Question

fans of the science fiction television show space cruiser zeta are hosting a convention. to figure out how much space the event will require, an employee at the convention center looked at the space requirements for previous conventions. for each convention, the employee recorded the number of attendees, x, and the floor space required (in square meters), y. attendees floor space (in square meters) 791 4,582 3,646 6,697 5,821 7,106 6,040 18,852 7,622 14,041 7,852 15,957 round your answer to the nearest thousandth. r = save answer

Explanation:

Step1: Calculate the means of \(x\) and \(y\)

Let \(x = [791,3646,5821,6040,7622,7852]\) and \(y=[4582,6697,7106,18852,14041,15957]\)
\(\bar{x}=\frac{791 + 3646+5821+6040+7622+7852}{6}=\frac{31772}{6}\approx5295.333\)
\(\bar{y}=\frac{4582 + 6697+7106+18852+14041+15957}{6}=\frac{67235}{6}\approx11205.833\)

Step2: Calculate the numerator and denominator of the correlation coefficient formula

The formula for the correlation coefficient \(r=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}}\)

Calculate \((x_{i}-\bar{x})(y_{i}-\bar{y})\) for each \(i\):

• For \(i = 1\): \((791 - 5295.333)(4582-11205.833)=(- 4504.333)(-6623.833)\approx29837777.7\)
• For \(i = 2\): \((3646-5295.333)(6697 - 11205.833)=(-1649.333)(-4508.833)\approx7436777.7\)
• For \(i = 3\): \((5821-5295.333)(7106 - 11205.833)=(525.667)(-4100.833)\approx - 2156777.7\)
• For \(i = 4\): \((6040-5295.333)(18852-11205.833)=(744.667)(7646.167)\approx5696777.7\)
• For \(i = 5\): \((7622-5295.333)(14041 - 11205.833)=(2326.667)(2835.167)\approx6608777.7\)
• For \(i = 6\): \((7852-5295.333)(15957-11205.833)=(2556.667)(4751.167)\approx12147777.7\)

\(\sum_{i = 1}^{6}(x_{i}-\bar{x})(y_{i}-\bar{y})\approx29837777.7+7436777.7-2156777.7 + 5696777.7+6608777.7+12147777.7\approx60571000\)

Calculate \((x_{i}-\bar{x})^{2}\) for each \(i\):

• For \(i = 1\): \((791 - 5295.333)^{2}=(-4504.333)^{2}\approx20289000\)
• For \(i = 2\): \((3646-5295.333)^{2}=(-1649.333)^{2}\approx2710000\)
• For \(i = 3\): \((5821-5295.333)^{2}=(525.667)^{2}\approx276000\)
• For \(i = 4\): \((6040-5295.333)^{2}=(744.667)^{2}\approx554000\)
• For \(i = 5\): \((7622-5295.333)^{2}=(2326.667)^{2}\approx5413000\)
• For \(i = 6\): \((7852-5295.333)^{2}=(2556.667)^{2}\approx6536000\)

\(\sum_{i = 1}^{6}(x_{i}-\bar{x})^{2}\approx20289000+2710000 + 276000+554000+5413000+6536000\approx35778000\)

Calculate \((y_{i}-\bar{y})^{2}\) for each \(i\):

• For \(i = 1\): \((4582-11205.833)^{2}=(-6623.833)^{2}\approx43875000\)
• For \(i = 2\): \((6697-11205.833)^{2}=(-4508.833)^{2}\approx20329000\)
• For \(i = 3\): \((7106-11205.833)^{2}=(-4100.833)^{2}\approx16817000\)
• For \(i = 4\): \((18852-11205.833)^{2}=(7646.167)^{2}\approx58464000\)
• For \(i = 5\): \((14041-11205.833)^{2}=(2835.167)^{2}\approx8038000\)
• For \(i = 6\): \((15957-11205.833)^{2}=(4751.167)^{2}\approx22574000\)

\(\sum_{i = 1}^{6}(y_{i}-\bar{y})^{2}\approx43875000+20329000+16817000+58464000+8038000+22574000\approx169097000\)

\(\sqrt{\sum_{i = 1}^{6}(x_{i}-\bar{x})^{2}\sum_{i = 1}^{6}(y_{i}-\bar{y})^{2}}=\sqrt{35778000\times169097000}\approx\sqrt{6.057\times10^{12}}\approx77830000\)

Step3: Calculate the correlation coefficient \(r\)

\(r=\frac{60571000}{77830000}\approx0.778\)

Answer:

\(0.778\)