mean of the distribution? Type as #.##

3) What is the variance of the distribution? Type as #.###

4) What is the standard deviation of the distribution? Type as #.###

Bernoulli distribution

Not Bernoulli distribution

Check Show answer

5.2 Poisson distribution

Poisson distribution

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The Poisson distribution gives the probability of independent, randomly occurring events happening over a period or area where events happen on average. The formula for the distribution is

The Poisson distribution is usually used for situations where the average count is fairly low. The distribution differs from the binomial and hypergeometric distributions in that right side of the distribution extends to in�nity. Ex: when �ipping a coin times, the probability of getting tails more than twenty times is exactly . However, when counting the number of raindrops falling in a bucket over a period of time, no theoretical upper limit exists. The probabilities for higher numbers become vanishingly small, but never exactly .

Example 5.2.1: Meteors per hour.

Under optimal conditions, a person can expect to see an average of about meteors per hour at night. What is the probability of a hiker seeing meteors in an hour one night?

Solution

Each meteor can be considered a random event, allowing the scenario to be modeled by a Poisson distribution with and . The probability of seeing the given number of

meteors is found by using the formula .

PARTICIPATION ACTIVITY 5.2.1: Using a Poisson distribution.

1) A particular highway has an average of potholes per mile. What is the

probability of a driver encountering potholes in a given mile? Type as #.###

k λ

P(k) = e−λ λk

20 0

10 15

λ = 10 k = 15

e−λ λk

P(k)

P(15)

= e−λ λk

= e−10 1015

15! = 0.035

8 7

Check Show answer

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2) A library has an average of books checked out per hour. What is the probability that books will be checked out during the next minutes? Type as #.###

3) An astronomer is taking images of a section of the night sky and keeping track of the number of cosmic ray strikes in each image. Given the frequencies of cosmic ray strikes in the table below, what is the probability that the next image will have cosmic ray strikes? Type as #.###

Cosmic rays Frequency

PARTICIPATION ACTIVITY 5.2.2: Identifying Poisson distributions.

Determine if each scenario can be modeled as a Poisson distribution.

3 10

Check Show answer

0 2

1 3

2 4

3 4

4 2

5 3

6 1

7 1

Check Show answer

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1) The number of earthworms living in a cubic meter of dirt.

2) The number of students in a class passing a test.

3) The number of radioactive particles being emitted from an element in a given time.

Cumulative probability

The cumulative probability that an event happens at most a certain number of times is the sum of the probabilities being considered. The cumulative probability that an event happens at least a certain number of times is found by summing the probabilities for fewer events and subtracting that from the total probability, which is .

Example 5.2.2: Dandelions in the lawn.

A square meter of lawn in a particular yard has an average of dandelions. What is the probability that a certain square meter has at most dandelions? What is the probability that a square meter has more than dandelions?

Solution

The probability of a square meter having at most dandelions is found using the formula

Poisson distribution

Not a Poisson distribution

Poisson distribution

Not a Poisson distribution

Poisson distribution

Not a Poisson distribution

11 3

P(k ≤ 3) = P(k)∑ k=0

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The probability of having three or fewer dandelions in a square meter of the yard is .

The probability of a square meter having more than dandelions is found by subtracting the probability of having or less from the total probability .

PARTICIPATION ACTIVITY 5.2.3: Selling lemonade.

A child is selling lemonade on the street corner. An average of customers come by every hour.

1) What is the probability that fewer than customers will stop by in an hour?

Type as #.###

2) What is the probability that at least customers will stop by in an hour? Type as #.###

P(3)

P(2)

P(1)

P(0)

P(k ≤ 3)

= = 0.0037e−11 113

= = 0.0010e−11 112

= = 0.0002e−11 111

= = 0.0000e−11 110

= P(k) = 0.0037 + 0.0010 + 0.0002 + 0.0000 = 0.0049∑ k=0

0.0049

3 3 1

P(k > 3) = 1 − P(k ≤ 3)

= 1 − 0.0049

= 0.9951

Check Show answer

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Shape of the Poisson distribution

The Poisson distribution is not symmetric, since the probability of fewer than zero events occurring is always zero, but the probabilities to the right of the mean only asymptote to zero. The shape of the distribution is therefore skewed to the right.

As might be expected, since is the average number of times an event happens in the given interval, the mean of the Poisson distribution is . The variance is , so the standard deviation is

The Poisson distribution applies to many situations across several disciplines, from the number of cosmic rays in a given time to the number of people with a given disease in a population. The distribution can also be used to test whether a set of events is random, called hypothesis testing, or to generate random numbers.

PARTICIPATION ACTIVITY 5.2.4: Shape of a Poisson distribution.

Example 5.2.3: Comparing a distribution to a Poisson distribution.

A hospital administrator keeps track of the number of patients coming into the emergency department every hour over a hour period. Given the table below, is the distribution of patients random?

# of patients Frequency

Check Show answer

μ = λ = λσ2

σ = λ −−√

Animation content:

undefined

Animation captions:

1. The shape of the Poisson distribution is skewed to the right. The mean is and the standard deviation is .

2. As gets larger, the shape of the distribution approaches the shape of a binomial distribution.

μ = λ σ = λ

−−√ λ

0 5

1 3

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Solution

The distribution can be graphed by dividing each frequency by the total number of hours and plotting the probabilities against the number of patients.

The mean number of patients per hour is . Comparing the distribution of patients to a Poisson distribution with the same mean suggests that the patient distribution is not random.

2 5

3 2

4 3

5 3

6 2

7 1

= 2.70865 24

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PARTICIPATION ACTIVITY 5.2.5: Shape of a Poisson distribution.

1) Which of the following distributions with is most likely to be Poisson?

2) Which of the following sets of numbers was most likely generated from a Poisson distribution with ?

μ = 2

λ = 7

7, 7, 6, 9, 7, 7, 4, 5, 13, 7

9, 3, 9, 10, 7, 2, 2, 5, 12, 11

2, 5, 1, 3, 7, 8, 4, 9, 10, 6

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5.3 Properties of continuous probability distributions

The probability density function of a continuous random variable

A probability density function (pdf) describes the relative likelihood of all values for a continuous random variable. Ex: The amount of time for Casey to do his chores is a random variable , where all values between hour and hours are equally likely. The notation is typically used for the pdf. For Casey's chores, for all values of between and and everywhere else.

