test for goodness of fit. the math department at a community college offers 3 classes that satisfy...
TRANSCRIPT
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*.
โข 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 ฮฑ = .05.
Option to work alone and check your answer
If youโd like to try this problem on your own and just check your answer when youโre done go ahead.
When youโre ready to check your answer click on the genius to the right.
If youโd 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)
Itโs 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, weโll 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, weโll 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, weโll 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 รท 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 รท 3 = 360
Step 1: State the hypotheses and identify the claim (if there is one).
The claim is that the studentshave no preference---thereโs no math symbol for this, so weโll 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 ฮฑ =.05.
5.991
The critical value is 5.991.
Adding the critical value to the pictureLetโs add this to the picture.
.05
5.991
Step 3:
Calculate the test value.
Formula for the test value
ฮง 2=โ (๐โ๐ธ )2
๐ธ
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
ฮง 2=โ (๐โ๐ธ )2
๐ธ
ยฟ ยฟโ๐ถ๐๐๐๐๐๐ ๐ด๐๐๐๐๐๐
+ ยฟโ๐๐ก๐๐ก๐๐ ๐ก๐๐๐
+ ยฟโ๐น๐๐๐๐ก๐ h๐๐๐ก
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
ฮง 2=โ (๐โ๐ธ )2
๐ธ
ยฟ(354โ360 )2
360โ๐ถ๐๐๐๐๐๐ ๐ด๐๐๐๐๐๐
+ ยฟโ๐๐ก๐๐ก๐๐ ๐ก๐๐๐
+ ยฟโ๐น๐๐๐๐ก๐ h๐๐๐ก
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
ฮง 2=โ (๐โ๐ธ )2
๐ธ
ยฟ(354โ360 )2
360โ๐ถ๐๐๐๐๐๐ ๐ด๐๐๐๐๐๐
+(480โ360 )2
360โ๐๐ก๐๐ก๐๐ ๐ก๐๐๐
+ ยฟโ๐น๐๐๐๐ก๐ h๐๐๐ก
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
ฮง 2=โ (๐โ๐ธ )2
๐ธ
ยฟ(354โ360 )2
360โ๐ถ๐๐๐๐๐๐ ๐ด๐๐๐๐๐๐
+(480โ360 )2
360โ๐๐ก๐๐ก๐๐ ๐ก๐๐๐
+(246โ360 )2
360โ๐น๐๐๐๐ก๐ h๐๐๐ก
Final calculation of test value
ฮง 2=โ (๐โ๐ธ )2
๐ธ
ยฟ(354โ360 )2
360โ๐ถ๐๐๐๐๐๐ ๐ด๐๐๐๐๐๐
+(480โ360 )2
360โ๐๐ก๐๐ก๐๐ ๐ก๐๐๐
+(246โ360 )2
360โ๐น๐๐๐๐ก๐ h๐๐๐ก
ยฟ76.2
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
๐ป0
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.
Hereโs a quick summary โฆ
SummaryEach click will show you one step. Step (*) is broken up into two clicks.
Step 1 ๐ป0 :ยฟ
h๐ ๐๐ ๐ก๐ข๐๐๐๐ก๐ h๐๐ฃ๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐ h๐ก ๐h๐ก ๐๐๐๐๐๐๐ ๐ ๐๐ .(๐๐๐๐๐)
ยฟ ยฟ
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