Ap Bio Lab 7

AP Bio Lab 7: Fly Lab Max Oltersdorf With Emma Horowitz and Erica Lee AP Bio with Michael Santos, 4th Period

AP Bio Lab 7 11/13/2009 - 11/17/2009 Part 1 with Emma Horowitz Part 2 with Erica Lee AP Bio with Michael Santos

Part 1: “Goodness Fit” The Fun way Background: In order to do this lab, we, the experimenter, need to know the difference between hypothesis and null hypothesis. The difference, with examples pertaining to this lab, is as follows: Hypothesis: “Differences in measured ratios are not due to chance (but perhaps do to faulty workmanship).” What is the hypothesis? The hypothesis is what the experimenter believes and is attempting to test. Null Hypothesis: “If the Mars Co. M&M sorters are doing their job correctly, then there should be no difference in M&M ratios between actual store-bought bags of M&Ms and what the Mars Co. claims are the actual ratios and any differences found are due to chance.“ What is the Null Hypothesis? The null hypothesis is a hypothesis that is the reverse of what we actually believe. At the end of the lab, we will either accept or reject the null hypothesis. If we accept the null hypothesis then any differences we do see are by chance. If we reject the Null Hypothesis, then we will know that the differences are not due to chance but rather foul play. We will determine if this Null hypothesis is valid or viable. 1

AP Bio Lab 7: Fly Lab Max Oltersdorf With Emma Horowitz and Erica Lee AP Bio with Michael Santos, 4th Period

Background Continued Chi Square: Chi-square is a quantitative measure used to determine whether a relationship exists between two categorical variables. Purpose: •

The purpose of this lab is to practice using chi square analysis with M&M personal bags and distinguish between null hypothesis and hypothesis.

Objectives: • • •

The first objective is to learn how to use deviation to determine the probability of getting a defined percent away from expected data. The second objective is to determine the probability that the difference between the observed and expected values occur simply by chance. The third objective is to accept or reject the null hypothesis.

Procedure: 1. The experimenter should obtain a 1.69 Oz bag of regular M&Ms at his or her own expense. 2. The experimenter separates the M&Ms into colors. 3. The experimenter records the number of M&Ms for each color in Chart 7.2, Row 1. 4. Calculate deviation, deviation squared, and chi square as outlined in calculations. Record in Chart 7.3. 5. The null hypothesis now should be accepted or rejected. Refer to Chart 7.4 to decide whether to accept or reject it. *For flow grams, refer to working lab notebook, page 78*

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AP Bio Lab 7: Fly Lab Max Oltersdorf With Emma Horowitz and Erica Lee AP Bio with Michael Santos, 4th Period

Data: Chart 7.1: Statistics from the Mars company on expected M&M color

Brown Yellow Red Blue Orange Green Chart 7.2: Actual occurrences of M&M Color for class (Group data on row 1) (All data in M&Ms)

Group 1 2 3 4 5 6 7 8 9 10 11 12 13 Total

Color Green

Yellow

Red

Orange

Blue

Brown

7 9 11 17 8 8 13 13 10 19 8 12 12 147

6 8 5 10 4 12 5 11 8 4 4 11 7 95

7 3 15 8 6 8 9 8 7 12 8 8 9 108

18 13 13 5 8 10 13 10 16 28 14 8 11 167

12 14 9 11 16 9 8 7 3 20 13 12 15 149

4 7 2 4 12 8 8 9 9 14 8 5 2 92

*See Calculations for calculations*

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AP Bio Lab 7: Fly Lab Max Oltersdorf With Emma Horowitz and Erica Lee AP Bio with Michael Santos, 4th Period

Data Continued

Number of M&Ms recorded as a group

Number of M&Ms

Observations: Orange had the most overall, then blue, followed closely by green, then red yellow and brown. Blue and orange had the most, as they should. However, according to our expected data, blue should have more than orange, and instead it was the other way around. Analysis: We do not yet know whether the numbers were off enough for us to accept the null hypothesis. However, the numbers that we received vary pretty Colors of M&Ms significantly from the expected. This is probably a result of one group not using the 1.69 oz bag, but could also stem from bad workmanship at the factory, lack of quality control in regards to M&M dispersal ratios or hungry students.

