# AP Biology Unit 0: Scientific Inquiry

The Scientific Method

• The necessary components of any experiment are a hypothesis, a control group, the independent variable, the dependent variable and data to analyze the conclusion. So basically, it all depends on the experiment what is needed for changing IV and tools to measure DV.

• Design a controlled experiment to test a fertilizer which claims that makes plants grow bigger

Be specific in the control group and controlled (fixed) variable.

The group being tested would be of one plant with the supposed growth inducing fertilizer. There would be a control group of one normal fertilizer plant. Plant without fertilizer is the control group. The controlled variables would be the same type of plant and the same plant pot and the same age and the same height. They also have to be watered at the same time and put next to each other. The plants have to be regularly measured to properly collect data.

• Controlled experiment

An experiment where the dependent variable is manipulated so that the independent variable can be compared. Every factor besides this is the same.

• Controlled variable

The item or person that is used as a constant unchanging standard to the experiment.

• What kind of data did you collect for the effect of fertilizer experiment?

The data being collected in this experiment are the measurements of the plant heights. These data samples are being compared to form an analysis.

Egg Osmosis Controlled experiment:

What could be a good hypothesis for this experiment?

If the egg is added to a hypertonic solution, then the egg's mass will decrease.

For the egg osmosis experiment,determine the IV,DV, control group, Experimental group and constant variable

The independent variable is the substance either hypertonic or hypotonic that the egg is added to. The dependent variable is the mass of the egg. The experimental group is the egg added to the hypertonic solution. The constant variables are the container that the egg is put in, and the amount of liquid put into each container. The control group is the egg in no solution or an isotonic solution.

Valid Experiments

1) A student wanted to determine if slugs preferred green leaf lettuce leaves over purple cabbage leaves for food. Pieces of both leaves were cut. One piece of each type of leaf and one slug were placed in each of ten containers. After three days, the surface area of each leaf section was measured and the results were recorded in a data table. State one reason that the results of this experiment might be considered invalid.

- This experiment may be invalid since the slugs may be different ages or different species. Although the experiment seems to be close to valid, these subtle details make a difference in determining the validity of the experiment.

2) A television commercial for a weight-loss pill claims that it has been "scientifically tested." The advertisement includes statements from 10 people who say that the pill worked for them. State two reasons why someone should question the claims made in this advertisement.

- The sample audience is too small. This should be tested on a larger group of people, maybe 500 people. - There should be data provided along with the statement that the pill worked. The weights of every participant should be recorded before and after the pill.

Chi Squared Analysis

1. A zookeeper hypothesizes that changing the intensity of the light in the primate exhibits will reduce the amount of aggression between the baboons. In exhibit A, with a lower light intensity, he observes 36 incidences of aggression over a one month period. In exhibit B, with normal lights, he observes 42 incidences of aggression. Should he support or reject his hypothesis?

·Identify Null Hypothesis

If the light intensity is changed in the primates exhibit, then the amount of aggression will not change.

·Identify Alternative Hypothesis

If the light intensity is lowered in the primates exhibit, then the amount of aggression will decrease.

·Calculate expected values:

36 + 42 = 78

78 / 2 = 39

·Make a table

 Exhibit Observed Expected Difference Difference ^2 (O-E)^2/expected A 36 39 -3 9 0.23077 B 42 39 3 9 0.23077 Totals 78 0.46154

·Calculate Chi Square value

x^2 statistic = 0.46154

·Calculate Degrees of freedom

Category - 1 = 2 - 1 = 1

·Compare your calculated value to critical value, and accept of reject null hypothesis

x^2 statistic = 0.46154

Critical value = 0.05 =>3.84

x^2(0.05) = 12.706

x^2 value is smaller than the critical value, so the null hypothesis is accepted.

Discuss the significance of rejecting or accepting null hypothesis.

Rejecting or accepting the null hypothesis is significant since it reflects upon the results that your experiment has concluded on. This states whether or not the null hypothesis accurately represents your conducted experiment.

Calculating Standard Deviation

The standard deviation is used to tell how far on average any data point is from the mean.  The smaller the standard deviation, the closer the scores are on average to the mean.  When the standard deviation is large, the scores are more widely spread out on average from the mean.

The standard deviation is calculated to find the average distance from the mean.

Practice Problem #1: Calculate the standard deviation of the following test data by hand. Use the chart below to record the steps.

Test Scores: 22, 99, 102, 33, 57, 75, 100, 81, 62, 29

Total: 660

Mean: 66                          n: 10

 Test Score(x) Difference from the mean(x –xm) (Difference from the mean) (x – xm)^2 22 22-66= -44 (-44)^2= 1946 99 33 1089 102 36 1296 33 -33 1089 57 -9 81 75 9 81 100 34 1156 81 14 196 62 -4 16 29 -37 1369 Sum of (Difference from the mean) ∑(x – xm)^28319

Sum of (Difference from the Mean) divided by degrees of freedom (n – 1 ):9 This is called variance.

∑〖(x-x ̅)〗^2/((n-1))= 8319/9 = 924.333

Final Step: Standard deviation = square root of  what you just calculated (variance).

Square root of 924.333 Standard deviation =  30.402845