Continuous Probability Distributions

Example 1: The amount of time it takes to assemble a computer is normally distributed, with a mean of 50 minutes and a standard deviation of 10 minutes.. Much like the standard normal

Continuous Probability

Distributions

Chapter 7

A continuous random variable has an

uncountable infinite number of values in the

P(a ≤ x ≤ b )

The probability that a continuous variable X will assume

any particular value is zero

The probability that x falls between ‘a’ and ‘b’ is the area under the graph of f(x) between ‘a’ and ‘b’

A random variable X is said to be

A random variable X is said to be uniformly uniformly

distributed if its density function is

with

12

) a b

( )

X (

V 2

b

a E(X)

b x

a a

b

1 )

x ( f

Example 7.1: The daily sale of gasoline is uniformly

distributed between 2,000 and 5,000 gallons Find the

probability that sales are: Between 2,500 and 3,000 gallons

A random variable X with mean µ and variance σ 2is normally distributed if its probability density function is

We denote a normal distribution by N(µ, σ 2 )

Normal distributions range from minus infinity to plus infinity

71828

2 e

and

14159

3 where

x

e 2

1 )

x ( f

2

x ) 2 / 1 (

=

= π

2 e

and

14159

3 where

x

e 2

1 )

x ( f

2

x ) 2 / 1 (

=

= π

µ

−

−

7.2 Normal Distribution

The Shape of the Normal Distribution

bell shaped, and symmetrical around µ

µ

Increasing the mean shifts the curve to

the right…

Increasing the standard deviation “ flattens” the curve…

Two facts help calculate normal probabilities:

- The normal distribution is symmetrical.

- Any random variable (rv) X having N(µ , σ2 ) ) can be transformed into a rv Z having

This shifts the mean of X to zero…

0

This changes the shape of the curve…

0

Example 1: The amount of time it takes to

assemble a computer is normally distributed, with

a mean of 50 minutes and a standard deviation of

10 minutes What is the probability that a

computer is assembled in between 45 and 60

minutes?

•Solution

Let X denote the assembly time of a computer.

We seek P(45<X<60).

The curve for σ =5%

The curve for σ = 10%

Example 2: The standard deviation of the rate of return (X) is now 10% What is the probability of losing money?

Using Excel to Find Normal Probabilities

For P(X<k) enter in any empty cell:

What percentage of the standard normal population

is located to the left of z .30 ?

Example 4: Determine z not exceeded by 5% of the population; that is, z is exceeded by 95% of the population.

Solution: Because of the symmetry of the normal distribution it is the negative value of z .05

0

0.95 0.05

-1.645

The Student t density function

ν (nu) is called the degrees of freedom

E(t) = 0 V(t) = n/(n – 2)

2 / ) 1 ( 2

t 1 )]!

2 [(

)]!

1

[(

)t ( f

+ ν

−

ν

=

2 / ) 1 ( 2

t 1 )]!

2 [(

)]!

1

[(

)t ( f

+ ν

−

ν

=

(for n > 2)

Much like the standard normal distribution, the Student

about its mean of zero:

As the number of degrees of freedom increases, the t distribution approaches the standard normal distribution.

The student t distribution is used extensively in statistical inference Table 4 in Appendix B lists values of ; that is, values of a Student t

random variable with degrees of freedom such that:

The values for A are pre-determined “critical”

values, typically in the 10%, 5%, 2.5%, 1% and

1/2% range.

For example, if we want the value of t with 10

degrees of freedom such that the area under the Student t curve is 05:

Area under the curve value (t A ) : COLUMN

Degrees of Freedom : ROW

t.05,10

t .05,10 =1.812

Chi squared values can be found from the chi

squared table or from Excel.

The χ2 -table entries are the χ2 values of the right hand tail probability (A), for which P(χ2 n > c 2 A ) = A.

0 5 10 15 20 25 30 35

A

χ2 A

Degrees of freedom

1 0.0000393 0.0001571 6.6349 7.87944

.

10 2.15585 2.55821 18.307 23.2093 25.1882

Degrees of freedom

1 0.0000393 0.0001571 6.6349 7.87944

.

.99 0 χ 2

.05 χ 2

.0 1 0

χ 2 005

A=.05

χ 2 Α

To find χ 2 for which

To find the point in a chi-squared distribution with 8 degrees of freedom, such that the area to the right is 05,

Look up the intersection of the 8 d.f row with the

column, yielding a value of 15.5

To find the point in a chi-squared distribution with

8 degrees of freedom, such that the area to the

left is 05,

=2.73 =15.5

F Distribution…

F > 0

is the “numerator” degrees of freedom and

is the “denominator” degrees of freedom.

0

F F

1

F 2

2 2

2 2

2 )

F

(

f

2 2

1

2

2 2

2

1 2

1

2 1

2 1

1 1







 ν

0

F F

1

F 2

2 2

2 2

2 )

F

(

f

2 2

1

2

2 2

2

1 2

1

2 1

2 1

1 1







 ν

!

!

!