Data Continued

Chart 7.3 Chart regarding calculations of recorded data – Class data

Green

Yellow

Color Red Orang

Blue

Brown Total 4

AP Bio Lab 7: Fly Lab Max Oltersdorf With Emma Horowitz and Erica Lee AP Bio with Michael Santos, 4th Period

Observ ed Expect ed (e) Deviati on (d) Deviati on square d (d2) (d2/e)

147 M&Ms 121 M&Ms

95 M&Ms 106 M&Ms

108 M&Ms 99 M&Ms

e 167 M&Ms 152 M&Ms

26

11

9

15

33

7

0

676

121

81

225

1089

49

0

5.59

1.14

.819

1.48

5.98

.49

0

χ2 = ∑(d2/e)

149 M&Ms 182 M&Ms

92 M&Ms 99 M&Ms

759 M&Ms 759 M&Ms

15.49 9

*See Calculations for calculations* Observations: As we go down the chart and do more equations, a tiny little difference in the deviation can end up being a fairly sizable change by the time we get down to (d2/e). For example, brown and green M&Ms have a difference in deviation of 19, or about 3.2 times the value of brown, but by the time we get down to the bottom, green is more than 11 times what brown is. Analysis: The above fact accounts for the especially high Chi Square value. Because a little increase in deviation produces a large (d2/e) value, the sum of all of the (d2/e) values will be a lot higher than if we had just added the relative deviations in the same way.

Chart 7.4: Should I accept or reject the null hypothesis?

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AP Bio Lab 7: Fly Lab Max Oltersdorf With Emma Horowitz and Erica Lee AP Bio with Michael Santos, 4th Period

Accept the null hypothesis←

Reject the null → hypothesis

Degrees of Freedo m 1

Probability 0.90

0.50

0.25

0.10

0.05

0.01

0.016

0.46

1.32

2.71

3.84

6.64

2

0.21

1.39

2.77

4.61

5.99

9.21

3

0.58

2.37

4.11

6.25

7.82

11.35

4

1.06

3.36

5.39

7.78

9.49

13.28

5

1.61

4.35

6.63

9.24

11.07

15.09

Degrees of freedom = number of independent pieces of information – number of steps = 6 M&M colors – 1 step = 5 degrees of freedom

Observations: This graph is directly out of the lab. It tells us whether we should accept or reject our null hypothesis based on the Chi Square value. Analysis: Our Chi Square value is 15.499 and our degrees of freedom is 5, making it off the chart on the right-hand side, implying that we should reject the null hypothesis.

Calculations: Chart 7.2 6

AP Bio Lab 7: Fly Lab Max Oltersdorf With Emma Horowitz and Erica Lee AP Bio with Michael Santos, 4th Period

Total = sum of total for green, yellow, red, orange, blue, and brown Total Green = 7+9+11+17+8+8+13+13+10+19+8+12+12 = 147 M&Ms Total Yellow = 6+8+5+10+4+12+5+11+8+4+4+11+7 = 95 M&Ms Total Red = 7+3+15+8+6+8+9+8+7+12+8+8+9 = 108 M&Ms Total Orange = 18+13+13+5+8+10+13+10+16+28+14+8+11 = 167 M&Ms Total Blue = 12+14+9+11+16+9+8+7+3+20+13+12+15 = 149 M&Ms Total Brown = 4+7+2+4+12+8+8+9+9+14+8+5+2 = 92 M&Ms

Chart 7.3 Expected (e) = (% as a decimal expected for a particular color) x 759 Expected Green = .16 x 759 M&Ms = 121 M&Ms Expected Yellow = .14 x 759 M&Ms = 106 M&Ms Expected Red = .13 x 759 M&Ms = 99 M&Ms Expected Orange = .20 x 759 M&Ms = 152 M&Ms Expected Blue = .24 x 759 M&Ms = 182 M&Ms Expected Brown = .13 x 759 M&Ms = 99 M&Ms

