In this exercise, you will solve three questions, where you will be asked to cal
ID: 3125934 • Letter: I
Question
In this exercise, you will solve three questions, where you will be asked to calculate a linear correlation coefficient and determine whether there is a linear correlation between the two given variables. Solve the following problems: Listed below are baseball team statistics, consisting of the proportions of wins and the result of this difference: Difference (number of runs scored) - (number of runs allowed). The statistics are from a recent year, and the teams are NY—Yankees, Toronto, Boston, Cleveland, Texas, Houston, San Francisco, and Kansas City. Difference 163 55 –5 88 51 16 –214 Wins 0.599 0.537 0.531 0.481 0.494 0.506 0.383 Construct a scatter plot, find the value of the linear correlation coefficient r, and find the critical values of r from Table VI, Appendix A, p. A-14, of your textbook Elementary Statistics. Use = 0.05. Is there sufficient evidence to conclude that there is a linear correlation between the proportion of wins and the above difference? Chirps in 1 Min 882 1188 1104 864 1200 1032 960 900 Temperature(°F) 69.7 93.3 84.3 76.3 88.6 82.6 71.6 79.6 A classic application of correlation involves the association between temperature and the number of times a cricket chirps in a minute. Listed below are the numbers of chirps in 1 minute and the corresponding temperatures in °F: Construct a scatter plot, find the value of the linear correlation coefficient r, and find the critical values of r from Table VI, Appendix A, p. A-14, of your textbook Elementary Statistics. Use = 0.05. Is there a linear correlation between the number of chirps in 1 minute and the temperature?
Explanation / Answer
Solution :
1st Part :
The formula for correlation coefficient is:
r = (n (xy)-xy)/ ((n(x2)-(x) 2) (n (y2)-(y) 2)
This looks like a big frightening equation, but let's break it down one piece at a time:
x = the sum of the first set
y = the sum of the second set
(xy) = the sum of the products of each pair of #s
(x2) = the sum of the squares of the first set
(y2) = the sum of the squares of the second set
n = the number of (x, y) pairs
I'm going to make a table to make it easy to calculate these values. The first columns are the given values, and at the bottom is the sum of each column.
Difference (x) wins (y) x2 y2 xy
163 0.599 26569 .358801 91.117
55 0.537 3025 .288369 29.535
-5 0.531 25 .281961 -2.655
88 0.481 7744 .231361 42.328
51 0.494 2601 .244036 25.194
16 0.506 256 .256036 8.096
214 0.383 45796 .146689 81.962
x=592 y=3.531 (x2)=86016 (y2)=1.807253 (xy)=275.58
n=7
(Go to full screen or reduce your text size if it is not coming out right.
Now that we have these 6 values, we can plug them into our formula
r = (n (xy)-xy)/ ((n(x2)-(x) 2) (n (y2)-(y) 2)
r = (7(275.577)-(592)(3.531))/((7(86016)-(5922))(7(1.807253)-(3.5312)
r = (1929.039-2090.352)/ ((602112-350464) (12.650771-12.467961)
r = -161.313/ (251648*.18281)
r = -161.313/46003.77088
r = -161.313/214.4848966
r = -.7521
This reveals a fairly strong negative correlation of r=-.7521.
To determine if this is significant at =.05, you can consult the appendix in your book for critical r values. Check under df=6 because df=n-1. My table shows a critical r of .707. Since the calculated r exceeds the critical r, we can conclude that there is a linear correlation between the two variables. (We only care about the absolute value of our r for this part so .7521>.707)
2nd Part :
Chirps in 1 Min 882, 1188, 1104, 864, 1200, 1032, 960, 900
Temperature (°F) 69.7, 93.3, 84.3, 76.3, 88.6, 82.6, 71.6, 79.6
(a)
Number of cases = 8
X=8130
Y=646
X^2=8391204
Y^2=52626.6
XY=663245.4
X Y = 663245.4
r = ( XY - X Y / n) / SQRT { [ X^2 - (X) ^2 / n] [ Y^2 - (Y) ^2 / n]}
Numerator of r = (663245.4) - (8130)(646) / 8 = 6747.9
Denomination of r = SQRT[8391204-(8130)^2/8]*SQRT[52626.6-(646.0)^2/8]
= SQRT [129091.5] * SQRT [462.1]
r = 6747.9 / [359.29306] * [21.49651]
Correlation coefficient r = 0.8737
Critical value of r using a = 0.05 = 0.707
(b)
Since the correlation of 0.8737 > critical value, the difference is significant. There is a linear correlation between the number of chirps in 1 min and the temperature.
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