Elizabeth Gutierrez PBI- Indoor Air Quality Project Wednesday ... · Elizabeth Gutierrez PBI-...

Elizabeth Gutierrez PBI- Indoor Air Quality Project

Wednesday & Thursday; Week 4

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PBI: PBI: PBI: PBI: INDOOR INDOOR INDOOR INDOOR AIR QUALITY PROJECTAIR QUALITY PROJECTAIR QUALITY PROJECTAIR QUALITY PROJECT AUTHORSAUTHORSAUTHORSAUTHORS’ NAMES: NAMES: NAMES: NAMES: Liz Gutierrez

TITLE OF THE LESSON: TITLE OF THE LESSON: TITLE OF THE LESSON: TITLE OF THE LESSON: Survey Analysis, Chi Square, and Scientific Paper.

TECHNOLOGY LESSONTECHNOLOGY LESSONTECHNOLOGY LESSONTECHNOLOGY LESSON: : : : No

DATE OF LESSON: DATE OF LESSON: DATE OF LESSON: DATE OF LESSON: Wednesday & Thursday; Week 4

LENGTH OF LESSON: LENGTH OF LESSON: LENGTH OF LESSON: LENGTH OF LESSON: 90 minutes

NAME OF COURSE: NAME OF COURSE: NAME OF COURSE: NAME OF COURSE: PBI-High School Classroom

TEKS ADDRESSED: TEKS ADDRESSED: TEKS ADDRESSED: TEKS ADDRESSED:

TEKS: 111.23 (11) Probability and statistics. The student understands that the way a set of data is displayed influences its interpretation. The student is expected to:

(A) select and use an appropriate representation for presenting and displaying relationships among collected data, including line plot, line graph, bar graph, stem and leaf plot, circle graph, and Venn diagrams, and justify the selection; and (B) Make inferences and convincing arguments based on an analysis of given or collected data.

(12) Probability and statistics. The student uses measures of central tendency and variability to describe a set of data. The student is expected to:

(A) Describe a set of data using mean, median, mode, and range; and (B) Choose among mean, median, mode, or range to describe a set of data and justify the choice for a particular situation.

PERFORMANCE OBJECTIVES: PERFORMANCE OBJECTIVES: PERFORMANCE OBJECTIVES: PERFORMANCE OBJECTIVES:



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SWBAT… Correctly perform Chi-Square on Survey results and write up a mini-scientific paper to display/describe results. RESOURCES: RESOURCES: RESOURCES: RESOURCES: None

SAFETY CONSIDERATIONS: SAFETY CONSIDERATIONS: SAFETY CONSIDERATIONS: SAFETY CONSIDERATIONS: None

SUPLEMENTARY MATERIALS, HANDOUTS: SUPLEMENTARY MATERIALS, HANDOUTS: SUPLEMENTARY MATERIALS, HANDOUTS: SUPLEMENTARY MATERIALS, HANDOUTS: • Chi Square Worksheet… similar to survey data with different data per

group. o (3 options calculated)

• Chi Square Chart ENGAGEMENTENGAGEMENTENGAGEMENTENGAGEMENT Time: Time: Time: Time: 15 minutes15 minutes15 minutes15 minutes What the TeacWhat the TeacWhat the TeacWhat the Teacher Will Doher Will Doher Will Doher Will Do Probing QuestionsProbing QuestionsProbing QuestionsProbing Questions Student ResponsesStudent ResponsesStudent ResponsesStudent Responses

Potential MisconceptionsPotential MisconceptionsPotential MisconceptionsPotential Misconceptions Say:

We have the data from

the survey and we analyzed it by looking and

calculating the what?

• The Mean • The Mode

Are we done? Is that all we wanted to know? What

do we do with the data now?

• Yes, we’re done • Analyze it again • Analyze it better

Why do you think we are

done or not done? We’re done because we

already know which option



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was chosen the most by looking at the mean and

the mode!

We’re not done, that didn’t seem like enough.

Well let’s see, what did we do before we

conducted the surveys?

• We hypothesized the results

• We created the survey

Yes! We created the survey and we

hypothesized the results. How did those hypothesis

statements begin?

• -------- option will be chosen the most out of the 3 because….

Right so we hypothesized that ONE of the options would be chosen more than the others do we

know that yet?

Yes. The mean and the mode told us that.

Yes, absolutely the mean and the mode both tell us which option was chosen

the most. But is there

• I don’t know • Yes • No



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some kind of test that we can conduct to see if our

hypothesis was statistically correct?