The pdf can be written as a function, but often a graphical representation is the most descriptive. The area under portions of the curve given by the pdf provide the probabilities. A pdf must be non-negative and the total area under the curve must be . Ex: The probability Casey spends between and hours doing chores is the area under the curve for values of between and . Calculus concepts such as integration are often required to �nd areas.

PARTICIPATION ACTIVITY 5.3.1: A probability density function.

PARTICIPATION ACTIVITY 5.3.2: Probability density functions.

Riley manages an auto repair shop and is modeling the time required for a car service. Four potential pdfs for this situation are shown below. Match each pdf with the best description of the model.

X 1 2 f(x)

f(x) = 1 x 1 2 0

1 1 1.5 f(x) = 1 x 1 1.5

Animation captions:

1. The amount of time Casey takes to do chores is a random variable where all values between hour and hours are equally likely.

2. for , and for all other values of . 3. The area under between and is the probability Casey spends between

and hours on chores. The area of the rectangle is .

X 1 2

f(x) = 1 1 ≤ x ≤ 2 0 x f(x) x = 1 x = 2 1

2 1 ⋅ 1 = 1

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Lowest chance of extreme (high or low) service times

Not a valid pdf

More longer service times than shorter service times

Same probability for a time from and as and

The cumulative distribution function of a continuous random variable

pdf c pdf d pdf b pdf a

0 1 1 2

Reset

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A cumulative distribution function (cdf) of a continuous random variable is the probability that for any number , the observed value of the random variable will be at most or . Ex: When Casey does housework, the cdf describes the probability of Casey �nishing in time less than or equal to any value such that the probability is less than or equal to . The notation is typically used for the cdf of , in contrast to lower-case for the pdf. Ex: is read "the probability is less than or equal to ". As with discrete random variables, the cdf always starts at

and ends at and never decreases as the value of increases. The cdf may approach the limits of and in cases where the possible values of are in�nite.

PARTICIPATION ACTIVITY 5.3.3: A cumulative distribution function.

PARTICIPATION ACTIVITY 5.3.4: Continuous cumulative distribution function (cdf).

Riley manages an auto repair shop and is modeling the time required for a car service. Four potential cdfs for this situation are shown below.

x x p(X ≤ x)

x X 1.5 F(x) X f(x) F(1.5) = P(X ≤ 1.5)

X 1.5 0 1 X 0 1 x

Animation captions:

1. The amount of time Casey requires to do chores is a random variable where all values between hour and hours are equally likely.

2. A continuous density function can be constructed from the probability density function .

3. The probability Casey takes hours or fewer to complete the chores is . 4. In the pdf, for all , and the area under totals . In the cdf, for all

X 1 2

F(x) f(x)

1.5 0.5 f(x) ≥ 0 x f(x) 1 F(x) = 1

x > 2

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1) The cdf which is not valid is:

2) If the correct model is cdf d, the probability of a service time less than 2 hours is:

3) According to the model depicted in cdf a, the probability of a service time completion in the �rst hour is _____ the probability the service is completed in the second hour.

4) Jeff takes his car in for service. He hopes to pick the car up no more than

cdf a

cdf b

cdf d

near 0

0.2

0.5

lower than

the same as

higher than

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hours later. Which cdf should Jeff hope is the correct model?

Mean, variance, and standard deviation of a continuous random variable

In a previous section, the mean, variance, and standard deviation for a discrete random variable were de�ned. The interpretation of the three measures is similar for continuous random variables.

The mean or expected value of a continuous random variable is a measure of the center of the distribution. The mean is a weighted average of the possible values of the random variable, with the pdf providing the weights. Graphically, the mean is where a pivot is placed so that the pdf balances. The variance of a continuous random variable is a measure of the spread of a distribution. The variance, like the mean, is a weighted average. The variance averages the squared distance of each possible value of from the mean, with weights provided by the pdf. The standard deviation is another measure of the spread of the distribution. The standard deviation is the square root of the variance, .

Integral calculus is required to compute the three quantities in the case of continuous random variables. The details are not discussed in this material.

PARTICIPATION ACTIVITY

5.3.5: The mean, variance, and standard deviation of a continuous random variable.

PARTICIPATION ACTIVITY

5.3.6: Mean, variance, and standard deviation of a continuous random variable.

cdf a

cdf c

cdf d

μ E(X) X

σ2 X

X σ

σ = σ2 −−√

Animation captions:

1. Let be the percentage of correct responses for a test. A pdf of the test results can be constructed.

2. The mean is the center of the distribution which balances the density function. The density is greater below percent than above.

3. The correct mean percentage is . 4. The variance and standard deviation measure the spread of the data. A smaller variance and

standard deviation correspond to a distribution with values closer to the mean.

0.5 0.33

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A measure of the center of a distribution

Measure involving the squared differences of from

A measure of spread with units the same as the random variable

Variance Mean Standard deviation

X E(X)

Reset

5.4 Normal distribution

Normal distribution

The normal distribution is a continuous probability distribution characterized by a bell-shaped probability distribution function and is symmetric around the mean . The normal distribution is also referred to as the Gaussian distribution. The normal distribution is pervasive because the distribution is a model for quantities that are computed as sums (totals) or averages. Often data is summarized using either the total or average. The normal distribution also occurs in many settings, including exam scores and heights or other physical measurements.

The normal distribution

Models: Averages, totals, many natural phenomenon such as measurement errors, test scores, body measurements. Notation: , read " has a normal distribution with parameters and ." The distribution is known as the standard normal distribution

. Parameters:

: Mean of the distribution : Standard deviation of the distribution

: Variance of the distribution

X ∼ N(μ, )σ2 X μ σ2 N(0, 1)

Z ∼ N(0, 1)

μ σ

σ2

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Possible values: All real numbers

Figure 5.4.1: Graphs of normal distributions with different values of and .

Empirical rule

Unimodal and symmetric distributions, such as the normal distribution, follow a de�nite pattern useful for obtaining probabilities and interpreting outcomes. A unimodal distribution is a distribution with exactly one mode. In such distributions, the mean, median, and mode are equal.

The empirical rule states that for any unimodal and symmetric distribution: (1) of the data fall within one standard deviation of the mean, (2) of the data fall within two standard deviations of the mean, and (3) of the data fall within three standard deviations of the mean. Mathematically,

A rule of thumb is that quantities or observations within two standard deviations of the mean are considered common or usual. Quantities outside of two standard deviations are considered uncommon or unusual.

The empirical rule is illustrated in the animation below.

