HFIT-565

Assessment & Evaluation of Health Fitness Parameters

Fall Semester 2009

Dr. Marc Schaeffer

mschaef@american.edu

Using a table to determine the significance of Pearson Product-Moment r

The first example comparing five students' shoe sizes with their respective heights

the data are repeated below

why are we comparing shoe size and height?

• we have an experimental hypothesis
• we hypothesize that there is an association between shoe size and height
• we believe that the association depicts increasing heights as a function of increasing shoe size
• thus, we will test a null hypothesis that there is no relationship between height and shoe size
• if we fail to reject this null hypothesis, we must reject the experimental hypothesis (this would be bad, because we probably could not publish our findings)
• in contrast, if we reject the null hypothesis, we will accept the experimental hypothesis and we probably can publish our findings
• however, by accepting the experimental hypothesis, we canNOT say we have PROVED our point or proved our hypothesis
• how do we go about rejecting or failing to reject our NULL HYPOTHESIS
• we have Excel compute the value of r
• we have used either the PEARSON OR CORREL functions and we have....

r = 0.827

df = number of pairs minus 2

there are 5 pairs - so df = 3 = 5 - 2

thus we have a test statistic r = 0.827 and the accompanying df = 3

now we must turn to the table below to assess one of three levels of significance

• p > 0.05 with this option we fail to reject our null hypothesis
• p < 0.05 with this option we reject our null hypothesis
• p < 0.01 with this option we reject our null hypothesis (the relationship is stronger here than when p < 0.05)

we use this table of r values and accompanying df to determine the significance of our computed correlation coefficient r = 0.827

• as the table suggests, it is a table of correlation coefficients
• this is important - it is NOT a table of p-values, but we determine p < 0.05 and/or p < 0.01 from using the table
• the objective is to find the proper row and column in this table to determine whether p < 0.05 or p < 0.01
• if p < 0.01 it is also < 0.05 - this is the same idea as when a number is less than one, it is also less than 5
• to find the proper row, we simply find the df associated with the number of pairs we used to generate r = 0.???
• we have 5 pairs and consequently 3 df in the problem where we obtained r = 0.827
• thus, we go to the row for df = 3
• on this row we have two values of r, including the one in 0.05 column = 0.878 and the one in the 0.01 column = 0.959
• our computed value of r must equal or exceed a tabled r value to reject the null hypothesis
• assuming we wish to reject the null hypothesis, we want our computed value of r to be as close to +1.000 or -1.000 as possible
• if you do not remember, go back to the first graphic in Lecture #5 and note that the values of r are normally distributed and ...
• ...the very largest (as well as the very smallest) r values are in the small areas of the tails
• also, in the above table, notice that the value of r must be greater to reject the null hypothesis at 0.01 level relative to the 0.05 level

with our r = 0.827, this r is not greater than the tabled r (= 0.878) required to reject the null hypothesis at the p = 0.05 level (i.e., 0.827 < 0.878)

• we can now act on the null hypothesis - we fail to reject the null hypothesis... or symbolically, p > 0.05
• we must reject our experimental (alternate) hypothesis
• our computed r was not big enough to support a significant relationship among our five ordered pairs
• this result does NOT mean that there is never a relationship between height and shoe size, simply not in these data

• if we obtained the same computed r ( = 0.827) with 10 ordered pairs or df = 8 ...
• ..our computed value 0.827 is greater than the tabled value 0.632 (for p < 0.05) and also greater than tabled value 0.765 (p < 0.01)
• in this hypothetical situation with 10 cases, we WOULD reject the null hypothesis because p < 0.01

• using the table can be tricky, but it does facilitate an understanding of the process of evaluating a null hypothesis
• we will soon start using a different data analysis tool to obtain the p-value automatically, but this will not completely remove all sources of confusion
• the take home message is an understanding of table use will make the automated process more comprehensible

back to Lecture 5

send email to Dr. Schaeffer

this page last modified by M Schaeffer

on September 17, 2009