19d. For Time Spent on Task: SS_yy = (use 2 dec. places during calculation; report whole number) __________ Q19 [a] A researcher measures the relationship between the number of interruptions in classand time spent on task [in minutes]. Compute the Pearson correlation coefficient for the databelow. Use your own paper to calculate the correlation coefficient; hang onto to it just in case.Note: The etext table is incorrect because of faulty formatting. Use the r distribution availablein Blackboard.Below you will enter:2.1 Mean for number of interruptions (x)2.2 SS, for number of interruptions2.3 Mean for time spent on task (y)2.4 SS,for time spent on task2.5 Sum of Products (SP, SSæy)2.6 Correlation coefficient (r) Number ofTime Spent onInterruptionsTask8183406.20232

The correlation are used to understand the strength of association or relationship between the two numerical variable. It is denoted as r and it value ranges from -1 to +1. If the value is +1, then it shows prefect positive correlation. If the value is -1, then it shows perfect negative correlation. If the value is 0, there exist no correlation between the variables. Given : The number of interpretation and the time spent on the task Interruption (x) Time spent (y) x2 y2 xy 8 18 64 324 144 3 40 9 1600 120 6 20 36 400 120 2 32 4 1024 64 ∑x=19 ∑y=110 ∑x2=113 ∑y2=3348 ∑xy=448 From the table, Mean of x : x¯=∑xn=194=4.75 SSx is 113 Mean of y : y¯=∑yn=1104=27.5 SSy is 3348 Sum of product is 448Using the values obtained from the calculations, the correlation is calculated as follows : r=n∑xy-∑x∑yn∑x2-∑x2n∑y2-∑y2=4×448-19×1104×113-1924×3348-1102=1792-2090452-36113392-12100=-29891×1292=-298342.8877=-0.869 Thus the correlation value is -0.869 Hence there exist negative strong correlation between the interruptions and the time spend on the tast.