μ σ2

68% 95%

99.7%

68 percent of data are located on the interval

95 percent of data are located on the interval

99.7 percent of data are located on the interval

[μ − σ,μ + σ]

[μ − 2σ,μ + 2σ]

[μ − 3σ,μ + 3σ]

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PARTICIPATION ACTIVITY 5.4.1: Empirical rule.

PARTICIPATION ACTIVITY 5.4.2: Empirical rule.

In 2015, the College Board SAT mathematics score of high school seniors, which follow an approximately normal distribution, had a mean of and a standard deviation of .

1) What percentage of seniors who took the test in 2015 scored between and ?

2) What percentage of seniors who took the test in 2015 scored above ?

3) What percentage of seniors who took the test in 2015 scored between and ?

4) Should a score of be considered unusual?

Animation captions:

1. A distribution that is unimodal and symmetric is centered around the mean, . 2. The standard deviation determines the spread of the distribution. 3. Approximately of the data is within standard deviation from the mean. 4. Approximately of the data is within standard deviations from the mean. 5. Approximately of the data is within standard deviations from the mean. 6. The percentage of data within speci�c intervals can be obtained as a consequence of the

empirical rule.

μ σ

68% 1 95% 2 99.7% 3

511 1201

271 751

50%

68%

95%

631

16%

68%

84%

271 511

68%

16%

47.5%

770

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-scores

A -score is a signed value that indicates the number of standard deviations a quantity is from the mean. A positive -score indicates that the quantity is above the mean and a negative -score indicates that the quantity is below the mean. A -score with high absolute value implies that the quantity is farther from the mean, and thus more unusual.

The -score is given by

where is the raw score, is the mean, and is the standard deviation.

-scores are particularly important for determining whether a data point is an outlier and for comparing quantities from different unimodal, symmetric distributions.

Example 5.4.1: Earnings surprise.

Earnings per share (eps) is a one of the major indicators of a company's pro�tability. Of interest, therefore, is the difference between expected earnings and reported earnings of a company. In the �scal quarter ending April 2016, Walmart had an eps of , compared to a consensus eps forecast of . The earnings surprise as a percentage is

The table below lists the earnings surprise percentage of Walmart's competitors in the �scal quarter ending in August 2016 or earlier.

Company Earnings surprise

Walmart

Target

Kroger

Whole Foods

Costco

Yes

z z z

z = x − μ

x μ σ

0.98 0.882

⋅ 100% = 11.36% 0.98 − 0.88

0.88

(%)

11.36

7.89

1.96

−3.12

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Supervalu

Analysis

To determine whether the eps of Walmart compared to the other companies is unusual, the sample mean and sample standard deviation should be computed.

Computing the -score for Walmart's earnings surprise percentage yields

Thus, Walmart's earnings surprise percentage is not considered too unusual in the grocery business because the -score is less than .

PARTICIPATION ACTIVITY 5.4.3: Using -scores to compare data from two different distributions.

According to the United States Department of Health and Human Services, the mean height for Americans is m for men and m for women . The standard deviation is

m for men and m for women.

1) What -score corresponds to a man who is m tall? Type as: #.###

2) What -score corresponds to a woman who is m tall? Type as: #.###

3) Is the man or the woman taller with respect to their gender? Type as: man or woman

−9.52

z = = 1.32 11.36 − 1.43

7.53

z 2

1.757 1.618 3

0.074 0.069

z 1.853

Check Show answer

z 1.758

Check Show answer

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PARTICIPATION ACTIVITY

5.4.4: The normal distribution.

The GRE (Graduate Record Exam) scores for both verbal and quantitative reasoning are approximately normally distributed and scaled to have a mean of with a standard deviation of .

1) What is the probability a randomly selected student scored higher than

on verbal reasoning?

2) What is the probability the quantitative reasoning score for a randomly selected student is between and

3) What is the approximate probability of scoring at least or higher on verbal reasoning?

4) What is the probability of scoring between and ?

PARTICIPATION ACTIVITY 5.4.5: Finding a -score given a probability.

150 8.754

150

0.5

0.68

0.95

132.5 167.5

0.5

0.68

0.95

159

0.16

0.32

0.5

165 170

Less than the probability of a score between and .150 155

The same as the probability of a score between and .150 155

Greater than the probability of a score between and .150 155

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The GRE (Graduate Record Exam) scores for both verbal and quantitative reasoning are approximately normally distributed and scaled to have mean with standard deviation of

1) Below what score do of all scores fall? Type as: #.###

2) Above what score do of all scores fall? Type as: #.###

Sampling distribution for sample means

Often, statisticians look at the distribution of a test statistic, such as the sample mean. Suppose a sample of size is taken from a population and the sample mean is computed. Repeating this process for multiple samples and creating a relative frequency plot of the obtained test statistic yields a sampling distribution.

The sampling distribution of the mean, denoted by , is the distribution of sample means when taking random samples of the same size. The mean of the sample means, denoted by , is the population mean. That is, . The standard error is the standard deviation of the sampling distribution, denoted by , when sampling with replacement. That is, . The

standard deviation requires a correction factor when sampling without replacement. This correction

factor is , where is the population size.

Example 5.4.2: Guessing squash weights.

At a carnival booth, the population is the weights of squash with unknown distribution. To play the game, a child weighs randomly selected squash and uses the sample mean weight of the selected squash to guess the population mean weight.

Squash A B C D E

Weight

a. Find the population mean .

150 8.75

40%

Check Show answer

20%

Check Show answer

X ¯ ¯¯̄

μ X ¯ ¯¯̄¯

= μμ X ¯ ¯¯̄¯ (SE)

σ X ¯ ¯¯̄¯ =σ

X ¯ ¯¯̄¯

n√

N−n N−1

− −−− √ N

5 2

5 6 7 9 13

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b. Find the distribution of the sample means for a sample size of . c. Find the mean of the sample means for a sample size of . d. Find the standard deviation of the sample means given .

Solution

a. The population mean is .

b. samples of size exist.

