Review of Basic Statistics

Definitions• Population - The set of all items of interest in a statistical

probleme.g. - Houses in Sacramento

• Parameter - A descriptive measure of a populatione.g. - Mean (average) appraised value of all houses

• Sample - A set of items drawn from a populatione.g. - 100 randomly selected homes

• Statistic - A descriptive measure of a samplee.g. - Mean appraised value of selected homes

• Statistical inference - The process of making an estimate, prediction, or decision based upon sample data

Types of Data

• Qualitative - Categorical, i.e., data represents categoriese.g. - Existence of an attached garage

• Quantitative - Data are numerical valuesDiscrete(countable) - Counts of thingse.g. - Number of bedroomsContinuous(interval) - Measurementse.g. - Appraised value or square footage

• Cross-sectional - Observations in a sample are collected at the same time.e.g. - Our sample of 100 homes; most surveys

• Time series - Data is collected at successive points in timee.g. - Housing starts, recorded monthly from July 1985 to June 1997

Numerical Descriptive Measures:Notation

• N = Size of Population• n = size of sample• = Population Mean• = sample mean• = Population Variance• = Population Standard Deviation• s2 = sample variance• s = sample standard deviation

x2

Sample Mean,

• where x i = i th observation, and

• n = sample size

x

xx

n

ii

n

1

Sample Variance, s2

sx x

n

ii

n

2

2

1

1

( )

Example

• Find the mean and variance of the following sample values (in years):

3.4, 2.5, 4.1, 1.2, 2.8, 3.7

Random Variables• Definition - A numerical variable whose value is

determined by chance!e.g. - For a randomly selected house:Let X = Appraised value Y = Number of bedrooms

W =

Then X, Y, and W are all random variables. (Why?)

1

0

for attached garage

otherwise

Note - Let X be a random variable, then , S2 and S are also random variables

What is the difference betweenX, , S2, S and x, , s2, s ?

X

X x

Probability Distributions• Definition - A probability distribution for a

random variable describes the values that the random variable can assume together with the corresponding probabilities.

• Importance - Its probability distribution describes the behavior of a random variable. Therefore, questions concerning a random variable cannot be answered without reference to its probability distribution.

Mean,Std. dev.

,

Normal Distributions

X

dens

ity

-3 -2 -1 1 2 30

0.1

0.2

0.3

0.4

Empirical (68, 95, 99.7) Rule• For a normally distributed random variable:

i) Approx. 68% of the values lie within 1 standard deviation, of the mean i.e., P(-< X < +) = 0.68

ii) Approx. 95% lie within 2 of P(-2 < X < ) = 0.95

iii)Approx. 99.7% lie within 3 of. P(-3 < X < ) = 0.997

Standard Normal Distribution

Mean,Std. dev.

0,1

Z

dens

ity

-3 -2 -1 0 1 2 30

0.1

0.2

0.3

0.4

Examples

Determine the following:

a. P(0 < Z < 1.46)

b. P(Z > 1.46)

c. P(1.28 < Z <1.46)

d. P(Z < -1.28)

Solutions

Using a table or Excel:

a. P(0 < Z < 1.46) = 0.4279

b. P(Z > 1.46) = 0.5 - 0.4279 = 0.0721

c. P(1.28 < Z <1.46) = 0.4279 - 0.3997 = 0.0282

d. P(Z < -1.28) = P(1.28 < Z) = 0.5 - 0.3997 = 0.1003

Example

Use a table or Excel, find and interpret z

P(Z > z0.05 ) = 0.05

Ans. z0.05 = 1.645 because P(Z > 1.645) = 0.05

z – scores andstandardized random variables

For a random variable X with mean and standard deviation ,

the number of standard = deviations above or below the mean x is.

is the Standardized Random Variable for X

zx

ZX

the Distribution of (the Sampling Distribution of the Mean)

Properties of : Let = mean of all sample means of size n

= variance of all sample means of size n

Then:

i) =

ii) =

X

X X

2

X

2

n

X

2

X

the Central Limit TheoremI. Central Limit Theorem - If a large sample is

drawn randomly from any population, the distribution of the sample mean, , is at least approximately normal!

II. Properties of 1.

2.

3. If X is normally distributed, then is normal regardless of the size of the sample!

X

X

X

22

X n

X

Example (filling problem)

Suppose that the amount of beer in a 16 oz bottle is normally distributed with a mean of 16.2 oz and a standard deviation of 0.3 oz. Find the probability that a customer buys

a. one bottle and the bottle contains more than 16 oz.

b.four bottles and the mean of the four is more than 16 oz .

Let X = amount of beer in a bottle.

a.

b.

P X PX

( ).

..

.

16

16 20 3

16 16 20 3

P Z 23 0 2487 0 5 0 7487. . .

P X P X( ) ..

..

16 162

034

16 16203

4

P Z 43 0 4082 0 5 0 9052. . .

Suppose you randomly selected 36 bottles and, after carefully measuring the amount of beer each contains, you determine that the mean amount for the sample is less than 16 oz. What would you conclude? Why?

P X P X( ) ..

..

16 162

0 336

16 1620 3

36

P Z 4 0

Inference-Confidence Intervals

Let X be a random variable with mean and standard deviation .

