Question

For science class, a student recorded the high and low temperatures, in Fahrenheit, over a ten day period in September. The data is shown in the table.

A) What is the correlation for a linear model of this data? Round to the nearest hundredth.

r =

B) Ten high and low temperatures in October are recorded and have a correlation coefficient of 0.89. In which month is there a stronger correlation between the high and low temperatures? Explain your answer.

Answer 1

The correlation of linear model of a data is given by

$r= \frac{n(\Sigma xy)-(\Sigma x)(\Sigma y)}{\sqrt{[n\Sigma x^2-(\Sigma x)^2][n\Sigma y^2-(\Sigma y)^2]}}$

The extended table for the data set is given as follows:

$\begin{tabular} {|c|c|c|c|c|} x&y&xy&x^2&y^2\\[1ex] 26&49&1,274&676&2,401\\ 28&50&1,400&784&2,500\\ 30&57&1,710&900&3,249\\ 32&54&1,728&1,024&2,916\\ 34&60&2,040&1,156&3,600\\ 35&58&2,030&1,225&3,364\\ 37&64&2,368&1,369&4,096\\ 38&66&2,508&1,444&4,356\\ 41&63&2,583&1,681&3,969\\ 45&72&3,240&2,025&5,184\\[1ex] \Sigma x=346&\Sigma y=593&\Sigma xy=20881&\Sigma x^2=12284&\Sigma y^2=35635 \end{tabular}$

Thus the correlation coefficient is given by:

$r= \frac{10(20,881)-(346)(593)}{\sqrt{[10(12,284)-(346)^2][10(35,635)-(593)^2]}} \\ \\ = \frac{208,810-205,178}{\sqrt{(122,840-119716)(356,350-351,649)}} = \frac{3,632}{\sqrt{(3,124)(4,701)}} \\ \\ = \frac{3,632}{\sqrt{14,685,924}} = \frac{3,632}{3,832.22} =\bold{0.95}$



Part B:

Given that the correlation of 10 high and low temperature in October is 0.89 and we found that the correlation of 10 high and low temperatures in September is 0.95.

Therefore, September has a stronger correlation between the high and low temperatures.