Sample Weights Sample mean Probability

A, B

A, C

A, D

A, E

B, C

B, D

B, E

C, D

C, E

D, E

The sampling distribution of the mean is

Only when squash C and D are picked would the child guess the population's mean weight correctly. Thus, the child has a in chance of correctly guessing the population mean .

c. The mean of the sample means is

as expected. The mean of the sampling distribution is always equal to the population mean.

n = 2 n = 2

σ = 2.83

μ = = 8 5 + 6 + 7 + 9 + 13

5 10 n = 2

x̄

5, 6 5.5 1/10

5, 7 6 1/10

5, 9 7 1/10

5, 13 9 1/10

6, 7 6.5 1/10

6, 9 7.5 1/10

6, 13 9.5 1/10

7, 9 8 1/10

7, 13 10 1/10

9, 13 11 1/10

X ¯ ¯¯̄

x̄ 5.5 6 6.5 7 7.5 8 9 9.5 10 11

P( )x̄ 1/10 1/10 1/10 1/10 1/10 1/10 1/10 1/10 1/10 1/10

1 10 μ

= 8 5.5 + 6 + 6.5 + 7 + 7.5 + 8 + 9 + 9.5 + 10 + 11

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d. Because the same squash cannot be chosen twice, the game involves sampling without replacement. Thus, a correction factor is needed when using the formula for standard deviation.

PARTICIPATION ACTIVITY 5.4.6: Guessing squash weights.

Suppose that squash instead of are used to guess the the population mean weight of squash. The possibilities and the corresponding sampling distribution are given below.

Sample Weights Sample mean Probability

A, B, C, D

B, C, D, E

A, B, D, E

A, B, C, E

A, C, D, E

1) Can the child correctly guess the mean population weight if squash are chosen? Type as: yes or no

2) What is the mean of the sample means?

= ⋅ = ⋅ = 1.733σ X ¯ ¯¯̄¯

N − n

N − 1

− −−−−− √ σ

n −−√

5 − 2

5 − 1

− −−−− √ 2.83

2 –√

4 2 5

x̄

5, 6, 7, 9 6.75 1/5

6, 7, 9, 13 8.75 1/5

5, 6, 9, 13 8.25 1/5

5, 6, 7, 13 7.75 1/5

5, 7, 9, 13 8.5 1/5

x̄ 6.75 8.75 8.25 7.75 8.5

P( )x̄ 1/5 1/5 1/5 1/5 1/5

Check Show answer

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3) What is the correction factor for the standard deviation of the sampling distribution? Type as: #.#

4) What is the standard deviation of the sampling distribution? Type as: #.###

Central Limit Theorem

Previously, �nding the probabilities for normally distributed random variables was discussed. However, most situations do not involve parameters that are normally distributed. Luckily, the distribution of sample means can be assumed to be approximately normal because of a powerful result of the Central Limit Theorem (CLT). The CLT is the basis for assuming averages and totals follow the normal distribution and underlies many of the tests and results used in data analysis. The CLT states that as the sample size drawn from the population with distribution becomes larger, the

sampling distribution of the means approaches that of a normal distribution . Thus,

the -score is

Central Limit Theorem: Assumptions and conditions

Randomness assumption - samples must be randomly selected. Independence condition - sample values must be independent from each other. Sample size assumption - sample size must be large enough. A rule of thumb is that sample sizes should be at least .

condition - sample size must be at most of the population size.

Example 5.4.3: Sampling distribution for sample means of an exponential

Check Show answer

X ¯ ¯¯̄

N (μ, )σ n√

z = − μx̄ σ

n√

30 10% 10%

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distributed random variable.

Consider an exponentially distributed random variable . samples of

size are collected. In the �gure below, the sampling distribution are shown for

, and . Although is not normally distributed, the shape of approaches the shape of the normal distribution as gets larger, which is consistent with the CLT.

Example 5.4.4: Applying the Central Limit Theorem.

An airline studies the mean arrival time of �ights from city A to city B. Arrival times (in hours) follow an exponential distribution with hours and hours. Suppose the study involves randomly selected �ights out of the overall number �ights in the time period. What is the probability that the mean �ight time is greater than hours?

Solution

X ∼ Exp(λ = 2) 2000

n X ¯ ¯¯̄

n = 1, 5, 25 50 X X¯ ¯¯̄

T = 3.57μT = 0.59σT 36 123, 501

3.8

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Although does not follow a normal distribution, the normal distribution can be used to �nd the mean �ight time because the situation satis�es all conditions for the CLT. Thus, the

sampling distribution for approaches the normal distribution .

The mean and standard deviation of the sampling distribution are

The -score that corresponds to is

Thus, the probability that the mean arrival time is greater than hours is

. On average, approximately in �ights from city A to city B have arrival times of at least hours.

PARTICIPATION ACTIVITY 5.4.7: Weights of adult males.

The weights of adult individuals in a certain country are normally distributed with a population mean of pounds and a population standard deviation of pounds. Suppose individuals are sampled.

1) What is the mean of the sampling distribution of the means?

2) What is the standard deviation of the sampling distribution of the means? Type as: #.###

T ¯¯̄̄

N (μ = ,σ = )μT σTn√

μ = = 3.57μT

σ = = = 0.098 σT

n−−√

0.59

36−−√

z = 3.8T¯¯̄̄

z = = = = 2.347 −t¯ μT σT

n√

− μt¯

3.8 − 3.57

0.098

3.8

P( ≥ 3.8) = P(z ≥ 2.34) = 0.0096T¯¯̄̄ 1 104 3.8

μ = 172 σ = 29 n = 36

Check Show answer

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3) What is the probability that randomly selected individuals will have a mean weight of at least pounds? Type as: #.###

Sampling distribution for sample proportion

Sometimes, a study deals with binary categorical variables instead of continuous variables. A binary categorical variable is a random variable that can only take on two possible names or labels. Ex: A free throw shot results in success or failure. A widget produced by a company is either functional or defective. For samples drawn from a population with a binomial distribution, the Central Limit Theorem for proportions apply.

The Central Limit Theorem for proportions states that if where is the number of trials and is the probability of success, then the sampling distribution for proportions follows a

normal distribution . Thus, the -score is

Central Limit Theorem for proportions: Conditions

Example 5.4.5: Applying the Central Limit Theorem for proportions.

Check Show answer

n = 36

180

Check Show answer

X ∼ B(n, p) n p p̂

N (p, )p(1−p) n

− −−−− √ z

z = − pp̂

p(1−p) n

− −−−− √

np ≥ 5 n(1 − p) ≥ 5

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A power tools manufacturer reviews the production history for all drill bits produced and found that of the drill bits made are defective. A change in the manufacturer's quality assurance process reduced the percentage of defective drill bits to of the drill bits sampled. If the quality assurance process was not changed, what is the probability that at most of drill bits are defective? The population proportion is and the sample proportion is .

