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QuestionDetermine the correlation coefficient rr, regression line, and predict weight for a 124-inch alligator. Is it valid for 20 inches?

Studdy Solution

STEP 1

Assumptions1. The length of the alligator is represented by the variable xx. . The weight of the alligator is represented by the variable yy.
3. We are assuming a linear relationship between the length and weight of the alligators.
4. We have the following data points\begin{tabular}{|c|c|} \hline xx & yy \\ \hline119 &675 \\ \hline112 &671 \\ \hline105 &551 \\ \hline107 &661 \\ \hline72 &410 \\ \hline118 &573 \\ \hline88 &544 \\ \hline77 &429 \\ \hline129 &672 \\ \hline97 &560 \\ \hline\end{tabular}

STEP 2

First, we need to calculate the means of xx and yy.
xˉ=∑xn\bar{x} = \frac{\sum x}{n}yˉ=∑yn\bar{y} = \frac{\sum y}{n}where nn is the number of data points.

STEP 3

Plug in the values for xx and yy to calculate the means.
xˉ=119+112+105+107+72+118+88+77+129+9710\bar{x} = \frac{119 +112 +105 +107 +72 +118 +88 +77 +129 +97}{10}yˉ=675+671+551+661+410+573+544+429+672+56010\bar{y} = \frac{675 +671 +551 +661 +410 +573 +544 +429 +672 +560}{10}

STEP 4

Calculate the means.
xˉ=102610=102.6\bar{x} = \frac{1026}{10} =102.6yˉ=574610=574.6\bar{y} = \frac{5746}{10} =574.6

STEP 5

Next, we need to calculate the standard deviations of xx and yy.
sx=∑(x−xˉ)2n−1s_x = \sqrt{\frac{\sum (x - \bar{x})^2}{n -1}}sy=∑(y−yˉ)2n−1s_y = \sqrt{\frac{\sum (y - \bar{y})^2}{n -1}}

STEP 6

Plug in the values for xx, yy, xˉ\bar{x}, and yˉ\bar{y} to calculate the standard deviations.
sx=(119−102.6)2+(112−102.6)2+(105−102.6)2+(107−102.6)2+(72−102.6)2+(118−102.6)2+(88−102.6)2+(77−102.6)2+(129−102.6)2+(97−102.6)29s_x = \sqrt{\frac{(119 -102.6)^2 + (112 -102.6)^2 + (105 -102.6)^2 + (107 -102.6)^2 + (72 -102.6)^2 + (118 -102.6)^2 + (88 -102.6)^2 + (77 -102.6)^2 + (129 -102.6)^2 + (97 -102.6)^2}{9}}sy=(675−574.6)2+(671−574.6)2+(551−574.6)2+(661−574.6)2+(410−574.6)2+(573−574.6)2+(544−574.6)2+(429−574.6)2+(672−574.6)2+(560−574.6)29s_y = \sqrt{\frac{(675 -574.6)^2 + (671 -574.6)^2 + (551 -574.6)^2 + (661 -574.6)^2 + (410 -574.6)^2 + (573 -574.6)^2 + (544 -574.6)^2 + (429 -574.6)^2 + (672 -574.6)^2 + (560 -574.6)^2}{9}}

STEP 7

Calculate the standard deviations.
sx=3202.49=19.95s_x = \sqrt{\frac{3202.4}{9}} =19.95sy=23556.49=51.37s_y = \sqrt{\frac{23556.4}{9}} =51.37

STEP 8

Now, we can calculate the correlation coefficient rr.
r=∑(x−xˉ)(y−yˉ)(n−1)sxsyr = \frac{\sum (x - \bar{x})(y - \bar{y})}{(n -1)s_xs_y}

STEP 9

Plug in the values for xx, yy, xˉ\bar{x}, yˉ\bar{y}, sxs_x, and sys_y to calculate rr.
r=(119−102.6)(675−574.6)+(112−102.6)(671−574.6)+(105−102.6)(551−574.6)+(107−102.6)(661−574.6)+(72−102.6)(410−574.6)+(118−102.6)(573−574.6)+(88−102.6)(544−574.6)+(77−102.6)(429−574.6)+(129−102.6)(672−574.6)+(97−102.6)(560−574.6)9×19.95×51.37r = \frac{(119 -102.6)(675 -574.6) + (112 -102.6)(671 -574.6) + (105 -102.6)(551 -574.6) + (107 -102.6)(661 -574.6) + (72 -102.6)(410 -574.6) + (118 -102.6)(573 -574.6) + (88 -102.6)(544 -574.6) + (77 -102.6)(429 -574.6) + (129 -102.6)(672 -574.6) + (97 -102.6)(560 -574.6)}{9 \times19.95 \times51.37}

STEP 10

Calculate rr.
r=0.922r =0.922

STEP 11

Now, we can find the equation of the regression line.
y=a+bxy = a + bxwhere a=yˉ−bxˉa = \bar{y} - b\bar{x} and b=rsysxb = r \frac{s_y}{s_x}.

STEP 12

First, calculate bb.
b=rsysx=0.92251.3719.95b = r \frac{s_y}{s_x} =0.922 \frac{51.37}{19.95}

STEP 13

Calculate bb.
b=2.37b =2.37

STEP 14

Next, calculate aa.
a=yˉ−bxˉ=574.6−2.37×102.6a = \bar{y} - b\bar{x} =574.6 -2.37 \times102.6

STEP 15

Calculate aa.
a=−175.74a = -175.74

STEP 16

So, the equation of the regression line isy=−175.74+2.37xy = -175.74 +2.37x

STEP 17

Now, we can use this equation to predict the weight of an alligator that is124 inches long.
y=−175.74+2.37×124y = -175.74 +2.37 \times124

STEP 18

Calculate yy.
y=117.86y =117.86So, an alligator that is124 inches long is predicted to weigh117.86 pounds.

STEP 19

To answer the last question, it would not be meaningful to use this model to predict the weight of an alligator that is inches long because the data we have does not include alligators of that length. This is an extrapolation beyond the range of our data, and it may not be accurate.

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