Suppose that X is normally distributed OR the

a sample is large (n > 30), then is at least

approximately normal with mean

and standard deviation

X

X

X n

A. Logic

Mean,Std. dev.,

Distribution of

0

0.1

0.2

0.3

0.4

x 2 x 3 x x 2 x 3 x

x

X

A. Logic

Mean,Std. dev.,

Distribution of

0

0.1

0.2

0.3

0.4

x x

x

X

A. Logic

Mean,Std. dev.,

Distribution of

0

0.1

0.2

0.3

0.4

x x

x

X

0.6834

A. Logic

Mean,Std. dev.,

Distribution of

0

0.1

0.2

0.3

0.4

2 x 2 x

x

X

A. Logic

Mean,Std. dev.,

Distribution of

0

0.1

0.2

0.3

0.4

2 x 2 x

x

X

0.9544

A. Logic

Mean,Std. dev.,

Distribution of

0

0.1

0.2

0.3

0.4

3 x 3 x

x

X

A. Logic

Mean,Std. dev.,

Distribution of

0

0.1

0.2

0.3

0.4

3 x 3 x

x

X

0.9974

A. Logic

Mean,Std. dev.,

Distribution of

0

0.1

0.2

0.3

0.4

x

X

Area = 1 -

Area = /2Area = /2

z2

z2

Confidence Interval for ( known)(when the Central Limit Theorem

applies)A (1 - )100% confidence interval for is

given by

= = x zn

2

x zX

2( )

x zn

x zn

2 2,

Student’s t Distributions(for 1 and 30 degrees of freedom)

1 30

-6 -4 -2 0 2 4 6

Deg. of freedom130

Student's t Distribution

x

dens

ity

-6 -4 -2 0 2 4 60

0.1

0.2

0.3

0.4

The Distribution of when is unknown

If X is normally distributed with mean then the

Studentized Random Variable

has a t Distribution with n - 1 degrees of freedom.

T XS

n

X

Student's t Distribution

x

dens

ity

-4 -3 -2 -1 0 1 2 3 40

0.1

0.2

0.3

0.4

df =

5Area = 1 -

Area = Area =

t2

t2

Confidence Interval for unknown)

(when X is normal or n > 30)A (1 - )100% confidence interval for is

given by

x t sn

2

Example

The general manager of a fleet of taxis surveys taxi drivers to determine the number of miles traveled by a total of 41 randomly selected customers.

If = 7.7 miles and s = 2.93 miles, estimate the mean distance traveled with 95% confidence.

x

Solution

(1 - )100% = 95%, therefore, 1 - = 0.95, so = 0.05 (and /2 = 0.025)

Since n = 41, we have n - 1 = 40 degrees of freedom.

The critical value is , so a 95% CI for the mean distance traveled is given by

or (6.78, 8.62)

t tn2

1 0 025 40 2021, , ..

770 2021 29341

770 092. . . . .

Hypothesis Tests for( Known)

Assumptions:

• X has mean • X is normally distributed OR the sample is

large, i.e., n > 30

Hypothesis Testing: Tests for the Population Mean Assumptions: X is normal or n > 30, is known Steps: 1. Identify the Hypotheses (the competing claims). Null Hypothesis, HO - often the claim of no

difference or no change. (Includes =) (Note: We always test HO. It is the defendant in our trial.) Alternative Hypothesis, HA - The competing

claim. (Note: We will identify tests as left-tailed, right-tailed, or two-tailed based upon HA.)

2. Select , the “significance level of the test,” based upon the consequence of making the error of incorrectly rejecting HO when in fact it’s True.

3. Draw a picture that sums up the test. 4. Divide the picture into regions, rejection (or

“critical”) vs. acceptance and use a table or Excel to find the z value(s) separating the regions. (These are the critical values.)

5. Take an SRS and calculate the Test Statistic,

zx

n

.

6. Reject HO if z lies in the critical region; otherwise

accept (or “fail to reject”) HO.

Hypothesis Testing: Tests for the Population Mean

Assumption: X is normal or n > 30

Steps: 1. Identify the Hypotheses (the competing claims). Null Hypothesis, HO - often the claim of no

difference or no change. (Includes =) (Note: We always test HO. It is the defendant in our trial.) Alternative Hypothesis, HA - The competing

claim. (Note: We will identify tests as left-tailed, right-tailed, or two-tailed based upon HA.)

2. Select , the “significance level of the test,” based upon the consequence of making the error of incorrectly rejecting HO when in fact it’s True.

3. Draw a picture which sums up the test.

4. Divide the picture into regions, rejection (or “critical”) vs. acceptance and use a table or Excel to find the t value(s) separating the regions. (These are the critical values.)

5. Take an SRS and calculate the Test Statistic,

t xs

n

6. Reject HO if t lies in the critical region; otherwise accept (or “fail to reject”) HO.

Example

You own a factory producing sulfuric acid.

The current output = 8,200 liters/hour,

normally distributed. To test a new process, 16

hours of output are obtained with the following

results: and

Can we conclude that the new process is less

efficient than the current process?

x 8110, s 2705.

P - Values (Probability Values)

Definition - The p-value is the smallest significance level at which you would reject Ho. (the p-value represents a tail probability.)

Using p-values in Hypothesis Tests:

• If p-value < , then Reject Ho

• If p-value >, thenaccept (fail to reject) Ho

We reject Ho for small p-values!