Solution

The population proportion is . Both and are greater than or equal to . Thus, the conditions of the

CLT are satis�ed and the shape of the binomial distribution approaches that of the normal

distribution .

The standard deviation is

The corresponding -score is

Thus, the probability that at most of the drill bits are defective is or .

PARTICIPATION ACTIVITY 5.4.8: Nearsightedness.

Nearsightedness affects of children in a certain country. The eyesight of randomly selected children are checked for nearsightedness.

1) What is the mean of the sampling distribution of proportions? Type as #.##

8% 5% 300

5% p = 0.08 = 0.05p̂

p = 0.08 np = 300(0.08) = 24 n(1 − p) = 300(0.92) = 276 5

N (p, )p(1−p) n

− −−−− √

= = 0.016 p(1 − p)

− −−−−−− √ 0.08(1 − 0.08)

300

− −−−−−−−−−−− √

z = = = −1.875 − pp̂

p(1−p) n

− −−−− √

0.05 − 0.08

0.016

5% P( ≤ 0.05) = P(z ≤ −1.875) = 0.0303p̂ 3.03%

8% 256

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2) What is the standard deviation of the sampling distribution of proportions? Type as: #.###

3) What is the probability that randomly selected children will have a proportion of nearsightedness of at least ? Type as #.###

Calculators

PARTICIPATION ACTIVITY 5.4.9: Probability to Z-score calculator.

Check Show answer

n = 256

Check Show answer

Enter each input above, then press Build graph.

Ex: 0.6 Ex: 0.1 Less than Ex: 0.5

Build graph

Mean: Standard deviation: Probability:

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PARTICIPATION ACTIVITY 5.4.10: Z-score to probability calculator.

References

(*1) "Total Group Pro�le Report SAT." College Board, 2015, https://research.collegeboard.org/programs/sat/data/archived/cb-seniors-2015.

(*2) "Walmart Inc. Earnings Surprise." NASDAQ, www.nasdaq.com/symbol/wmt/earnings-surprise

(*3) US Department of Health and Human Services. "Anthropometric Reference Data for Children and Adults: United States, 2011-2014 ." Center for Disease Control Vital and Health Statistics, Series 3 Number 39, August 2016, www.cdc.gov/nchs/data/series/sr_03/sr03_039.pdf

Enter each input above, then press Build graph.

Ex: 0.6 Ex: 0.1 Less than Ex: 0.5

Build graph

Mean: Standard deviation: X:

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(*4) "Guide to the Use of Scores." Educational Testing Service Graduate Record Exam, 2018, www.ets.org/s/gre/pdf/gre_guide.pdf

5.5 Student's t-Distribution

-distribution

The central limit theorem provides a useful tool to calculate probabilities associated with non-normal distributions assuming the sample sizes are su�ciently large. In practice, obtaining a large enough sample may not be possible or the population standard deviation is unknown. In both cases, the sample standard deviation divided by the square of the sample size can be used in place of the population standard deviation.

The Student's -distribution or -distribution is used in place of the normal distribution in situations where the sample size is too small or the population standard deviation is unknown. The -distribution has one parameter, degrees of freedom or , which is equal to . As the sample size (and consequently ) increases, the -distribution approaches the normal distribution with a mean of and standard deviation of .

Figure 5.5.1: The -distribution has the same shape as the normal distribution, but has a wider spread because of a slightly larger standard deviation.

t t t

df n − 1 n df t 0

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The -statistic is obtained from a sample assumed to have a -distribution and involves the population mean and a larger variability from estimating the population standard deviation. The same process applies to the computation of probabilities involving the -distribution as shown in an earlier section with normal distribution. The formula for the -statistic is

where is the sample mean, is the sample standard deviation, and is the sample size.

Example 5.5.1: Finding probability using the -statistic.

The United States Census Bureau determined that the mean number of children in an American household is . Suppose households are polled and the sample mean is found to be and the standard deviation is found to be . What is the probability of another household sample with a sample mean of at least ?

Solution

The -statistic is

Thus, the probability of another household sample with a sample mean of at least is

t t

t = − μx̄̄̄ s

n√

x̄̄̄ s n

1.86 50 2.1 1.57

50 2.1

t = = ≈ 1.081 − μx̄̄̄ s

50√

2.1 − 1.86 1.57

50√

50 2.1

P( ≥ 2.1) = P(t ≥ 1.081) ≈ 0.143x̄̄̄

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PARTICIPATION ACTIVITY 5.5.1: Finding probabilities associated with a -distribution.

The United States Census Bureau determined that the mean number of children in an American household is . Polls of households were conducted in a certain city.

1) What is the number of degrees of freedom?

2) A poll of households had a sample mean of children and a sample standard deviation of children. What is the -statistic? Type as: #.###

3) What is the probability of �nding households with a sample mean of

children or less? Type as: #.###

4) A different poll of households had a sample mean of children and a sample standard deviation of children. What is the -statistic? Type as: #.###

5) What is the probability of �nding households with a sample mean between and children and a standard deviation of ? Type as: #.###

1.86 15

Check Show answer

15 3.26

2.12 t

Check Show answer

3.26

Check Show answer

15 2.72

2.12 t

Check Show answer

2.72 3.26 2.12

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PARTICIPATION ACTIVITY 5.5.2: Finding a percentile of a -distribution.

1) Find the th percentile of the - distribution with degrees of freedom. Typeas #.###

2) Find the th percentile of the - distribution with degrees of freedom. Type as #.###

Check Show answer

25 t 30

Check Show answer

60 t 4

Check Show answer

5.6 Con�dence intervals

Estimation and con�dence intervals

Estimation is the process of obtaining information about a parameter by using a statistic. An estimator is a statistical method used to calculate an estimate based on observable data. A good estimator gives estimates that are both accurate and precise. Accuracy is measured in terms of bias. Numerically, bias is the distance between the mean of the sampling distribution and the population mean. Precision is measured in terms of standard error.

PARTICIPATION ACTIVITY 5.6.1: Accuracy and precision.

Animation captions:

1. Accuracy is a measure of closeness to the true value of a parameter. Precision is a measure of closeness to the average value of the measurements.

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Two types of estimates exist: point estimates and interval estimates. A point estimate is a single value estimate for a parameter. An interval estimate is a range of values that is likely to contain the parameter being estimated. Combined with a probability statement, an interval estimate is called a con�dence interval. The percentage in which the con�dence interval contains the parameter is called the con�dence level, which is denoted by .

