{"id":618,"date":"2020-05-13T21:09:38","date_gmt":"2020-05-14T01:09:38","guid":{"rendered":"http:\/\/pressbooks.library.upei.ca\/montelpare\/?post_type=chapter&p=618"},"modified":"2020-09-03T09:53:45","modified_gmt":"2020-09-03T13:53:45","slug":"computing-the-pearson-product-moment-correlation-coefficient","status":"publish","type":"chapter","link":"https:\/\/pressbooks.library.upei.ca\/montelpare\/chapter\/computing-the-pearson-product-moment-correlation-coefficient\/","title":{"raw":"Computing the Pearson Product Moment Correlation Coefficient","rendered":"Computing the Pearson Product Moment Correlation Coefficient"},"content":{"raw":"

Defining the Correlation Coefficient for Continuous Data: the Pearson Product Moment Correlation Coefficient<\/h1>\r\nPearson\u2019s Product Moment Correlation Coefficient is aptly named for the mathematical computational steps that are used to produce the outcome.\u00a0 First, the computation is based on the multiplication of variables, which result in generating a product. Second, the variables, labelled x and y, are computed as independent statistical moments, whereby each score is evaluated against the variable\u2019s algebraic measure of centrality of the group scores, a.k.a. the mean. In this way, the correlation coefficient provides the researcher with an estimate of a relationship between two dependent variables.\r\n\r\nThe outcome estimate of the Pearson Product Moment Correlation Coefficient does NOT imply cause or causality between the two variables. Rather, the outcome estimate is merely an estimate of how closely two variables describe independent responses for a sample. The correlation coefficient is calculated by combining the estimates of variance on each of the separately measured variables of interest.\r\n\r\nSpecifically, the correlation coefficient is a ratio score that ranges from -1.00 to +1.00 and is created from a cross product:\r\n\r\n\"\"\r\n\r\nas the numerator term and then adjusted through division by the error term in the denominator, which is the square root of the product of the sum of squares for the x variable and the sum of squares for the y variable, as shown here:\"\"\r\n\r\nThe formula to compute the Pearson product-moment correlation coefficient is shown here:\r\n\r\n\"\"\r\n\r\nWhen presented graphically, the paired measures of the x and y variables for the set of scores are represented as a single point within the graphing space. The calculation of the correlation coefficient describes the mathematical relationship between the x and y variable pairs are associated within a geometric space.\u00a0 The following three graphs illustrate the extremes of the Pearson Product Moment Correlation Coefficients for the relationships between x and y variable pairs. In Figure 1, the relationship illustrates a Pearson Product Moment Correlation Coefficient as extreme positive with an r value of 1.00. In Figure 2, the relationship illustrates an extreme negative r value of -1.00.\u00a0 In Figure 3, the relationship is shown as a Pearson Product Moment Correlation Coefficient of zero, or no mathematical relationship between each participant\u2019s paired x and y scores.\r\n\r\nFigure 1. Representation of the Pearson Product Moment Correlation Coefficient as r=1.00\r\n\r\n\"\"\r\n\r\nIn the figure above,\u00a0 the cluster of paired scores on the variable x and the variable y are presented in a positive direction within the graphing space. For each individual\u2019s score on the x variable there is an equal score on the y variable relative to the scale of scores. This means that a correspondingly high score on the y variable matches a high score on the x variable; and similarly a correspondingly low score on the y variable matches a low score on the x variable.\u00a0 When the x and y variables as represented as a single point in the graphing space the resulting image suggests a positive correlation between the variables x and y for the set of participants\u2019 scores.\r\n\r\nFigure 2. Representation of the Pearson Product Moment Correlation Coefficient as r= -1.00\r\n\r\n\"\"\r\n\r\nIn the figure above, the cluster of paired scores on the variable x and the variable y are presented in a negative direction within the graphing space. For each individual\u2019s score on the x variable, there is an inverse score on the y variable relative to the scale of scores. This means that a correspondingly high score on the y variable matches a low score on the x variable, and similarly, a correspondingly low score on the y variable matches a high score on the x variable.\u00a0 When the x and y variables as represented as a single point in the graphing space the resulting image suggests a negative correlation between the variables x and y for the set of participants\u2019 scores.\r\n\r\nFigure 3. Representation of the Pearson Product Moment Correlation Coefficient as r= 0.00\r\n\r\n\"\"\r\n\r\nIn Figure 3, above, the cluster of paired scores on the variable x and the variable y are presented as having no definitive direction within the graphing space. For each individual\u2019s score on the x variable, there is no corresponding score on the y variable relative to the scale of scores. In this instance, the x and y variables are simply represented as pairs of scores within the graphing space. No relationship exists when there is complete randomness in the scoring by an individual on the x and on the y variables and thereby do not depict a relationship between the two variables for the set of participants\u2019 scores.\r\n\r\nThe correlation coefficient provides the researcher with an estimate of a relationship between two variables. The Pearson Product Moment Correlation Coefficient does\u00a0NOT<\/strong>\u00a0imply cause or causality between the two variables. Rather, the measure is merely an estimate of how closely two variables describe independent responses for a sample. The correlation coefficient is calculated by combining the measures of variance on each of the separately measured variables, as shown in the following formula:\r\n\r\n\"\"\r\n\r\n
\r\n\r\n