This density function generates a rich family of

distributions, depending on the values of n 1 and n 2

The F Distribution

ν1 = 5, ν2 = 10

ν1 = 50, ν2 = 10

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008

ν1 = 5, ν2 = 10

ν1 = 5, ν2 = 1

For example, what is the value of F for 5% of the

area under the right hand “tail” of the curve, with

a numerator degree of freedom of 3 and a

denominator degree of freedom of 7?

Numerator Degrees of Freedom : COLUMN

Denominator Degrees of Freedom : ROW

F.05 , 3 , 7

There are different tables

for different values of A.

Make sure you start with

the correct table !!

F .05,3,7 =4.35

Sampling Distributions

Chapter 8

40

Samples are random, so the sample

statistic is a random variable.

As such it has a

As such it has a sampling distribution sampling distribution .

42

8.1 Sampling Distribution of the Mean

Example 1: A die is thrown infinitely many

times Let X represent the number of

spots showing on any throw The

Suppose we want to estimate µ from the mean

of a sample of size n = 2.

What is the distribution of ?

44

x

Throwing a die twice – sample mean

The distribution of when n = 2

Calculating the relative frequency of each value

of we have the following results

) 25

( 1167

5 3

25 n

2 x

2 x

x

σ

=

= σ

= µ

=

) 10

( 2917

5 3

10 n

2 x

2 x

x

σ

=

= σ

= µ

=

) 5

( 5833

5 3

5 n

2 x

= µ

=

As the sample size changes, the mean of the sample mean does not change!

) 25

( 1167

5 3

25 n

2 x

2 x

x

σ

=

= σ

= µ

=

) 10

( 2917

5 3

10 n

2 x

2 x

x

σ

=

= σ

= µ

=

) 5

( 5833

5 3

5 n

2 x

= µ

=

As the sample size increases, the

variance of the sample mean decreases!

Compare the range of the population

to the range of the sample mean. Let us take samples of two observations

2

The Central Limit Theorem

If a random sample is drawn from any population, the

sampling distribution of the sample mean is:

– Normal if the parent population is normal,

normal, provided the sample size is sufficiently large The larger the sample size, the more closely the

sampling distribution of will resemble a normal

distribution.

50

x

2 x

Example 2: The amount of soda pop in each bottle

is normally distributed with a mean of 32.2

ounces and a standard deviation of 3 ounces

Find the probability that a bottle bought by a

customer will contain more than 32 ounces.

0.7486 67)

P(z

) 3

32.2

32 σ

μ

x P(

Find the probability that a carton of four bottles will

have a mean of more than 32 ounces of soda per bottle.

9082

0 )

33

1 z

( P

) 4

3

2 32 32

x ( P )

32 x

P( >

Example 3: The average weekly income of B.B.A graduates one year after graduation is $600

Suppose the distribution of weekly income has a standard deviation of $100

What is the probability that 35 randomly selected graduates have an average weekly income of less than $550?

57

0.0015 2.97)

P(z

) 35 100

600

550 σ

μ

x P(

550) x

58

Chapter 9

59

Statistical inference is the process by which we acquire information and draw conclusions about populations from samples.

There are two procedures for making inferences:

The objective of estimation is to determine the

approximate value of a population parameter on the basis

We saw earlier that point probabilities in

continuous distributions were virtually zero The probability of the point estimator being correct is zero

An

An interval estimator interval estimator draws inferences about a

population by estimating the value of an unknown parameter using

parameter using an interval an interval .

That is we say (with some _% certainty) that the population parameter of interest is between some lower and upper bounds.

For example, suppose we want to estimate the mean summer income of a class of business students For

n = 25 students, is calculated to be 400 $/week.

point estimate interval estimate

An alternative statement is:

The mean income is between 380 and 420 $/week.

Estimating when is known…

From Chapter 9, the sampling distribution of is approximately normal with mean µ and standard deviation

X

n /

σ

n /

X Z

σ

µ

−

=

Thus, substituting Z produces

With a little bit of algebra,

With a little bit of different algebra we have

x z

z x

The confidence interval

Lower confidence limit (LCL) =

Upper confidence limit (UCL) =

The probability 1 – α is the

The probability 1 – α is the confidence level confidence level, which is a

measure of how frequently the interval will actually include µ.

Four commonly used confidence levels

Example 2: Doll Computer Comp found that the

demand over the lead time is normally

distributed with a standard deviation of 75

Estimate the expected demand over the lead time

at 95% confidence level Assume N=25 and x = 370 16

[ 340 76 , 399 56 ] 40

29 16

.