A con�dence interval is accurate if the con�dence interval contains the true population parameter. A con�dence interval's precision refers to the width of the con�dence interval.

A con�dence interval is constructed by looking at the sample statistic and margin of error. A margin of error, denoted by , is the range of values above and below the point estimate. Numerically,

where the critical value, which depends on and the underlying distribution of the statistic, is the number of standard errors to be added to the point estimate. Thus,

The resulting interval is referred to as the con�dence interval. That is, of the time, the true value of the population parameter will be in the con�dence interval when the same estimator is used, as shown in the animation below.

PARTICIPATION ACTIVITY 5.6.2: A con�dence interval for a population parameter.

Example 5.6.1: How millennials acquire news.

2. Measurements that are neither accurate nor precise are neither close to the true value nor tightly clustered.

3. Measurements that are accurate but not precise are loosely spread out around the true value. 4. Measurements that are precise but not accurate are tightly clustered around an incorrect

value. 5. Measurements that are both accurate and precise are tightly clustered around the true value.

m = (critical value)(standard error)

estimate ± m = estimate ± (critical value)(standard error)

c(100)% c(100)% c(100)%

95%

Animation captions:

1. Suppose the true value of the population parameter is . 2. A sample is drawn from the population. The sample statistic is . 3. A con�dence interval of implies that if samples are drawn repeatedly, the

probability that the sample statistic is in the interval is . 4. The sample statistic is one possible outcome of repeated drawing of samples. 5. The size of the con�dence interval is such that on average, only of every potential

con�dence intervals will not include the true value of the population parameter.

4 3.5

95% [1, 6] [1, 6] 95%

1 20 95%

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A survey by the Media Insight Project asked a random sample of American adults between the ages of 18 and 34 questions about technology use when acquiring news and frequency . The results of the survey have a margin of error of percentage points.

Percent of millennials who ...

Say keeping up with the news is at least somewhat important

Get news daily

Regularly follow �ve or more 'hard' news topics

Usually see diverse opinions through social media

Pay for at least one news-speci�c service, app, or digital subscription

a. What is the point estimate for the percent of millennials who get their news daily? b. What is the con�dence interval for the percent of millennials who get their news

daily? c. What is the interpretation of this interval?

Solution

a. b. c. A con�dence exists that the true population mean percentage of millennials who

get news daily is between and . The represents the probability that the true population mean is contained in a con�dence interval if the estimator is used repeatedly.

PARTICIPATION ACTIVITY 5.6.3: Estimation and con�dence intervals.

1) A con�dence interval estimate of a population parameter is constructed around the sample statistic.

2) A con�dence interval always gives a correct estimate of the population parameter.

1, 045

1 ±3.8

85%

69%

45%

86%

40%

95%

69% 69% ± 3.8%

95% 65.2% 72.8% 95%

True

False

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3) The true population parameter is . An estimator that produces a

con�dence interval of is accurate.

4) A poll has a margin of error of . The width of the corresponding con�dence interval is .

PARTICIPATION ACTIVITY 5.6.4: Americans' view of socialism and capitalism.

According to the results of a Gallup poll dated May 2-4, 2016 on American views involving Capitalism and Socialism, young people view the federal government and socialism more positively than older Americans do .

The table below gives the percentage of people who view the following positively according to age group: small business, entrepreneurs, free enterprise, capitalism, big business, federal government, and socialism.

Just off the top of your head, would you say you have a positive or negative image of each of the following?

18-29 years 30-49 years 50-64 years 65+ years

Small business

Entrepreneurs

Free enterprise

Capitalism

Big business

Federal government

Socialism

True

False

p = 0.5 [0.42, 0.49]

True

False

±2.5%

True

False

98 94 96 93

90 87 87 83

78 84 89 91

57 54 69 63

57 49 52 53

58 43 38 40

55 37 27 37

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The poll has a margin of error at a con�dence level.

Point estimate for the percent of Americans ages 18-29 who view the federal government positively

Point estimate for the percent of Americans ages 18-29 who view socialism positively

con�dence interval for the percent of Americans ages 18-29 who view the federal government positively

con�dence interval for the percent of Americans ages 18-29 who view socialism positively

References

(*1) The Media Insight Project. "How Millennials Get News: Inside the Habits of America's First Digital Generation." NORC at the University of Chicago, American Press Institute, 2017, mediainsight.org/Pages/how-millennials-get-news-inside-the-habits-of-americas-�rst-digital- generation.aspx

(*2) Newport, Frank. "Americans' Views of Socialism, Capitalism Are Little Changed." Gallup, American Press Institute, 6 May 2016, news.gallup.com/poll/191354/americans-views-socialism-capitalism- little-changed.aspx

±4% 95%

58% [54%, 62%] 55% [51%, 59%]

95%

Reset

5.7 Con�dence intervals for population means

con�dence intervalsz

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Choosing an appropriate estimator for the population mean depends on what is known about the population. Although the population mean is usually unknown, the con�dence interval that quanti�es the range within which the true population mean lies can be calculated by sampling the population. Ex: A con�dence interval means that the interval calculated has a probability of containing the population mean. If such a population is sampled times, of the calculated intervals are expected to contain the population mean.

PARTICIPATION ACTIVITY 5.7.1: The -con�dence interval.

The critical values for common con�dence levels are shown in the following table.

Table 5.7.1: Critical values for common con�dence levels.

Con�dence level Critical value

Example 5.7.1: Con�dence interval for the mean �nal grade.

Suppose the mean �nal grade is estimated for all introductory statistics classes taught at a

95% 95% 20 19 20

Animation content:

undefined

Animation captions:

1. A con�dence interval for the population mean of a normal distribution is calculated from the sample mean , standard deviation , and number of samples .

2. The margin of error for the population mean is equal to .

3. The con�dence level is the area under the normal distribution curve in the interval .

4. The signi�cance level is the area under the curve outside the con�dence interval.

x̄̄̄ σ n m z∗ σ

n√

c [ − m, + m]x̄̄̄ x̄̄̄

z∗

c = 0.90 = 1.645z∗

c = 0.95 = 1.960z∗

c = 0.99 = 2.576z∗

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particular college. The population standard deviation is . The �nal grades for a randomly selected statistics class with people are: , , , , , , , , and . Find the

con�dence interval for the mean �nal grade.

Solution

The sample mean is

At the con�dence level, the margin of error is

Thus, the con�dence interval is .