A SAS Example<\/h2>\r\nConsider the following data set, with 6 participants measured on two separate variables.\u00a0 Your intention is to determine if the group measures on VAR_1 are similar to the group measures of VAR_2. For those of you that need more tangible labels, let\u2019s consider that VAR_1 represents IQ scores and that VAR_2 represents SHOE SIZE.\r\n\r\nSo that our query is simply, is there a relationship between IQ scores and SHOE SIZE. In order to analyze these data, you create the following SAS program which produces the output shown below.\r\n
\r\n

SAS Code to produce Pearson Correlation Coefficient and A Line of Best Fit in a Scatterplot<\/p>\r\n\r\n<\/header>\r\n

\r\n
\r\n

DATA CORREX1;<\/p>\r\n

\u00a0 TITLE 'CORRELATION BETWEEN 2 MEASURES IN SAME PERSON';<\/p>\r\n

INPUT ID VAR1 VAR2;<\/p>\r\n

LABEL VAR1='MEASURE FOR VARIABLE 1'<\/p>\r\n

\u00a0\u00a0\u00a0\u00a0\u00a0 VAR2='MEASURE FOR VARIABLE 2';<\/p>\r\n

DATALINES;<\/p>\r\n

001 55 7<\/p>\r\n

002 77 9<\/p>\r\n

003 88 10<\/p>\r\n

004 92 10.5<\/p>\r\n

005 100 11<\/p>\r\n

006 105 12<\/p>\r\n

;<\/p>\r\n

RUN;<\/p>\r\n

PROC UNIVARIATE; VAR VAR1 VAR2;<\/p>\r\n

PROC CORR PEARSON; VAR VAR1 VAR2;<\/p>\r\n

PROC SGPLOT NOAUTOLEGEND DATA= CORREX1;<\/p>\r\n

TITLE1\u00a0 \"SCATTER PLOT OF CORRELATION BETWEEN VAR1 AND VAR2\";<\/p>\r\n

TITLE2\u00a0 \"LINE OF BEST FIT ADDED TO SCATTERPLOT\";<\/p>\r\n

REG Y=VAR1 X=VAR2;<\/p>\r\n

RUN;<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nOutput from PROC CORR with Corresponding Line of Best Fit in a Scatterplot<\/strong>\r\n\r\nCorrelation between 2 measures in the same person using the SAS PROC<\/strong> CORR Procedure. Table 1. Simple Statistics\r\n<\/strong>\r\n

