Statistical Analysis Determining the Significance of Data Accepting or Rejecting the Null Hypothesis

Analysis of Variance (ANOVA) Test statistical method used to test general differences between two or more means. Allows you to make a general conclusion regarding your Null Hypothesis : If You Accept the Null using this test, all the means are the same or very close If you Reject the Null using this test, at least one of the means is significantly different at least one other mean.

Analysis of Variance (ANOVA) Test This statistical method provides you with a p-value (a probability). Based on the concept of sampling (gathering data on a sample that should represent the whole population). Need a cut-off point for the p-value Common “cut points”: 0.05, 0.01, .001 Most Biologists use 0.05 for a p-value A p-value < 0.05 means we are 95% or more confident our data IS significantly different, therefore we reject the Null Hypothesis. A p-value > 0.05 means our data is NOT significantly different, therefore we do not reject the Null Hypothesis.

Analysis of Variance (ANOVA) Test Let’s look at an example using Redi’s experiment which he used to disprove Spontaneous Generation. Table 1: # of flies present around Jars with different covers no cover mesh sealed Jar 1 36 30 1 Jar 2 42 45 Jar 3 38 44 Jar 4 47 2 Jar 5 33 mean St. Dev

Analysis of Variance (ANOVA) Test Finding Mean using Excel: Select the Cell next to “mean”. Type =average( select the data you want to average type ) press enter. Finding Standard Deviation (how spread out the #’s are): Select the Cell next to “St. Dev” Type =stdev( select the data you want press enter

Analysis of Variance (ANOVA) Test Enter labels for each column of data Enter data in all columns From top ribbon, select Data, then Data Analysis If your computer does not have Data Analysis, Select File, Select Options, Add-ins. Select Analysis Tool Pac, Select Go….Check Analysis Too Pac Now from the ribbon, select Data and then Data Analysis

Analysis of Variance (ANOVA) Test Select ANOVA: Single Factor Click on Input Range Icon (A) Highlight all raw data including labels in the 1st row (area in yellow) (do not include trials column) Click Input Range Icon again Check the box for Labels in First Row (B) Important: Click in the circle for Output Range (C) Click on a Cell where you want the table to appear Click OK (E) If P<0.05, then do T-tests to identify significance If P>0.05, then there is no significant difference and Null is accepted

T-Test: Statistical Significance A t-test’s indicates whether or not the difference between two groups’ means are significantly different in the population from which the groups were sampled A statistically significant t-test result is one in which a difference between two groups is unlikely to have occurred because the sample happened to be atypical. Statistical significance is determined by the size of the difference between the group averages, the sample size, and the standard deviations of the groups. For practical purposes statistical significance suggests that the two larger populations from which we sample are “actually” different.

T-Test: Statistical Significance A t-test also provides a p-value and compares an experimental group to the control group. We still use the following interpretation: A p-value < 0.05 means we are 95% or more confident our data IS significantly different, therefore we reject the Null Hypothesis. A p-value > 0.05 means our data is NOT significantly different, therefore we do not reject the Null Hypothesis. Is performed after receiving a p<0.05 in an ANOVA Test Allows one to determine which experimental group(s) is causing the significant difference

T-Test: Statistical Significance Let’s use Redi’s experiment again to practice running a T-test. Add a row and Label it “ttest” Click on the adjacent empty cell Click on Formulas in the tool bar Click on “More Functions” then “Statistical” then select T.Test from the menu bar (you should see a table similar to Table 2) For “array 1” select the first column of data to be compared For “array 2” select the second column of data to tbe compared For “tails” select 1 since we predict that one group will be higher than the other For “type” select 3 since the standard deviations are different for each group Add the information, hit return and the number generated is the P-value.

T-Test: Statistical Significance

Chi Square: Goodness of Fit Test The Chi Square Goodness of Fit Test determines if a set of data is significantly different from an expected outcome. Calculates a chi square value! Our Null says that the IV has NO effect on the DV If a Chi Square Test indicates you should reject the Null, then there IS a significant difference between the expected values and the experimental group values.

Chi Square: Goodness of Fit Test The Chi Square Table:

Chi Square: Goodness of Fit Test Degrees of Freedom (df) = # of groups minus one

Chi Square: Goodness of Fit Test Degrees of Freedom (df) = # of groups minus one If the calculated chi-square value is greater than or equal to this critical value, then the two groups ARE significantly different, and the null hypothesis is rejected. If the null hypothesis is rejected and we are 95% confident that there is significant difference. If the calculated chi-square value is less than this critical value, then the two groups are NOT significantly different, and the null hypothesis is not rejected/accepted.

Chi Square: Goodness of Fit Test Practice: Let’s look at the number of males and females in our class. Biologically, there is a 50/50 chance of a couple having a boy or a girl. Therefore, our expected number of males and females in a class of 40 students is . In a real class of students, there were 13 boys and 27 girls. Does this significantly differ from the expected values?

Chi Square: Goodness of Fit Test The Chi Square Table:

If Chi Square value is greater than or equal to critical value, reject the Null (is significantly different). If Chi Square value is less than the critical value, accept the Null (not significantly different).