370 25

75 96

1 16

370

25

75 z

16

370 n

± α

Comparing two confidence intervals with the same level of confidence, the narrower interval provides more information than the wider interval

The width of the confidence interval is calculated by

and therefore is affected by

• the population standard deviation (s)

• the confidence level (1-a)

• the sample size (n).

n

Z

2 n

z

x n

1 z

If the standard deviation grows larger, a longer

confidence interval is needed to maintain the

confidence level

Note what happens when σ increases to 1.5 σ

Example 1: Estimate the mean value of the distribution resulting from the 100 repeated throws of the die It is known that σ = 1.71

Use 90% confidence level:

Use 95% confidence level:

1 645

1 96

1

2 n

2 n

By increasing the sample size we can decrease

the width of the confidence interval while the

confidence level can remain unchanged.

n

z 2 width

Interval = /2 σ

α

There is an inverse relationship between the width of the interval and the sample size

The phrase “estimate the mean to within W

units”, translates to an interval estimate of the form

The required sample size to estimate the mean is

77

2 2

Example 4: To estimate the amount of lumber that can be harvested in a tract of land, the mean diameter of trees in the tract must be estimated to within one inch with 99%

confidence What sample size should be taken for the margin of error +/-1 inch? (assume diameters are normally distributed with σ = 6 inches).

78

The confidence level 99% leads to α = 01, thus zα/2 = z .005 = 2.575.

239 1

2.575(6) W

σ

z n

2

2 2

Introduction to

Hypothesis Testing

Chapter 10

• Is there statistical evidence that support the hypothesis that more than p% of all

potential customers will purchase a new products?

• Is the hypothesis that a certain drug is effective?

Step 1: Two hypotheses are defined.

H 0 : The null hypothesis specifies our current belief

about the parameter we test (µ = 170, p = 4, etc.) Must be a specific value.

H 1 : The alternative hypothesis specifies a range of

values for the parameter tested (µ > 170; p ≠ .4; etc.) effected by the belief.

11.1 Concepts of Hypothesis Testing

The Reject-Region method

H 0 Needed in the test Do not reject H 0

Example 1: The manager of a department determines

that new billing system will be cost-effective only if the mean monthly account is more than $170 A random

sample of 400 monthly accounts is drawn, for which the sample mean is $178 Assume a standard deviation of

$65 Can the manager conclude from this that the new system will be cost-effective?

Step 1: H 0 : µ = 170

H 1 : µ > 170

10.2 Testing the Population Mean when the Population Standard Deviation is Known

β= Probability of committing the type II error

Send an innocent person

Send an innocent person

Step 3: Determine the sample size n and hence the

sampling distribution.

Example 1:n=400; N(0,1)

Step 4: Depending on the sampling distribution , the H A , and the value of α, find the suitable critical value and the reject region

H A : μ > 170 (1-sided test)

1.645 Critical value

N(0,1)

α = 0 5

Reject region: Z > 1.645

Right-Tail Testing

Left-Tail Testing

Two–Tail Testing

Step 5: Collect data x = 178

Calculate the standard test statistic

46

2400

/65

170

178/− = − =

=

n

x z

σ

µ

H A : μ = 170 Example 1:

Reject H 0 : There is enough statistical evidence to

conclude that the alternative hypothesis is true.

Do not reject H 0: There is not enough statistical

evidence to conclude that the alternative hypothesis is true.

Step 6: If the test statistic is in the reject region, then reject H 0 , otherwise do not reject H 0 .

Example 1: Reject H 0 at α = 05

A Left Hand Tail Test

Reject H 0 if falls herex Critical

value

The SSA envelop plan example: The chief financial officer in FedEx believes that including a stamped self-addressed (SSA) envelop in the monthly

invoice sent to customers will decrease the amount

of time it take for customers to pay their monthly

bills Currently, customers return their payments in

22 days on the average, with a standard deviation of

6 days.

A random sample of 220 customers was selected

and SSA envelops were included with their invoice packs The mean time it took customers to pay their bill was 21.63

Can the CFO conclude that the plan will be

successful at 10% significance level?

/ 6

22 63

σ

µ

Critical value Reject H 0 if falls herex

Example 2: A statistician believes the monthly

mean of the long-distance bills for all AT&T

difference between AT&T’s bills and the

competitor’s bills (on the average)?

55

17

=

x

19

1 100

/ 87

3

09

17 55

σ

µ

−1.96 Critical value

1.96 Critical value

α /2 = 025 α /2 = 025

Chapter 11

Inference About a

Population

Recall: By the central limit theorem, when σ2 is known

is normally distributed if:

• the sample is drawn from a normal population, or

• the population is not normal but the sample is

sufficiently large.

When σ2 is unknown, we use s 2 instead, and has the

t-distribution

x

11.1 Inference About a Population Mean

When the Population Standard Deviation is Unknown

Example 1: The productivity of newly hired

trainees is studied It is believed that trainees can process and distribute more than 450 packages per hour within one week of hiring.

Can we conclude that this belief is correct, if the mean productivity observation of 50 trainees is 460.38 and the standard deviation is 38.83.

Step 4: Reject Region t ≥ tα,n-1 ≅ t. 05,50 = 1.676

Cf 1.645 for the Z-distribution

Step 5:

Step 6: Reject the null hypothesis

89

150

83

38

45038

x

1.676 Critical value

Confidence interval estimator of µ when σ 2 is unknown

1 n

f.

d n

s t

n

s t

Example 2: An investor is trying to estimate the return

on investment in companies that won quality awards last year A random sample of 83 such companies is

selected, yields

Construct a 95% confidence interval for the mean

return.

8.31 68.98

s

68.98 s