\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Variable<\/strong><\/td>\r\nN<\/strong><\/td>\r\nMean<\/strong><\/td>\r\nStd Dev<\/strong><\/td>\r\nSum<\/strong><\/td>\r\nMinimum<\/strong><\/td>\r\nMaximum<\/strong><\/td>\r\n<\/tr>\r\n<\/thead>\r\n
Measure for Var_1 (IQ scores)\r\n<\/strong><\/td>\r\n6<\/td>\r\n86.16667<\/td>\r\n18.10433<\/td>\r\n517.00000<\/td>\r\n55.00000<\/td>\r\n105.00000<\/td>\r\n<\/tr>\r\n
Measure for Var_2 Shoe Size\r\n<\/strong><\/td>\r\n6<\/td>\r\n9.91667<\/td>\r\n1.74404<\/td>\r\n59.50000<\/td>\r\n7.00000<\/td>\r\n12.00000<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\nTable 2. Matrix of Pearson Correlation Coefficients, N = 6, <\/strong>Prob > |r| under H0: Rho=0<\/strong>\r\n
\r\n\r\n\r\n\r\n\r\n\r\n\r\n
\u00a0<\/strong><\/td>\r\nIQ scores (Var_1)\r\n<\/strong><\/td>\r\nShoe Size (Var_2)<\/strong><\/td>\r\n<\/tr>\r\n<\/thead>\r\n
IQ scores (Var_1)<\/strong><\/td>\r\n1.00000\r\n\r\n <\/td>\r\nr= 0.99500\r\n\r\np <.0001<\/td>\r\n<\/tr>\r\n
Shoe Size (Var_2)<\/strong><\/td>\r\nr= 0.99500\r\n\r\np <.0001<\/td>\r\n1.00000\r\n\r\n <\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n\"\"\r\n\r\nThe results indicate that there is a strong correlation between IQ scores and shoe size as indicated by the correlation coefficient value in Table 2 of the SAS output in the scatterplot, above. The reported correlation coefficient (r value) was 0.995. which is considered statistically significant (p<0.0001)[footnote]Note: just because the p value extends beyond 0.01 you need only to report p<0.01[\/footnote]. The r value indicates that as a group the data on each variable are traveling in the same direction. In other words, low IQ scores were associated with small shoe sizes and high IQ scores were associated with larger shoe sizes.\r\n\r\n
\r\n\r\n