If yes… what test is that? • I don’t know

EXPLORATIONEXPLORATIONEXPLORATIONEXPLORATION Time: Time: Time: Time: 15 minutes15 minutes15 minutes15 minutes What the Teacher Will DoWhat the Teacher Will DoWhat the Teacher Will DoWhat the Teacher Will Do Probing Probing Probing Probing QuestionsQuestionsQuestionsQuestions Student ResponsesStudent ResponsesStudent ResponsesStudent Responses

Potential MisconceptionsPotential MisconceptionsPotential MisconceptionsPotential Misconceptions As a class:

Work through a quick and straightforward set of data in order to introduce the

definition of the Chi Square Test

That test is the Chi Chi Chi Chi Square Test.Square Test.Square Test.Square Test. The Chi-Square test is a statistical test which computes the probability that there is no significant difference between the expected frequency of an occurrence with the

What do you think that means?

• I don’t know • It’s going to

calculate the expected frequencies of something….



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observed frequency of that occurrence. Again… The Chi-Square test is a statistical test which computes the probability that there is no significant difference between the expected frequency of an occurrence with the observed frequency of that occurrence.

What do you think it means to compute the

probability that THERE IS NO SIGNIFICANT

DIFFERENCE between the EXPECTED frequency of an occurrence with the OBSERVED frequency of

that occurrence?

• It’s going to calculate the difference between our hypothesized results and the results we actually got.

Can anybody think of something that has an

easily discernable expected and observable

frequency?

• Tossing a coin, has 50% expected and observed frequency because there are only 2 possibilities…

Just to review… Where does our data come from?

• The results from our surveys

So what do you think our OBSERVED frequencies

are?

• The results from our survey, so the frequency in which each option was



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chosen

So what do you think our EXPECTED

FREQUENCY is?

• Our hypothesis about that one of the options would be chosen more than the others.

Say… Although our hypothesis consists of which option WE expected to be chosen the most the Chi Square Test is a tests of FAIRNESS…

In other words, in an ideal and fair world how often

would each option be chosen?

• They would all be chosen the same amount of times… they’d be equal

Right! If the chi square test, tests for fairness than the test has to know what would be fair…. Now, for our surveys, what do you think our EXPECTED FREQUENCIES WOULD BE?

• 25% for each option, which means each option would have been chosen 25 times in our survey



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How did you come up with those numbers?

Come up to the board to

show the class.

Since the chi square is a test of fairness, and we said that in order for the test to know what is fair,

we must tell it…

I calculated the numbers as follows…

We surveyed 100 people and had 4 options. So in order for it to be fair each option would be chosen

the same number of times…. So each option

would be chosen 25 times….

25(%) +25(%) +25(%) +25(%) = 100(%)

Good, so in order to calculate your survey

statistics… what is your EXPECTED

FREQUENCY?

25 people or 25%.

And your OBSERVED FREQUENCY?

That’s the percentage that you actually got from your

survey.



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EXPLANATIONEXPLANATIONEXPLANATIONEXPLANATION Time: Time: Time: Time: 30 m30 m30 m30 minutesinutesinutesinutes What the Teacher Will DoWhat the Teacher Will DoWhat the Teacher Will DoWhat the Teacher Will Do Probing QuestionsProbing QuestionsProbing QuestionsProbing Questions Student Student Student Student ResponseResponseResponseResponse

Potential MisconceptionsPotential MisconceptionsPotential MisconceptionsPotential Misconceptions Divide students into

groups and Pass out Chi Square worksheets.

Students will work in their groups to calculate the

Chi Square of their group data… each group will have different data all

probing questions will be directed to specific data…

probing questions referring to specific data is for illustration purposes.

Students will present their conclusions to the class

for discussion.

Say… Now in order for you to

get more familiar with the Chi Square Test we’re going to work through a

problem together.



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Briefly introduce the problem from the

worksheet.

There is a river with 3 large species of fish

gumpies, sticklebarbs, and spotheads, and we want to perform a Chi Square test to see the

frequency in which they inhabit the river.

In order to conduct this test a random sample of 300 fish was taken from

the river. How many species of fish

are we looking at?

So what is our EXPECTED

FREQUENCY for each fish?

• 33.3%

How did you calculate that?

• Since there are only 3 types of fish… and our EXPECTED FREQUENCY is



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that each fish will be found at the same frequency we have…

• 33.3% +33.3%+ 33.3%= 99.9%

So how many fish are EXPECTE from each

type?

• 33 • 100

Why do you say 33? Because that’s our

frequency…

So 33 is 33% of a sample of 300 fish?

No

Why do you say 100? Because 100 is 33.3% of 300.

Of the sample of 300 fish:

89 Gupies 120 Sticklebarbs

91 Spotheads Were retrieved.

(Each group will have

different data from different “lakes”)



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So at what frequency were the Gupies found? And how do we calculate

the frequency?

