T-test calculations

Compare the incident rate between two industries. We took a sample of 10 workplaces for each industry to compare. See Table 1 Ho: the average incident rate for Ind 1 = Incident rate for Ind 2 H1: the average incident rate for Ind 1 ≠ Incident rate for Ind 2

Table 1 Ind 1 Ind 2 1 5 2 6 3 7 4 8 9 10 Mean (M) 6.4 6.1 S 1.577621 2.514403 α = 0.05 N1 = N2 = 10 Dregree of freedom = N1+N2-2 df = 18

Correlation The measure of the strength of linear relationship between two variables The correlation coefficient ranges from +1 to -1. General rules for strength of relationship: r < + 0.2 Weak r < + 0.4 Moderate r < + 0.6 Moderate to Strong r < + 0.8 Strong r > + 0.8 Very Strong

Correlation Table http://www.andrews.edu/~calkins/math/edrm611/edrm13.htm

PHI Correlation Both variables are dichotomous nominal As an example, consider the following data organized by gender and employee classification (faculty/staff). Check for correlation between gender and employee classifications

Contingency table 2x2

phi = (25-100)/sqrt(15•15•15•15) = -75/225 = -0.33, indicating a slight correlation

Chi Square χ2 2 x 2 Contingency Table Use Phi example Χ2 = 30 x 0.333332 = 3.333 df = (No. of Raw -1)(No. of Column – 1) df = (2 -1)(2 -1) = 1 http://ocw.jhsph.edu/courses/FundEpiII/PDFs/Lecture17.pdf

Point Biserial Correlation Example Categorical variable: Yes-No, F-M Ratio or interval variable: No. of incidents, lost days, or grade

Formula

Example

Hypothesis Setup Ho: There is no relationship between respondent gender and earned score H1: There is a relationship between respondent gender and earned score Use an Alpha Level=.05 n = 20

Is the Correlation Significant? Now we need to determine if the correlation coefficient of -0.23 is significant. This is done by performing a t-test.

T-test for Correlations df = 20-2 = 18 r = 0.23 t calculated = 1.00 To interpret the , compare the 1.00 to the critical score. If the obtained score is greater than the critical score, reject the Null and accept the alternative. The critical score from the t-table at .05 and dF = 18 is 2.1 (NOTE: On a T-table, use the .025 column since .025 at one end and .025 at the other end gives you .05).

T-Table The critical score from the t-table at .05 and DF = 18 is 2.1. (NOTE: On a T-table, use the .025 column since .025 at one end and .025 at the other end gives you .05).

Conclusions Since 1.0 is less than 2.1, We fail to Reject the Null Hypothesis and conclude the relationship between the variables is not significant.

Rank Biserial Correlation Variable 1: Nominal Variable 2: Ordinal

Rank Biserial Correlation Example A researcher wishes to determine if a significant relationship exists between ratings on job satisfaction and gender Question 1: Your gender Question 2 asks “How satisfied are you with your job”   1 2 3 4 5 6 7 8 9 10 Very dissatisfied Neutral Very satisfied

Step 1: Data Setup F 2 7 4 6 M 10 9 Case Question 1 Question 2 1 3 5 8 X Y Case Question 1 Question 2 1 F 2 7 3 4 6 5 M 10 8 9

Step 1: Data Setup 2 10 7 6 4 9 8 20/5 = 4 35/5 = 7 Case Female (Yo) Average 20/5 = 4 35/5 = 7

Formula rrb = 2(Y0 – Y1)/n Yo: average in group “o” Y1: average in group “1” n: total cases or subjects

Calculations: Yo = 4 Y1 = 7 N = 10 r = 2(4-7)/10 = -0.6

t-test Calculations: t = 0.6/sqrt((1-0.62)/8) t = 2.12 Critical t from tables: t = 2.3 at α = 0.05/2 df = 8 Since t calculated is less than t critical, then we fail to reject H0 and we conclude that the relationship is not significant.

Pearson Correlation when the sets of data to be correlated represent either interval or ratio scales. Pearson Correlation is best suited for samples of 150 or more. It is typically the most stable measure of correlation. Employee # Age # of unsafe acts 1 27 30 2 22 26 3 15 25 4 35 36 5 33 6 52 7 58 32 8 40 54 9 50 10 43

Pearson Correlation

Spearman Correlation Coefficient The Spearman correlation coefficient (rs )is used to find the relationship when using Ordinal Data or when the data is ranked. It is sometimes referred to as the Rank correlation coefficient. Where: d = difference scores n = number of cases

Example of Spearman Coefficient: Let's consider a hypothetical situation in which two employees (Joe & Fred) rank the comfort of five types of safety glasses. The five glasses are listed as a, b, c, d, and e based on the manufacturer. The comfort scale is 1-5 with 1 being the least comfort. When we compare their findings we see that they ranked the glasses in terms of comfort in exactly the opposite order, perfect disagreement. What is the relationship?