Another SAS Working Example<\/h2>\r\nConsider the following comparison of VO2<\/sub> maximum tests to illustrate the computations for the Pearson product-moment correlation coefficient.\r\n\r\nAn individual\u2019s ability to perform maximal re-synthesis of ATP is often determined from their estimated VO2<\/sub> maximum. Yet the true estimate of VO2<\/sub> maximum requires an individual to undergo an extensive performance test in a \u201cclinical or laboratory\u201d setting.\r\n\r\nFurther, because physical fitness is often established from the VO2<\/sub> maximum, researchers strive to develop approaches that can provide a suitable estimate for VO2<\/sub> maximum without requiring the laboratory test. To this end, several \u201cfield tests\u201d have been presented as reliable and valid proxies for the laboratory estimate of VO2<\/sub> maximum.\r\n\r\nThe validity of field-tests for VO2<\/sub> maximum as effective measures of true physiological functioning within this system are based on the linear relationship between heart rate and oxygen consumption. Since the changes in heart rate match the changes in aerobic metabolism, an individual\u2019s ability to respire can be predicted indirectly by heart rate rather than the direct determination of oxygen utilization from a controlled laboratory setting.\r\n\r\nLaboratory tests for VO2<\/sub> maximum typically require that the individual runs or walks on a treadmill, or cycles on a standard bicycle ergometer while expired air is collected and the oxygen concentration in the expired air, is measured. The field tests, however, can use several stimulus modalities (e.g. walking, running, stepping, and cycling) and rather than measure the oxygen concentration in expired air, the technician simply records heart rate at specific stages of the exercise stress test. The relationship between field-tests and a declared \u201cgold standard\u201d is determined by comparing the statistical probability of the association between the predictive nature of the field test and a baseline reference standard measured in the laboratory.\r\n\r\nIn determining the accuracy of a clinical or diagnostic test, researchers compare the performance of subjects on the selected diagnostic (field) test against the \"gold standard\u201d. In the following example, this clinical epidemiological approach was used to determine the accuracy of field tests for predicting maximal oxygen consumption against a laboratory standard treadmill test (Bruce protocol). The field tests included the Cooper's 12-minute run, the Astrand-Rhyming bicycle test, the Canadian Aerobic Fitness Test, and the one-mile walking test. A sample of the results for each test is presented in the table, below.\r\n
Table of VO2 Maximum Estimates For A Sample Of Ten Individuals On 5 Different Procedures<\/h5>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Subject<\/td>\r\ntreadmill\r\nVO2<\/sub>\u00a0max\r\nml\/kg\/min<\/td>\r\n12-minute run\r\nVO2<\/sub>\u00a0max\r\nml\/kg\/min<\/td>\r\nBicycle ergometer\r\nVO2\u00a0<\/sub>max\r\nml\/kg\/min<\/td>\r\nStep test\r\nVO2<\/sub>\u00a0max\r\nml\/kg\/min<\/td>\r\n1-mile walk test\r\nVO2<\/sub>\u00a0max\r\nml\/kg\/min<\/td>\r\n<\/tr>\r\n
01<\/td>\r\n47<\/td>\r\n33<\/td>\r\n44<\/td>\r\n57<\/td>\r\n45<\/td>\r\n<\/tr>\r\n
02<\/td>\r\n36<\/td>\r\n19<\/td>\r\n53<\/td>\r\n53<\/td>\r\n43<\/td>\r\n<\/tr>\r\n
03<\/td>\r\n55<\/td>\r\n25<\/td>\r\n49<\/td>\r\n61<\/td>\r\n47<\/td>\r\n<\/tr>\r\n
04<\/td>\r\n39<\/td>\r\n41<\/td>\r\n23<\/td>\r\n56<\/td>\r\n44<\/td>\r\n<\/tr>\r\n
05<\/td>\r\n57<\/td>\r\n27<\/td>\r\n50<\/td>\r\n58<\/td>\r\n45<\/td>\r\n<\/tr>\r\n
06<\/td>\r\n46<\/td>\r\n39<\/td>\r\n52<\/td>\r\n61<\/td>\r\n53<\/td>\r\n<\/tr>\r\n
07<\/td>\r\n43<\/td>\r\n36<\/td>\r\n55<\/td>\r\n57<\/td>\r\n44<\/td>\r\n<\/tr>\r\n
08<\/td>\r\n29<\/td>\r\n31<\/td>\r\n42<\/td>\r\n53<\/td>\r\n42<\/td>\r\n<\/tr>\r\n
09<\/td>\r\n54<\/td>\r\n28<\/td>\r\n44<\/td>\r\n56<\/td>\r\n44<\/td>\r\n<\/tr>\r\n
