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: Identify variables

Let \( x \) be the month of the year (\( 9, 10, 11, 12 \)) and \( y \) be the average hours of daylight (\( 12.37, 11.37, 10.57, 10 \)).

Step2: Calculate means

Mean of \( x \): \( \bar{x}=\frac{9 + 10 + 11 + 12}{4}=\frac{42}{4} = 10.5 \)
Mean of \( y \): \( \bar{y}=\frac{12.37 + 11.37 + 10.57 + 10}{4}=\frac{44.31}{4}=11.0775 \)

Step3: Calculate deviations and products

For each data point:

• \( (9 - 10.5)(12.37 - 11.0775)=(-1.5)(1.2925)= - 1.93875 \)
• \( (10 - 10.5)(11.37 - 11.0775)=(-0.5)(0.2925)= - 0.14625 \)
• \( (11 - 10.5)(10.57 - 11.0775)=(0.5)(-0.5075)= - 0.25375 \)
• \( (12 - 10.5)(10 - 11.0775)=(1.5)(-1.0775)= - 1.61625 \)

Sum of products: \( \sum (x_i - \bar{x})(y_i - \bar{y})=-1.93875 - 0.14625 - 0.25375 - 1.61625=-3.955 \)

Step4: Calculate standard deviations

For \( x \):
\( \sum (x_i - \bar{x})^2=(-1.5)^2+(-0.5)^2+(0.5)^2+(1.5)^2 = 2.25 + 0.25 + 0.25 + 2.25 = 5 \)
Standard deviation of \( x \): \( s_x=\sqrt{\frac{5}{4 - 1}}=\sqrt{\frac{5}{3}}\approx1.291 \)

For \( y \):
\( \sum (y_i - \bar{y})^2=(12.37 - 11.0775)^2+(11.37 - 11.0775)^2+(10.57 - 11.0775)^2+(10 - 11.0775)^2 \)
\( =(1.2925)^2+(0.2925)^2+(-0.5075)^2+(-1.0775)^2 \)
\( =1.67055625 + 0.08555625 + 0.25755625 + 1.16100625 = 3.174675 \)
Standard deviation of \( y \): \( s_y=\sqrt{\frac{3.174675}{4 - 1}}=\sqrt{1.058225}\approx1.0287 \)

Step5: Calculate correlation coefficient

\( r=\frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{(n - 1)s_xs_y} \) (or using \( n \) in formula, here \( n = 4 \))
\( r=\frac{-3.955}{(4 - 1)\times1.291\times1.0287}=\frac{-3.955}{3\times1.328}\approx\frac{-3.955}{3.984}\approx - 0.993 \)

Answer:

\(-0.993\)