Analysis

The population mean �nal grade is generally unknown. The obtained interval means that a con�dence exists that contains the true population mean �nal

grade.

PARTICIPATION ACTIVITY 5.7.2: Con�dence interval.

The weights of squash (in pounds) are , and . The sample weights have a mean of . The accepted population standard deviation for this type of squash is .

1) What is the margin of error at the con�dence level? Type as #.###

2) What is the margin of error at the con�dence level? Type as: #.###

4 9 76 80 82 83 83 85 85 87 88

95%

= = 83.222x̄ 76 + 80 + 82 + 83 + 83 + 85 + 85 + 87 + 88

95%

z∗

z∗ σ

n −−√

= 1.960

= 1.960 4

9–√ ≈ 2.613

95% [83.222 − 2.613, 83.222 + 2.613] = [80.609, 85.835]

95% [80.609, 85.835]

5 10, 17, 17.5, 18.5 19.5 = 16.5x̄

σ = 1.25

90%

Check Show answer

99%

Check Show answer

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3) What is the con�dence interval? Type as: [##.###, ##.###]

4) What is the con�dence interval? Type as: [##.###, ##.###]

Margin of error and sample size for means when is known

The width of the con�dence interval is twice the margin of error. Recall that the margin of error depends on the con�dence level and the standard error. Thus, given a con�dence level, the width of the con�dence interval changes by changing the standard error. Increasing the sample size decreases the standard error. Similarly, decreasing the sample size increases the standard error. The size of the sample needed to guarantee a con�dence interval with a speci�ed margin of error is given by the formula

Example 5.7.2: Microbeads in a water reservoir.

Scientists study the amount of microbeads (in g/liter) in a water reservoir. The scientists need to know how many water samples should be taken to be sure that their estimate

differs from the actual value by at most g/liter given a population standard deviation of g/liter. How many water samples should the scientists take?

Solution

The margin of error is and the population standard deviation is . Since the con�dence level is , the critical value is . Using the formula above,

Thus, the scientists should take at least water samples.

90%

Check Show answer

99%

Check Show answer

n = ( )σz ∗

μ 95%

x̄ μ 0.8 μ σ = 2.5 μ

m = 0.8 σ = 2.5 95% = 1.960z∗

n = = ≈ 37.516( )σz ∗

( )1.960 ⋅ 2.5 0.8

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PARTICIPATION ACTIVITY 5.7.3: Tax assessment.

A tax assessor assesses the mean property tax bill for all property owners in a certain city. A recent survey obtained a sample mean of dollars. The population standard deviation is known to be dollars.

1) How many tax records should be obtained at a con�dence level to have a margin error of dollars? Round up to the nearest whole number.

2) If the population standard deviation goes up to , would the margin of error be equal, greater than, or less than ? Type as: equal, greater than, or less than

con�dence intervals

The con�dence interval for a population mean is de�ned as

In practice, however, the population standard deviation is rarely known. Thus, con�dence intervals using the -distribution are more useful because the sample standard deviation can always be computed.

The -con�dence interval is given by

1400 1000

90% 100

Check Show answer

1500

100

Check Show answer

[ − , + ]x̄ z∗ σ n −−√

x̄ z∗ σ

n −−√

[ − , + ]x̄ t∗ s n −−√

x̄ t∗ s

n −−√

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where is the critical value that depends on the degrees of freedom and signi�cance level. The values for common signi�cance levels given selected degrees of freedom are shown in the table below.

Table 5.7.2: Critical values for selected values of and .

Example 5.7.3: Circumference of basketballs.

The mean circumference of basketballs produced in a manufacturing facility is supposed to be inches. A random sample of basketballs has a mean of inches with a sample standard deviation of inches. Find the con�dence interval at the signi�cance level.

Solution

At a signi�cance level of , the con�dence level is . The critical value that corresponds to a con�dence level and degrees of freedom is

from the table above. Thus, the margin of error is

Thus, the con�dence interval is

Analysis

t∗

df α

α = 0.1 α = 0.05 α = 0.01

df = 5 2.015 2.571 4.032

df = 10 1.812 2.228 3.169

df = 15 1.753 2.131 2.947

df = 24 1.711 2.064 2.797

29 25 29.1 0.217 α = 0.01

α = 0.01 99% 99% df = 25 − 1 = 24

= 2.797t∗

m = = 2.797 ≈ 0.121t∗ s

n −−√

0.217

25 −−√

99%

[ − m, + m] = [29.1 − 0.121, 29.1 + 0.121] = [28.979, 29.221]x̄ x̄

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The obtained con�dence interval suggests that a con�dence exists that the true population mean circumference of basketballs produced in the facility is between and inches. Since the con�dence interval contains the population mean of inches, insu�cient evidence exist to believe that the mean circumference of basketballs is not inches.

PARTICIPATION ACTIVITY 5.7.4: Weights of pumpkins.

The weights of pumpkins (in pounds) of pumpkins are and . The sample weights have a mean of and a standard deviation of .

1) What is the degrees of freedom?

2) What is the critical value when the signi�cance level is ? Type as: #.###

3) What is the margin of error? Type as: #.###

4) What is the -con�dence interval for pumpkin weights? Type as: [#.### , #.###]

Margin of error and sample size for means when is unknown

The -distribution is commonly used to estimate the necessary sample size for a survey. Since the population standard deviation is generally unknown, the -distribution should be used instead. For a

99% 28.992

29.208 99% 29

6 5, 7, 7.5, 8, 8.5, 8.75 = 7.458x̄ s = 1.245

Check Show answer

α = 0.1

Check Show answer

90% t

Check Show answer

z t

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situation with an unknown population standard deviation, the sample standard deviation should be used. Thus,

where is the critical value at a speci�ed con�dence level, is the sample standard deviation based on a preliminary study, and is the margin of error.

However, depends on the degrees of freedom , which is also in terms of . To �nd , is used to �nd a preliminary sample size . That is,

Then can be calculated using the formula above with .

Example 5.7.4: Quality control.

A quality control department in a company wants to estimate the average length of a part to within a margin of error cm at con�dence interval. A preliminary pilot study found that the sample standard deviation of the length of parts is cm. How many parts should the department use?

Solution

The formula above uses , and thus a preliminary sample size is needed. To �nd , is used in the formula instead of . Since at the con�dence level,

The critical value that corresponds to is , which yields

Thus, the quality control department should use a sample size of parts.