10<\/td>\r\n37<\/td>\r\n44<\/td>\r\n54<\/td>\r\n57<\/td>\r\n45<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe following computations illustrate the application of the Pearson product-moment correlation coefficient to the oxygen consumption data shown above.\r\n\r\nSTEP 1: Select two variables to compare\u00e0, for example, we will begin with VO2 measured with the treadmill test versus VO2 measured with the 1-mile walk test. The mean for the scores on the treadmill, which in this example we designate as the x variable is 44.3 (\u00b1 9.23), and the scores on the walk test, which in this example we designate as the y variable and is 45.2 (\u00b1 3.04).\r\n\r\n\"\"\r\n\r\nThe mean score for the variable x and the mean score for the variable y are used independently in Step 2: computing variance for each selected variable.\r\n\r\nSTEP 2:\u00a0Compute the sum of squares for the variance elements of each selected variable, separately, as shown below:\"\"\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Scores on \"x\"\r\ntreadmill<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/tr>\r\n
47<\/td>\r\n47 - 44.3 = 2.7<\/td>\r\n(47 - 44.3)2= 7.29<\/td>\r\n<\/tr>\r\n
36<\/td>\r\n36 - 44.3 = -8.3<\/td>\r\n(36 - 44.3)2= 68.89<\/td>\r\n<\/tr>\r\n
55<\/td>\r\n55 - 44.3 = 10.7<\/td>\r\n(55 - 44.3)2= 114.49<\/td>\r\n<\/tr>\r\n
39<\/td>\r\n39 - 44.3 = -5.3<\/td>\r\n(39 - 44.3)2= 28.09<\/td>\r\n<\/tr>\r\n
57<\/td>\r\n57 - 44.3 = 12.7<\/td>\r\n(57 - 44.3)2= 161.29<\/td>\r\n<\/tr>\r\n
46<\/td>\r\n46 - 44.3 = 1.7<\/td>\r\n(46 - 44.3)2= 2.89<\/td>\r\n<\/tr>\r\n
43<\/td>\r\n43 - 44.3 = -1.3<\/td>\r\n(43 - 44.3)2= 1.69<\/td>\r\n<\/tr>\r\n
29<\/td>\r\n29 - 44.3 = -15.3<\/td>\r\n(29 - 44.3)2= 234.09<\/td>\r\n<\/tr>\r\n
54<\/td>\r\n54 - 44.3 = 9.7<\/td>\r\n(54 - 44.3)2= 94.09<\/td>\r\n<\/tr>\r\n
37<\/td>\r\n37 - 44.3 = -7.3<\/td>\r\n(37 - 44.3)2= 53.29<\/td>\r\n<\/tr>\r\n
<\/td>\r\n= 0<\/td>\r\n\u00a0= 766.01<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Scores on \"y\"\r\nthe walk test<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/tr>\r\n
45<\/td>\r\n45 - 45.2= -0.2<\/td>\r\n(45 - 45.2)2= 0.04<\/td>\r\n<\/tr>\r\n
43<\/td>\r\n43 - 45.2 = -2.2<\/td>\r\n(43 - 45.2)2= 4.84<\/td>\r\n<\/tr>\r\n
47<\/td>\r\n47 - 45.2 = 1.8<\/td>\r\n(47 - 45.2)2= 3.24<\/td>\r\n<\/tr>\r\n
44<\/td>\r\n44 - 45.2 = -1.2<\/td>\r\n(44 - 45.2)2= 1.44<\/td>\r\n<\/tr>\r\n
45<\/td>\r\n45 - 45.2 = -0.2<\/td>\r\n(45 - 45.2)2= 0.04<\/td>\r\n<\/tr>\r\n
53<\/td>\r\n53 - 45.2 = 7.8<\/td>\r\n(53 - 45.2)2= 60.84<\/td>\r\n<\/tr>\r\n
44<\/td>\r\n44 - 45.2 = -1.2<\/td>\r\n(44 - 45.2)2= 1.44<\/td>\r\n<\/tr>\r\n
42<\/td>\r\n42 - 45.2 = -3.2<\/td>\r\n(42 - 45.2)2= 10.24<\/td>\r\n<\/tr>\r\n
44<\/td>\r\n44 - 45.2 = -1.2<\/td>\r\n(44 - 45.2)2= 1.44<\/td>\r\n<\/tr>\r\n
45<\/td>\r\n45 - 45.2 = -0.2<\/td>\r\n(45 - 45.2)2= 0.04<\/td>\r\n<\/tr>\r\n
<\/td>\r\n\u00a0= 0<\/td>\r\n= 83.6<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSTEP 3: Compute the cross products of\u00a0 x and y using: \u00a0\u00a0 [latex](x_i - \\overline{x})(y_i - \\overline{y})[\/latex]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
participant<\/td>\r\n\u00a0\u00a0 [latex](x_i - \\overline{x})(y_i - \\overline{y})[\/latex]<\/td>\r\n<\/tr>\r\n
01<\/td>\r\n2.7 * -0.2 = -0.54<\/td>\r\n<\/tr>\r\n
02<\/td>\r\n-8.3 * -2.2 = 18.26<\/td>\r\n<\/tr>\r\n
03<\/td>\r\n10.7 * 1.8 = 19.26<\/td>\r\n<\/tr>\r\n
04<\/td>\r\n-5.3 * -1.2 = 6.36<\/td>\r\n<\/tr>\r\n
05<\/td>\r\n12.7 * -0.2 = -2.54<\/td>\r\n<\/tr>\r\n
06<\/td>\r\n1.7 * 7.8 = 13.26<\/td>\r\n<\/tr>\r\n
07<\/td>\r\n-1.3 * -1.2 = 1.56<\/td>\r\n<\/tr>\r\n
08<\/td>\r\n-15.3 * -3.2 = 48.96<\/td>\r\n<\/tr>\r\n
09<\/td>\r\n9.7 * -1.2 = -11.64<\/td>\r\n<\/tr>\r\n
10<\/td>\r\n-7.3 * -0.2 =1.46<\/td>\r\n<\/tr>\r\n
\"\"= 94.4<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSTEP 4: Compute the denominator of the correlation coefficient using the square root of the product of the sum of squares for the x and the y variables.\r\n\r\n\"\"\r\n\r\nThe specific values can be taken from the computations above where the \"\"= 766.01 and the\u00a0 \"\"= 83.6 \u00a0 The denominator term is thus the square root of:\u00a0 SQRT(766.01 x 83.6) = SQRT(64038.44) = 253<\/strong>\r\n\r\nSTEP 5: Compute the correlation coefficient by substituting the values from that which was calculated, above.\r\n\r\n\"\"\r\n\r\nIn our example, the correlation coefficient computed for the relationship between the VO2\u00a0on the treadmill and the VO2\u00a0on the 1 mile walk test = 0.37\r\n\r\n
\r\n\r\n