We calculate the frequency by calculating

the average. So we take 89/300 and

we get that the frequency at which Gupies were

found is ~29.7% At what frequency were

the Sticklebarbs found in the river?

~40.0%

And the Spotheads? ~30.3%

The question of statistical significance in this situation is of the

observed frequency pattern of 89/120/91 versus the expected frequency pattern of

100/100/100. And the first step in answering the question is to devise a way for measuring the

degree to which the two patterns differ from each

other.



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How can we calculate the difference between the EXPECTED and the OBSERVED frequencies?

I don’t know.

How do we calculate the DIFFERENCE between any two things?

By subtracting.

A straightforward way of going about this would be

to take, for each of the three categories, the

difference between the observed frequency and the expected frequency,

and then divide that difference by the expected

frequency.

Why would we divide the difference by the expected frequency?

• In order to take the average again

• To compare the difference by the expected frequency

The outcome of this operation would be a

measure of the proportionate amount by

which each observed frequency deviates from

its corresponding expected frequency.



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Go ahead and take a few minutes to calculate the differences between the observed and expected

frequencies divided by the expected frequencies...

(Allow students time to calculate proportions)

Now let’s talk about our findings.

How does the frequency of Gupies in the sample differ from the expected

frequency?

• The observed population of Gupies is 11% smaller than the expected frequency.

How does the frequency

of Sticklebarbs in the sample differ from the expected frequency?

• The population of observed Sticklebarbs is 20% greater than the expected frequency.

And the Spotheads? • The population of Spotheads is 9% smaller than the expected



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frequency.

The advantage of this procedure is that you will find it easy to know what

is being measured.

Its limitation is that the proportionate differences measured for the several

categories—in the present example,

—.11, +.20, and —.09— will always sum to zero

and so will not be able to provide a measure of how much the observed and

expected patterns of frequencies differ from

each other overall.

How do you think we can overcome this problem?

• Squaring the difference.

How would squaring the

difference take care of the problem?

Because when we square a number we get rid of any negative numbers so our difference would no



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longer = 0.

The effect of this operation will be to get rid

of the minus signs and thus provide a set of

measures whose sum will reflect the aggregate

degree of difference that actually exists between

the observed and expected patterns of frequencies. For the

present example, where the patterns are 89/120/91

and 100/100/100.

Go ahead and take some time to redo your

calculations, but this time square the difference!

What was the squared difference for the Gupies?

1.21

Sticklebarbs? 4.0



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Spotheads? .81

And the Sum of the squared differences

comes out to?

6. 02

Briefly talk about Chi Square Notation:

= (OOOO—EEEE)2

EEEE

= 1.21 + 4.0 + .81

= 6.02

In order to finish up our calculations we must

determine the Degrees of freedom, dfdfdfdf, is simply an index of the amount of

random variability, mere chance coincidence that

can be present in a particular situation. Its

closest literal translation would be something along

the line of "degrees of arbitrariness."



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In our fish example the number of categorical

cells is three; hence dfdfdfdf=2. Now look at the chart on

the last page of your packet and write a quick paragraph depicting what you conclude from the fish

example, based on our Chi Square result of 6.02

with df=2.

Discuss within your survey groups and prepare to

present your lakes findings with the class.

ELABORATIONELABORATIONELABORATIONELABORATION Time: Time: Time: Time: 15 mi15 mi15 mi15 minutesnutesnutesnutes What the Teacher Will DoWhat the Teacher Will DoWhat the Teacher Will DoWhat the Teacher Will Do Probing QuestionsProbing QuestionsProbing QuestionsProbing Questions Student ResponsesStudent ResponsesStudent ResponsesStudent Responses

Potential MisconceptionsPotential MisconceptionsPotential MisconceptionsPotential Misconceptions Students will use Chi

Square test to conduct statistical analysis on their

survey data.

Hold individual workshops to help students with their



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calculations or check their finished calculations. Allow students time to

finish mini-scientific papers.

EVALUATIONEVALUATIONEVALUATIONEVALUATION THURSDAY/FRIDAYTHURSDAY/FRIDAYTHURSDAY/FRIDAYTHURSDAY/FRIDAY

What the Teacher Will DoWhat the Teacher Will DoWhat the Teacher Will DoWhat the Teacher Will Do Probing QuestionsProbing QuestionsProbing QuestionsProbing Questions Student ResponsesStudent ResponsesStudent ResponsesStudent Responses Potential MisconceptionsPotential MisconceptionsPotential MisconceptionsPotential Misconceptions

Evaluation of survey analysis/conclusions.

