Homework Assignment

1 / 66

# Homework Assignment - PowerPoint PPT Presentation

Homework Assignment. 10.17, 10.19, 10.23, 10.29, 10.32 Due in Class Dec 1. Last Time: Finished Contingency Tables Reviewed Basics on Linear Regression. Suppose I sample n many people:. How many observations do I expect to get in cell (i,j)?

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Homework Assignment' - elvina

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### Homework Assignment

10.17, 10.19, 10.23, 10.29, 10.32

Due in Class Dec 1

### Last Time:Finished Contingency TablesReviewed Basics on Linear Regression

Suppose I sample n many people:
• How many observations do I expect to get in cell (i,j)?
• If the Null Hypothesis holds, i.e., if the columns and rows are independent, then I expect the number of observations in cell (i,j) to be

How to compare??

Both are unknown!

Suppose I sample n many people:

How to compare??

Both are unknown!

Slight Change of Notation

Homogeneity of

parallel samples

Example

255

32

FIXED

Equivalent Equations

More Generally:

The rest is the same as in previous scenario,

i.e., we get the same Chi-square again.

The Square of a Standard Normal Random Variableis a Chi-Square Random Variablewith 1 degree of freedom.

### Today:From descriptive to inference statistics…Estimation and Hypothesis Testingfor Linear Regression

Statistical Inference (for a single variable)

Estimation: (Confidence Intervals)

Point

estimate

critical

value

Std. dev. of

point estimate

±

·

For instance: Confidence Interval for the mean:

±

·

Example 20 kindergarteners

1pt

2pts

3pts

“Popularity Score” = Average Score

Example 20 kindergarteners

“Social Competence Score”

Example 20 kindergarteners

“Popularity Score”

“Social Competence Score”

Statistical Inference (for two variables)

Example:

Children X: Popularity,

Y: Social competence

Goal: Explain (linear) relationship between X and Y

Y

1

X

Simple (linear) regression

Explain the (linear) relationship (if it exists)

between random variable X and random variable Y.

Four assumptions

4

For different

values of X,

the error terms

are uncorrelated

1

3

2

The error term

is a normally

distributed

random variable

No matter what

value X takes,

the error has

a mean of zero

Y

Error

X

Y

X

Y

X

We will sample data to estimate the parameters.

This leads to point estimates, confidence intervals

and hypothesis testing for each parameter,

in addition to a general test of the model as a whole.

Parameter Estimates:

Degrees of freedom

loose 1 df for X

loose 1 df for Y

Sampling Distributions

Don’t

Know!

Hypothesis Testing

Remember the general rule for Confidence Intervals:

Point

estimate

critical

value

Std. dev. of

point estimate

±

·

Confidence Intervals for Intercept and Slope

Point

estimate

critical

value

Std. dev. of

point estimate

±

·

Hypothesis Test on Slope

If p-value of the standardized statistic <  then reject H0

and conclude that there is indeed a linear relationship

p-value

< .001

Analysis of Variancefor Regression

How much

y differs

from mean

How much

predicted y

differs

from mean

residual

/

error

Involves

only data

Analysis of Variance for Regression

Sum Squares

Total

(SST)

Sum Squares

Error

(SSE)

Sum Squares

Model

(SSM)

Variability

in the

Data

Variability

unaccounted

for

Variability

accounted for

by the Model

Degrees of Freedom

DFT = n-1

Degrees of Freedom

DFE=n-2

Degrees of Freedom

DFM = 1

=

+

Analysis of Variance for Regression

Degrees of Freedom

DFT = n-1

Degrees of Freedom

DFE=n-2

Degrees of Freedom

DFM = 1

Mean Squares

Total

(MST)

Mean Squares

Error

(MSE)

Mean Squares

Model

(MSM)

Hypothesis Testingand the ANOVA Table

Mean Squares

Error

(MSE)

It can be shown that

the Null Hypothesis

implies that

MSM is also an unbiased

estimator of

Analysis of Variance Table

Table E

df in the numerator

df in the denominator

Analysis of Variance Table

p

p-value here: <.001

The Square of a t Random Variable with n-2 degrees of freedomis an F Random Variablewith 1 degree of freedom in the numerator andwith n-2 degrees of freedom in thedenominator.
Slight shift of perspective
• Treated the X variable as if it were fixed.
• Now, let’s think of X and Y as both being random variables (jointly distributed).
• Let’s assume they are both normally distributed (i.e. they are “jointly normal”)
• We can define a population correlation ρ
• We can use the sample correlation r as an estimate of ρ.

Question: Can we / can we not draw a line close to the data?

Answer: No, unless we provide sufficient evidence that we can.

loose 1 df for X

loose 1 df for Y

If p-value of the standardized statistic <  then reject H0

and conclude that there is indeed a linear relationship

One more thing…

Percent of Variance

explained by the model