Verifying Our Computations with SAS<\/h2>\r\nThe data set used in the example above included two columns of data: the VO2 max scores recorded for each participant on the treadmill test and the 1-mile walk test. The columns were separated by a tab character which we denote in our INFILE statement with the command: delimiter='09'x<\/strong> . The data set has three variables: id, treadmill scores and 1-mile walk test scores.\r\n
\r\n

raw data set for three variables<\/p>\r\n\r\n<\/header>\r\n

\r\n
\r\n

01\u00a0\u00a0\u00a0 47\u00a0\u00a0\u00a0 45<\/p>\r\n

02\u00a0\u00a0\u00a0 36\u00a0\u00a0\u00a0 43<\/p>\r\n

03\u00a0\u00a0\u00a0 55\u00a0\u00a0\u00a0 47<\/p>\r\n

04\u00a0\u00a0\u00a0 39\u00a0\u00a0\u00a0 44<\/p>\r\n

05\u00a0\u00a0\u00a0 57\u00a0\u00a0\u00a0 45<\/p>\r\n

06\u00a0\u00a0\u00a0 46\u00a0\u00a0\u00a0 53<\/p>\r\n

07\u00a0\u00a0\u00a0 43\u00a0\u00a0\u00a0 44<\/p>\r\n

08\u00a0\u00a0\u00a0 29\u00a0\u00a0\u00a0 42<\/p>\r\n

09\u00a0\u00a0\u00a0 54\u00a0\u00a0\u00a0 44<\/p>\r\n

10\u00a0\u00a0\u00a0 37\u00a0\u00a0\u00a0 45<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nThe SAS program written for SAS Studio to compute the correlation coefficient for the relationship between the scores on the treadmill and the 1-mile walk test is shown here:\r\n

\r\n

SAS Code to produce Pearson Correlation Coefficient for treadmill and the 1-mile walk test<\/p>\r\n\r\n<\/header>\r\n

