Bread Mold Worksheet PDF

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Biology lab...

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Name: _Jheanelle Clarke_______

Statistical Worksheet Question of the lab: Does temperature and/or humidity effect fungal growth? Brand of Bread? Publix Bakery fresh baked bread

Based on what you know, how do you think temperature will effect fungal growth? How do you think humidity will effect fungal growth? Note these are the general hypotheses (Think science fair). They will be the ones you focus on and report in your lab report. HA1: Humidity will increase the growth of fungus on the bread. HA2: Temperature will increase the growth of fungus on the bread.

We are using two treatments and comparing them to a control group. Write out the null hypotheses. Note: this is not to report in your lab report. H01: Humidity will not increase the growth of fungus. H02: Temperature will not increase the growth of fungus.

You can use the following table as an idea for how to write your data. Date:

Zip locks (sample #) R1 R2 R3 R4 R5 Average

Room Temperature Percent cover # of squares covered 100 100 98 85 100 96.6

Refrigerated Zip locks (sample #) F1 F2 F3 F4 F5 Average

Percent cover # of squares covered 0 0 0 0 0 0

Humidity Zip locks (sample #)

Percent cover # of squares covered

H1 H2 H3 H4 H5 Average

100 90 100 100 100 98

Observations: One of the bread in the humifity condition was the first to develop mold, then on day 6, 4 out of 5 of them had mold. While, only 2 from room temp. condition had mold day 6. Record all the averages you calculated before in the table below. This will be the table you use to make your graph. The results show amount of mold growth on each bread sample and compare it to the 1

number of days that have passed. Although you are free to use any software you like to produce your graph, most students will use Excel. For the time being, draw your graph in the space below. Day

Room Temp

Refrigerated

Humidity

0

0

0

0

3

9

0

16

6

26

0

44

9

61

0

74

12

78

0

87

15

84

0

89

18

90

21

96.6

0

94

0

98

100

80 Percent Fungi Growth(%)

60

40

20

3

6

9

12

15

18

Day

Room Temp

2

Refrigerated

Humidity

21

STOP THE REST OF THE WORKSHEET SHOULD BE DONE WITH YOUR TA OR WHILE WATCHING YOUR LECTURE.

Please copy the data given to you from your lab instructor into the table below. This is the information you will be using for your stats test. You will not be taking information from any other day. Note that you have a total of 5 samples. This value is known as N. So N=5. Calculate the average by dividing the sum (calculated in the table to the left) by N. Record this value in the ´ spaces given to the left of the table. Recall that in statistics, X refers to the value and X refers the mean. Therefore the sample # will be put in for X . Refrigerated

Humidity

Sample #

Room temperature Percent cover

Percent cover

Percent cover

1 2 3 4 5

79 75 76 77 73

0 0 0 0 0

93 87 89 85

Sum ()

380

0

444

Σ X Room Temperature 5 =____76_____________

´x RoomTemperature =

Σ X Refrigerated 5 =____0_____________

90

´x Refrigerated =

´x Humidity =

Σ X Humidity 5

= _______89____________

Now square each of those values, then determine the sum. Room temperature

Refrigerated

Humidity

Sample #

(Percent cover-Mean)^2

(Percent cover-Mean)^2

(Percent cover-Mean)^2

1 2 3 4 5

9 1 0 1 9

0 0 0 0 0

16 1 4 0 16

Sum ()

20

0

37

Now calculate the variance (S2) for each treatment as indicated. Record the values in the table ´ refers the mean. Therefore below. Recall that in statistics, X refers to the value and X the sample # will be put in for X . 3

S

2 RoomTemperature

S 2 Refrigerated

S

=

2 Humidity

=

∑ ([ X RoomTemperature −X´ RoomTemperature ] 2 ) 5−1

∑ ([ X Refridgerated−X´ Refridgerated] 2) 5−1

2 ([ X Humidity− ´X Humidity] ) ∑ = 5−1

Mean ´x environment Variance S 2 environment

Room Temperature 20

0

37

5

0

9.25

0

3

2.2 Standard Deviation (square root the Variance) S √ (¿¿ 2 environment ) ¿

Refrigerated

Calculate the T-statistic for each pairwise comparison: Refridgerated−¿ x´ RoomTemperature ∨

√

s

2 Refridgerated

¿ +s2RoomTeperature 5

¿ ´x¿ t Refridgerated , R oom Temperature =¿ Humidity−¿ x´ RoomTemperature ∨ ¿ x´ ¿

√

s

2 Humidity

¿ +s2RoomTeperature 5

Humidity

-20

= 6

=

t Humidity , Room Temperature =¿ The next step is to determine whether the t statistic exceeds the critical value for that test. You need two pieces of information to determine the critical value for the T test. First, we need to decide the significance level, or α (alpha). The significance level is the probability of rejecting the null hypothesis when it is true (i.e., saying there is an effect when there actually wasn’t one). By convention, in science, we use a significance level of 5% (or 0.05). In practical terms, this means we are willing to accept a 5% risk of mistakenly concluding that a difference exists when there isn’t one. Second, we need to determine the degrees of freedom of our test, which 4

Name: _Jheanelle Clarke_______

is the number of independent observations (or pairs of observations) being compared: in this case it is N-1.Thus, your degree of freedom is 4. You now have enough information to determine the critical value for t in this experiment. To do this find the row and column with your degrees of freedom and alpha level respectively. For example, if you have 10 degrees of freedom and have decided to use α=20%=0.20, your critical value would be 1.372. Use this approach to determine the critical value for t in your experiment. Circle it on this table!

Your critical value for t is ___________2.776______________________. 5

Now compare this critical value to the test statistics you calculated above. If your t value is > than the critical value in the chart, then your p value is below 0.05, and you can reject the null hypothesis.

T Statistic for Treatments VS Room Temperature

t Refrigerated, Room Temperature : t Humidity, Room Temperature :

t ≥ critical value Thus, p-value 0.05

Indicate for each comparison if the p-value is greater or less than 0.05. T refrigerated, room temperature: __p>0.05______________________________________ T humidity, room temperature: ________p0.05), thus the null hypothesis is not rejected. Also, for Humidity and Room temperature the t-value is 6, therefore, the p-value is less than 0.05 (p...