Calculations Continued Deviation (d) = Observed-Expected Deviation Green = 147-121 = 26 7

AP Bio Lab 7: Fly Lab Max Oltersdorf With Emma Horowitz and Erica Lee AP Bio with Michael Santos, 4th Period

Deviation Yellow =95-106 = 11 Deviation Red =108-99 = 9 Deviation Orange =167-152 = 15 Deviation Blue =149-182 = 33 Deviation Brown =92-99 = 7

Deviation Squared (d2) = Deviation2 Deviation Squared Green = 262= 676 Deviation Squared Yellow = 112=121 Deviation Squared Red = 92=81 Deviation Squared Orange = 152=225 Deviation Squared Blue = 332=1089 Deviation Squared Brown = 72=49

Calculations Continued D2/e = Deviation SquaredExpected D2/e Green = 676/121 = 5.59 D2/e Yellow = 121/106 = 1.14 8

AP Bio Lab 7: Fly Lab Max Oltersdorf With Emma Horowitz and Erica Lee AP Bio with Michael Santos, 4th Period

D2/e Red = 81/99 = .819 D2/e Orange = 225/152 = 1.48 D2/e Blue = 1089/182 = 5.98 D2/e Brown = 49/99 = .495

χ2= ∑(d2/e) = Sum of the(d2/e) for green, yellow, red, orange, blue, and brown Chi Square Total = 5.59+1.14+.819+1.48+5.63+.49 = 15.149 Chart 7.4 Degrees of freedom = number of independent pieces of information – number of steps Degrees of freedom = 6 M&M colors – 1 step = 5 degrees of freedom

Analysis 1. Based on our sample, should we accept or reject our null hypothesis? a. Based on our findings and chart 7.4, we should reject the null hypothesis, meaning that our findings conclude that the variations are not due to chance but due to manipulation or foul play or faulty workmanship.

Analysis Continued 2. If we reject the null hypothesis, what might be some possible explanations for this outcome? a. One possible explanation involves manipulation of M&M count at either the factory or in the classroom. If kids eat some of the M&Ms, it will 9

AP Bio Lab 7: Fly Lab Max Oltersdorf With Emma Horowitz and Erica Lee AP Bio with Michael Santos, 4th Period

result in a higher chi square value and cause us to reject the null hypothesis. Error could come from the fact that on our sheet, the chart in which to record data was vertical but on the board, the chart was horizontal. Maybe a group got confused and put their data the wrong way. Other errors could stem from the above mentioned snacking before recording, as well as simple arithmetic errors, rounding error with the calculations, and significant figure errors creating data that looks more accurate than it is. Conclusion The first part of lab 7 was a success, but our data was not as accurate as we would have liked. However, that did not take away from the educational value of doing this lab. During the course of this lab, I learned what chi analysis was and how to use it. I also learned how to use deviation and what that was. I came to the realization that a tiny shift in deviation can result in a huge shift in chi square. We used chi square to determine that we should reject our null hypothesis, due to the conclusion we reached that you chi square value was 15.499, well to the right of the cutoff point for five degrees of freedom. This value of 15.499 determined that the values that we saw for M&Ms were too far off to just have happened by chance and that, according to the hypothesis and null hypothesis, either the workmanship is sloppy or data was manipulated in the classroom (likely by snacking high school students!). The purpose of this lab was to practice using chi square analysis to distinguish between null hypothesis and hypothesis. Considering I did not know what a null hypothesis was, nor did I know what chi square was, I would say I accomplished that goal. Next time we do the experiment, I feel like we should discuss more what chi square is beforehand as well as what the data all means. I was a little iffy about that at the beginning. Although our data may have been a little off, myself, as well as the class learned a lot about chi square and null hypotheses, allowing us to broaden our knowledge and preparing us for the next time we have to use these skills.

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