Question

the table shows the average number of hours of daylight per day for the last four months of the year.
monthly daylight hours

month of the year	average hours of daylight per day
10	11.37
11	10.57
12	10

what is the correlation coefficient for the data in the table?
-0.993
-0.791
0.791
0.993

Explanation:

Step1: Identify Variables

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

Step2: Calculate Means

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

Step3: Compute Deviations

For \( x \): \( 9 - 10.5=-1.5 \), \( 10 - 10.5=-0.5 \), \( 11 - 10.5 = 0.5 \), \( 12 - 10.5 = 1.5 \)
For \( y \): \( 12.37 - 11.0775 = 1.2925 \), \( 11.37 - 11.0775 = 0.2925 \), \( 10.57 - 11.0775=-0.5075 \), \( 10 - 11.0775=-1.0775 \)

Step4: Calculate Products and Squares

\( \sum (x - \bar{x})(y - \bar{y})=(-1.5)(1.2925)+(-0.5)(0.2925)+(0.5)(-0.5075)+(1.5)(-1.0775) \)
\( = -1.93875 - 0.14625 - 0.25375 - 1.61625=-3.955 \)

\( \sum (x - \bar{x})^2=(-1.5)^2+(-0.5)^2+(0.5)^2+(1.5)^2=2.25 + 0.25 + 0.25 + 2.25 = 5 \)

\( \sum (y - \bar{y})^2=(1.2925)^2+(0.2925)^2+(-0.5075)^2+(-1.0775)^2 \)
\( \approx 1.6705 + 0.0856 + 0.2576 + 1.1610 = 3.1747 \)

Step5: Apply Correlation Formula

\( r=\frac{\sum (x - \bar{x})(y - \bar{y})}{\sqrt{\sum (x - \bar{x})^2 \sum (y - \bar{y})^2}}=\frac{-3.955}{\sqrt{5\times3.1747}} \)
\( \sqrt{15.8735}\approx3.984 \), so \( r=\frac{-3.955}{3.984}\approx - 0.993 \)

Answer:

\(-0.993\)