\r\n
\r\n

DATA CORR1(LABEL= 'CORRELATION COEFFICIENTS');<\/p>\r\n

TITLE 'PRACTICE DATA SET TO COMPUTE CORRELATION COEFFICIENTS';<\/p>\r\n

INFILE '\/FOLDERS\/DATASETS\/CORR1.DAT' DELIMITER='09'X ;<\/p>\r\n

* THE DATA IN THE FILE CORR1.DAT ARE TAB DELIMITED;<\/p>\r\n

INPUT ID 1-3 @4 TRDMILL ONEMLWLK;<\/p>\r\n

PROC CORR; VAR TRDMILL ONEMLWLK;<\/p>\r\n

RUN;<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

<\/a>Output from the SAS PROC CORR PROCEDURE<\/h3>\r\n
\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Variable<\/strong><\/td>\r\nN<\/strong><\/td>\r\nMean<\/strong><\/td>\r\nStd Dev<\/strong><\/td>\r\nSum<\/strong><\/td>\r\nMinimum<\/strong><\/td>\r\nMaximum<\/strong><\/td>\r\n<\/tr>\r\n<\/thead>\r\n
trdmill<\/strong><\/td>\r\n10<\/td>\r\n44.30000<\/td>\r\n9.22617<\/td>\r\n443.00000<\/td>\r\n29.00000<\/td>\r\n57.00000<\/td>\r\n<\/tr>\r\n
Onemlwlk<\/strong><\/td>\r\n10<\/td>\r\n45.20000<\/td>\r\n3.04777<\/td>\r\n452.00000<\/td>\r\n42.00000<\/td>\r\n53.00000<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n
\r\n\r\n\r\n\r\n\r\n\r\n\r\n
<\/td>\r\ntrdmill<\/strong><\/td>\r\nOnemlwlk<\/strong><\/td>\r\n<\/tr>\r\n<\/thead>\r\n
Trdmill<\/strong><\/td>\r\nr= 1.00000<\/td>\r\nr = 0.37301\r\n\r\np < 0.2884<\/td>\r\n<\/tr>\r\n
Onemlwlk<\/strong><\/td>\r\nr = 0.37301\r\n\r\np < 0.2884<\/td>\r\nr = 1.00000<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n\r\n
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Testing the significance of the sample correlation coefficient<\/h2>\r\nWhen we compute the r statistic (correlation coefficient) the relationship that we observe is specific to the sample from which the data were drawn. If we wish to compute the population parameter of a correlation coefficient, that is the extension of the statistic for the sample to the parameter in the population, then our next step is to compute the rho statistic \u2013 rho is the population parameter estimate of the correlation coefficient and it is determined using a t test with (n-2 degrees of freedom).\r\n\r\nTo determine if the computed correlation coefficient is equal to 0 as in the null hypothesis test:\u00a0 H0<\/sub>: r = 0 at\u00a0 p<0.05 we use the following equation:\r\n

[latex]t = r \\sqrt\\frac{n-2}{1-r^2}[\/latex]<\/p>\r\nwith the degrees of freedom (npairs - 2).\r\n\r\nTherefore, in the example presented above, the correlation coefficient was r=0.37, and the sample size was n=10. The t value to evaluate the population parameter is shown below with degrees of freedom = (npairs - 2) = (10 \u2013 2) = 8.\r\n\r\n\"\"\r\n\r\nThe t <\/em>critical value for t=1.126 with df=8 is 2.31 for p<0.05. Therefore, because our t statistic is less than the critical value, we would accept the null hypothesis at p<0.05 and suggest that in this sample, the estimate of the relationship between treadmill VO2 maximum scores with 1-mile walk test scores is not an estimate of the true population parameter (i.e. the rho score). Finally, in addition to computing the significance of the correlation coefficient, we can also determine the strength of the correlation coefficient.\u00a0 Typically,\u00a0 the decision rule concerning the strength of a correlation coefficient is as follows: if the absolute value of r is low (<0.7) then we say that is no correlation, but if the absolute value is (>0.7) then we say that there is a correlation between the scores on the variable x and the scores on the variable y.\r\n\r\n

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Strength of the Correlation<\/strong><\/h3>\r\nThe strength of the correlation is determined by squaring the value of r (the correlation coefficient) and multiplying the result by 100. Taken in this way, we are able to express the bivariate correlation (i.e. the relationship between two variables) as a percent of variance that is explained between the two variables.\r\n\r\nFor example, if we square the correlation coefficient computed for the relationship between the VO2<\/sub> on the treadmill and the VO2<\/sub> on the 1-mile walk test, as follows:\r\n\r\nr = 0.37 and r2<\/sup> = 0.139 x 100 = approximately 14%\r\n\r\nWe can then say that the variable x and the variable y have a very low relationship and when this relationship is expressed in terms of variance, then the variables x and y in this example share only about 14% of the variance between these two variables, leaving more than 86% of the variance between these two variables unexplained.\r\n\r\nConversely<\/strong>, if we had computed a correlation coefficient of 0.9 between the variable x and the variable y then we would say that there is a strong relationship and could express the variance shared by these two variables as follows:\r\n\r\nr = 0.9 and r2<\/sup> = 0.81 x 100 = approximately 81% explained variance.\r\n\r\n
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Here is what we covered in this chapter<\/h2>\r\nIn this chapter you were introduced to:\r\n