PARTICIPATION ACTIVITY 5.7.5: Machine output.

n = ( )st ∗

t∗ t s m

t∗ n − 1 n t∗ z∗

n∗

=n∗ ( )sz ∗

n df = − 1n∗

±0.35 95% 2.50

t∗ n∗ n∗ z∗

t∗ = 1.960z∗ 95%

= = = 196n∗ ( )sz ∗

( )1.960 ⋅ 2.50 0.35

t df = − 1 = 196 − 1 = 195n∗ = 1.972t∗

n = = ≈ 199( )st ∗

( )1.972 ⋅ 2.50 0.35

199

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A company that manufactures tools want to estimate the average output of a machine to within a margin of error of units with con�dence. A preliminary pilot study found that the sample standard deviation of the machine's output is units.

1) What is the preliminary estimate ?

2) What is the sample size ?

CHALLENGE ACTIVITY 5.7.1: Con�dence intervals for population means.

±2.1 90% 7.3

n∗

Check Show answer

Start

Suppose the mean height in inches of all 9th grade students at one high school is estimated. The population standard deviation is inches. The heights of randomly selected students are , , , , , , and .

= Margin of error at con�dence level =

con�dence interval = [

6 8 75 71 75 62 70 64 74 73

x̄

99%

99% , ] [smaller value, larger value]

Ex: 12.34

Ex: 1.23

Ex: 12.34 Ex: 12.34

Check Next

5.8 Con�dence intervals for population proportions

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Con�dence interval for population proportions

Constructing a con�dence interval for a population proportion is similar to constructing a con�dence interval for population means. Suppose the sample proportion and the number of samples are

known from sampling the population. The margin of error is given by , where

is the critical value corresponding to the desired con�dence level . The con�dence interval for the population proportion is given by .

Example 5.8.1: Con�dence interval for the proportion.

In a survey of randomly selected registered voters in a particular city, people are in favor of banning public smoking. State and interpret the con�dence interval for the population proportion of voters in favor of banning public smoking.

Solution

The sample proportion is

At the con�dence level, the margin of error is

Thus, the con�dence interval is

The con�dence level suggests that if the same estimator was used to construct the interval, then the true population parameter would be within the interval of the time. The obtained con�dence interval suggests a con�dence exists that the true proportion of voters who favor banning smoking is between and .

PARTICIPATION ACTIVITY 5.8.1: Con�dence interval for the proportion.

In a poll of randomly selected prospective voters in a local election, voters were in favor of a school bond measure.

1) What is the sample proportion? Type as: #.###

p̂ n

m m = z∗ (1− )p̂ p̂ n

− −−−− √ z∗

c p [ − m, + m]p̂ p̂

1200 348 95%

= = 0.29p̂ 348

1200

95%

m = = 1.960 ≈ 0.026z∗ (1 − )p̂ p̂

− −−−−−−− √ 0.29(1 − 0.29)

1200

− −−−−−−−−−−− √

95%

[ − m, + m] = [0.29 − 0.026, 0.29 + 0.026] = [0.264, 0.316]p̂ p̂

95% 95%

95% 26.4% 31.6%

1000 281

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2) What is the margin of error for the con�dence level? Type as: #.###

3) What is the margin of error for the con�dence level? Type as: #.###

4) What is the con�dence interval? Type as: [#.###, #.###]

Margin of error and sample size for proportions

Finding the necessary sample size given a con�dence level and margin of error is similar to �nding the sample size for a con�dence interval for population means. The size of the sample needed to guarantee a con�dence interval with a speci�ed margin of error is given by the formula

where is the critical value at a speci�ed con�dence level, is the proportion, and is the margin of error.

However, also depends on . Two possible options exist:

1. A preliminary value for can be used based on a previous study. 2. The worst case scenario with can be used, which provides the largest sample size

needed to satisfy the required margin of error at a speci�ed con�dence level.

Example 5 8 2: Six months after Brexit

Check Show answer

90%

Check Show answer

95%

Check Show answer

95%

Check Show answer

n = p(1 − p)( )z ∗

z∗ p m

p n

p p = 0.50

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Example 5.8.2: Six months after Brexit.

Brexit is a popular term for the successful referendum on June 2016 for the withdrawal of the United Kingdom from the European Union. Six months later, a poll conducted by CNN and ComRes resulted in of the respondents voting to leave the European Union had the referendum been held that day . If the margin of error is percentage points, �nd the sample size needed for a con�dence interval if

a. is used b. no previous estimate is given

Solution

a. The margin of error is . Since a previous estimate is known, the sample proportion is . Since the con�dence level is , the critical value is

. Using the formula above,

Thus, the pollsters sampled at least voting aged adults.

b. When no previous estimate is known, is used. Using the same formula gives,

Thus, the pollsters would have sampled voting aged adults if no previous estimate is given in the news story.

PARTICIPATION ACTIVITY 5.8.2: Margin of error and sample size for proportions.

A poll reported a approval rating for a politician with a margin of error of percentage point.

1) How many voters should be sampled for a con�dence interval? Round up to the nearest whole number.

2) How many voters should be sampled

47% 1 ±2.17

90%

p = 0.47

m = 0.0217 p = 0.47 90%

= 1.645z∗

n = p(1 − p) = (0.47)(1 − 0.47) ≈ 1432( )z ∗

( )1.645 0.0217

1432

p = 0.5

n = p(1 − p) = (0.50)(1 − 0.50) ≈ 1437( )z ∗

( )1.645 0.0217

1437

36% 1

90%

Check Show answer

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for a con�dence interval? Round up to the nearest whole number.

3) How many voters should be sampled for a con�dence interval if no previous estimate is given? Round up to the nearest whole number.

CHALLENGE ACTIVITY 5.8.1: Con�dence intervals for population proportions.

Critical values for quick reference during this activity.

Con�dence level Critical value

95%

Check Show answer

95%

Check Show answer

0.90 = 1.645z∗

0.95 = 1.960z∗

0.99 = 2.576z∗

Start

2 3 4

In a poll of randomly selected voters in a local election, voters were in favor of �re department bond measures. What is the sample proportion ? What is the margin of error for the con�dence level?

1000 737

p̂

m 90%

Ex: 0.123

Check Next

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References

(*1) Greene, Richard Allen. "Brexit poll: Six months on, Brits stand by EU referendum decision." CNN, 19 December 2016, www.cnn.com/2016/12/19/europe/cnn-brexit-poll/index.html

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