Test for Goodness of Fit. The math department at a community college offers 3 classes that satisfy...

Test for Goodness of Fit

The Problem, slide 1The math department at a community college offers 3 classes that satisfy the math requirement for transfer in majors that do not require calculus: College Algebra, Statistics, and Finite Math*.

• At Saddleback College, we no longer offer Finite Math, as this course proved to be significantly less popular than the other two.

College Algebra

Statistics

Finite Math

The problem, slide 2The math department chair is trying to determine how many sections of each class to offer. She claims that students show no preference for which class they take; if this proves to be so she will offer equal numbers of each class.

She looks at the number of students who enrolled in each class during the previous semester.

College Algebra students

Statisticsstudents

Finite Math students

The problem, slide 3 (data)She finds the following data:

College Algebra Statistics Finite Math

# students enrolled

354 480 246

Determine whether it is reasonable to suppose students have no preference between the three classes (and thus to offer the same number of sections of each.)

Use the goodness-of-fit test with α = .05.

Option to work alone and check your answer

If you’d like to try this problem on your own and just check your answer when you’re done go ahead.

When you’re ready to check your answer click on the genius to the right.

If you’d rather work through this problem together click away from the genius or hit the space bar or forward arrow key.

Set-upThe table tells us the observed frequency.

College Algebra Statistics Finite Math

# students enrolled

354 480 246

(Observed frequency)

It’s our job to calculate the expected frequency.

Adding a row for expected frequency to the table

College Algebra Statistics Finite Math# students enrolled

(Observed frequency)

354 480 246

Expected frequency

The need to calculate the total number of students

To do this, we’ll need to calculate the total number of students enrolled in all three classes.

College Algebra Statistics Finite Math

# students enrolled

(Observed frequency)

354 480 246

Expected frequency

Calculating the total number of studentsTo do this, we’ll need to calculate the total number of students enrolled in all three classes.

College Algebra

Statistics Finite Math Total

# students enrolled

(Observed frequency)

354 480 246

Expected frequency

354480

+ 2461080

Adding the total to the tableTo do this, we’ll need to calculate the total number of students enrolled in all three classes.

College Algebra

Statistics Finite Math Total

# students enrolled

(Observed frequency)

354 480 246 1080

Expected frequency

354480

+ 2461080

Preparing to calculate expected valuesNow we can calculate the expected frequency. If the students have no preference between the three classes, we would expect the students to be equally distributed between them.

College Algebra

Statistics Finite Math Total

# students enrolled

(Observed frequency)

354 480 246 1080

Expected frequency

How to calculate the expected values

College Algebra

Statistics Finite Math Total

# students enrolled

(Observed frequency)

354 480 246 1080

Expected frequency

Divide the total number of students by 3, the number of classes they can choose from.

Calculating the expected values

College Algebra

Statistics Finite Math Total

# students enrolled

(Observed frequency)

354 480 246 1080

Expected frequency

1080 ÷ 3 = 360

Adding expected frequencies to the table

College Algebra

Statistics Finite Math Total

# students enrolled

(Observed frequency)

354 480 246 1080

Expected frequency

360 360 360

1080 ÷ 3 = 360

Step 1: State the hypotheses and identify the claim (if there is one).

The claim is that the studentshave no preference---there’s no math symbol for this, so we’ll just say it in words.

The hypotheses

The students have no preference among the three classes. (claim)

Eeny, meeny, miney, moe

AlgebraStat Finite

This is the Null since the Null always states there is no difference between things.

The students have a preference among the three classes.

We care!

Step (*)

Draw the chi-square distribution and label the area in the right tail.

Hang on!

Can we use this

distribution?

Verifying we can use the chi-square distribution

Since all the expected frequencies are at least 5, we canuse the chi-square distribution!

Adding the area to the right tail

.05

Remember, the chi-square test is always right-tailed.

(In this case, so is the bull.)

Step 2Step 2: Mark off the critical value.

.05

The critical value is the boundary of the right tail. It will go here.

Table G

Any time we use the Chi-square distribution, we need to use table G.

Remember that the degrees of freedom will be one less than the number of categories (in this case, the number of classes.)

Calculating the degrees of freedom

College Algebra

Statistics

Finite Math

Since there were 3 classes, the degrees of freedom is

3-1 = 2

Finding the critical value on Table G

So we look in the row for d.f. = 2.

And the column for α =.05.

5.991

The critical value is 5.991.

Adding the critical value to the pictureLet’s add this to the picture.

.05

5.991

Step 3:

Calculate the test value.

