Question

the table shows the average number of hours of daylight per day for the last four months of the year. what is the correlation coefficient for the data in the table? monthly daylight hours month of the year average hours of daylight per day 9 12.37 10 11.37 11 10.57 12 10 -0.993 -0.791 0.791 0.993

Explanation:

Step1: Recall the formula for correlation coefficient

The formula for the correlation coefficient \(r=\frac{n\sum xy-\sum x\sum y}{\sqrt{[n\sum x^{2}-(\sum x)^{2}][n\sum y^{2}-(\sum y)^{2}]}}\)
Let \(x\) be the month (\(x = 9,10,11,12\)) and \(y\) be the average hours of daylight (\(y=12.37,11.37,10.57,10\))
\(n = 4\)
\(\sum x=9 + 10+11+12=42\)
\(\sum y=12.37+11.37+10.57+10 = 44.31\)
\(\sum xy=(9\times12.37)+(10\times11.37)+(11\times10.57)+(12\times10)=111.33+113.7+116.27+120 = 461.3\)
\(\sum x^{2}=9^{2}+10^{2}+11^{2}+12^{2}=81 + 100+121+144 = 446\)
\(\sum y^{2}=12.37^{2}+11.37^{2}+10.57^{2}+10^{2}=153.0169+129.2769+111.7249+100=494.0187\)

Step2: Substitute values into the formula

\(n\sum xy-\sum x\sum y=4\times461.3-42\times44.31=1845.2-1861.02=-15.82\)
\(n\sum x^{2}-(\sum x)^{2}=4\times446 - 42^{2}=1784-1764 = 20\)
\(n\sum y^{2}-(\sum y)^{2}=4\times494.0187-44.31^{2}=1976.0748 - 1963.3761=12.6987\)
\(\sqrt{(n\sum x^{2}-(\sum x)^{2})(n\sum y^{2}-(\sum y)^{2})}=\sqrt{20\times12.6987}=\sqrt{253.974}\approx15.9366\)
\(r=\frac{- 15.82}{15.9366}\approx - 0.993\)

Answer:

-0.993