(Thursday)

Evaluation of scientific paper. (Friday)

(Paper)

Filled in Table from Filled in Table from Filled in Table from Filled in Table from ONE ONE ONE ONE WorksheetWorksheetWorksheetWorksheet, (each group will get different data to work , (each group will get different data to work , (each group will get different data to work , (each group will get different data to work with)with)with)with)::::

gumpies sticklebarbs spotheads Totals

ObservedObservedObservedObserved frequency of cases

89 (29.7%)

120 (40.0%)

91 (30.3%)

300



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ExpectedExpectedExpectedExpected frequency of cases (MCE)

100 (33.3%)

100 (33.3%)

100 (33.3%)

300

Results from proportions of differences over expected frequency:Results from proportions of differences over expected frequency:Results from proportions of differences over expected frequency:Results from proportions of differences over expected frequency:

gumpies: (89—100)

100

= —.11

sticklebarbs:

(120—100)

100

= +.20

spotheads: (91—100)

100

= —.09

Results after squaring the differences:Results after squaring the differences:Results after squaring the differences:Results after squaring the differences:



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Name:_______________________________________ Date:__________________

CHICHICHICHI----SQUARE TESTSQUARE TESTSQUARE TESTSQUARE TEST WORKSHEETWORKSHEETWORKSHEETWORKSHEET

The ChiChiChiChi----Square testSquare testSquare testSquare test is a statistical test which computes the probability that there is no significant difference between the expected frequency of an occurrence with the observed frequency of that occurrence. For more than a century, the three species of large fish—

gumpies, sticklebarbs, and spotheads—that are native to a certain river have been observed to co-exist in equal proportions of one-third each. But now a random sample of 300 large fish drawn from a standard fish- sampling location has turned

gumpies: (89—100)2

100

= 1.21

sticklebarbs:

(120—100)2

100

= 4.0

spotheads: (91—100)2

100

= .81

sum = 6.02



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up numbers and proportions suggesting that something has occurred to upset the natural ecology of the river. If the three fish species still inhabited the river in equal proportions, we would expect to find about 100 instances of each in a sample of size N=300; whereas what we actually observe are 89 gumpies, 120 sticklebarbs, and 91 spotheads. Fill in the table with the appropriate numbers.

Gumpies Sticklebarbs Spotheads Totals

OBSERVED frequency

________ (_____%)

________ (_____%)

________ (_____%)

________

EXPECTED frequency

_________ (_____%)

________ (_____%)

_________ (_____%)

________

Chi Square Worksheet Continued...



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Proportion of Differences:Proportion of Differences:Proportion of Differences:Proportion of Differences:

Gupies:Gupies:Gupies:Gupies:

observed frequency—expected frequency

expected frequency

Sticklebarbs:Sticklebarbs:Sticklebarbs:Sticklebarbs:


expected frequency

Spotheads:Spotheads:Spotheads:Spotheads:


expected frequency

Sum of the Differences:Sum of the Differences:Sum of the Differences:Sum of the Differences:____________________________________________________________



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Chi Square Worksheet Continued...

Proportion of Proportion of Proportion of Proportion of Squared Differences:Squared Differences:Squared Differences:Squared Differences: Gupies:Gupies:Gupies:Gupies:

(observed frequency—expected frequency)2

expected frequency

Sticklebarbs:Sticklebarbs:Sticklebarbs:Sticklebarbs:


expected frequency

Spotheads:Spotheads:Spotheads:Spotheads:




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expected frequency

Sum of Squared Differences:_________Sum of Squared Differences:_________Sum of Squared Differences:_________Sum of Squared Differences:_____________________________ Chi Square Worksheet Continued... Partial Partial Partial Partial Table of Critical Values of ChiTable of Critical Values of ChiTable of Critical Values of ChiTable of Critical Values of Chi----SquareSquareSquareSquare

Level of Significance (non-directional test)

df

1 2

.05

3.84 5.99

.025

5.02 7.38

.01

6.63 9.21

.005

7.88

10.60

.001

10.83 13.82



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3 4 5 — 10 11 —

7.81 9.49

11.07 —

18.31 19.68

—

9.35 11.14 12.83

— 20.48 21.92

—

11.34 13.28 15.09

— 23.21 24.73

—

12.84 14.86 16.75

— 25.19 26.76

—

16.27 18.47 20.52

— 29.59 31.26

—

Illustration for df=2

If the observed value of chi-square is:

smaller than 5.99

equal to 5.99 greater than 5.99



etc.

Then it is:

non-significant significant at the .05 level significant beyond the .05 level significant at the .025 level significant beyond the .025 level significant at the .01 level significant beyond the .01 level etc.

Elizabeth Gutierrez PBI- Indoor Air Quality Project Wednesday ... · Elizabeth Gutierrez PBI-...

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