Formula for the test value

Χ 2=∑ (𝑂−𝐸 )2

𝐸

sum

observed frequency

expected frequency

College Algebra

Statistics Finite Math Total

# students enrolled

(Observed frequency)

354 480 246 1080

Expected frequency

360 360 360Refer back to this table to find the

observed and expected frequencies

The test value is a sum of 3 terms.

College Algebra

Statistics Finite Math Total

# students enrolled

(Observed frequency)

354 480 246 1080

Expected frequency

360 360 360

Χ 2=∑ (𝑂−𝐸 )2

𝐸

¿ ¿⏟𝐶𝑜𝑙𝑙𝑒𝑔𝑒 𝐴𝑙𝑔𝑒𝑏𝑟𝑎

+ ¿⏟𝑆𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐𝑠

+ ¿⏟𝐹𝑖𝑛𝑖𝑡𝑒 h𝑀𝑎𝑡

The term with info from college algebra

College Algebra

Statistics Finite Math Total

# students enrolled

(Observed frequency)

354 480 246 1080

Expected frequency

360 360 360

Χ 2=∑ (𝑂−𝐸 )2

𝐸

¿(354−360 )2

360⏟𝐶𝑜𝑙𝑙𝑒𝑔𝑒 𝐴𝑙𝑔𝑒𝑏𝑟𝑎

+ ¿⏟𝑆𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐𝑠

+ ¿⏟𝐹𝑖𝑛𝑖𝑡𝑒 h𝑀𝑎𝑡

Adding in the second term, corresponding to statistics

College Algebra

Statistics Finite Math Total

# students enrolled

(Observed frequency)

354 480 246 1080

Expected frequency

360 360 360

Χ 2=∑ (𝑂−𝐸 )2

𝐸

¿(354−360 )2

360⏟𝐶𝑜𝑙𝑙𝑒𝑔𝑒 𝐴𝑙𝑔𝑒𝑏𝑟𝑎

+(480−360 )2

360⏟𝑆𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐𝑠

+ ¿⏟𝐹𝑖𝑛𝑖𝑡𝑒 h𝑀𝑎𝑡

Adding in the third term, corresponding to Finite Math

College Algebra

Statistics Finite Math Total

# students enrolled

(Observed frequency)

354 480 246 1080

Expected frequency

360 360 360

Χ 2=∑ (𝑂−𝐸 )2

𝐸

¿(354−360 )2

360⏟𝐶𝑜𝑙𝑙𝑒𝑔𝑒 𝐴𝑙𝑔𝑒𝑏𝑟𝑎

+(480−360 )2

360⏟𝑆𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐𝑠

+(246−360 )2

360⏟𝐹𝑖𝑛𝑖𝑡𝑒 h𝑀𝑎𝑡

Final calculation of test value

Χ 2=∑ (𝑂−𝐸 )2

𝐸

¿(354−360 )2

360⏟𝐶𝑜𝑙𝑙𝑒𝑔𝑒 𝐴𝑙𝑔𝑒𝑏𝑟𝑎

+(480−360 )2

360⏟𝑆𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐𝑠

+(246−360 )2

360⏟𝐹𝑖𝑛𝑖𝑡𝑒 h𝑀𝑎𝑡

¿76.2

Adding the test value to the pictureNow add the test value to the picture.

.05

5.991 76.2

76.2 is (much) bigger than 5.991, so it goes to the right.

Step 4: Make the decision.

.05

5.991 76.2

The test value is in the critical region. Reject the Null!

The Null expresses disappointment

𝐻0

RATS! Rejected again!

We rats had nothing to do with it.

Step 5: Answer the question in plain English.

There is enough evidence to reject the claim that students have no preference among the three math classes.

Here’s a quick summary …

SummaryEach click will show you one step. Step (*) is broken up into two clicks.

Step 1 𝐻0 :¿

h𝑇 𝑒𝑠𝑡𝑢𝑑𝑒𝑛𝑡𝑠 h𝑎𝑣𝑒𝑛𝑜𝑝𝑟𝑒𝑓𝑒𝑟𝑒𝑛𝑐𝑒𝑎𝑚𝑜𝑛𝑔 h𝑡 𝑒h𝑡 𝑟𝑒𝑒𝑐𝑙𝑎𝑠𝑠𝑒𝑠 .(𝑐𝑙𝑎𝑖𝑚)

¿ ¿

Step (*).05

5.991Step 2 76.2 Step 3

Step 4 Reject the Null.

Step 5 There is enough evidence to reject the claim that the students have no preference among the three classes.

And there was much rejoicing