Question

lesson app 3.4 flying dinosaur or early bird? archaeopteryx is an extinct beast having feathers like a bird, but teeth and a long bony tail like a reptile. because the known specimens differ greatly in size, some scientists think they are different species rather than individuals from the same species. however, if the specimens belong to the same species and differ in size because some are younger than others, there should be a positive linear relationship between the lengths of a pair of bones from all individuals. an outlier from this relationship would suggest a different species. here are data on the lengths (in centimeters) of the femur (a leg - bone) and the humerus (a bone in the upper arm) for five specimens that preserve both bones. length of femur (cm) 38 56 59 64 74 length of humerus (cm) 41 63 70 72 84 1. make a scatterplot using length of femur as the explanatory variable. do you think that all five specimens come from the same species? explain your reasoning. 2. find the correlation r. explain how your value for r matches your graph in question 1. 3. suppose that a new fossil was discovered. if the femur is 70 centimeters and the humerus is 40 centimeters, do you think this specimen came from the same species? explain your reasoning. 4. what effect will the new fossil have on the correlation? explain your reasoning.

Explanation:

Step1: Making scatter - plot

We plot the length of the femur (x - values: 38, 56, 59, 64, 74) on the x - axis and the length of the humerus (y - values: 41, 63, 70, 72, 84) on the y - axis. If all specimens are from the same species, we expect a positive linear relationship. Visually, the points seem to follow a positive linear trend, suggesting they may be from the same species.

Step2: Calculating correlation coefficient $r$

The formula for the correlation coefficient $r$ is $r=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}}$
First, calculate the means:
$\bar{x}=\frac{38 + 56+59+64+74}{5}=\frac{291}{5}=58.2$
$\bar{y}=\frac{41 + 63+70+72+84}{5}=\frac{330}{5}=66$
Then calculate the numerator and denominators:
$\sum_{i = 1}^{5}(x_{i}-\bar{x})(y_{i}-\bar{y})=(38 - 58.2)(41 - 66)+(56 - 58.2)(63 - 66)+(59 - 58.2)(70 - 66)+(64 - 58.2)(72 - 66)+(74 - 58.2)(84 - 66)$
$=(- 20.2)\times(-25)+(-2.2)\times(-3)+0.8\times4 + 5.8\times6+15.8\times18$
$=505+6.6 + 3.2+34.8+284.4$
$=834$
$\sum_{i = 1}^{5}(x_{i}-\bar{x})^{2}=(38 - 58.2)^{2}+(56 - 58.2)^{2}+(59 - 58.2)^{2}+(64 - 58.2)^{2}+(74 - 58.2)^{2}$
$=(-20.2)^{2}+(-2.2)^{2}+0.8^{2}+5.8^{2}+15.8^{2}$
$=408.04+4.84 + 0.64+33.64+249.64$
$=706.8$
$\sum_{i = 1}^{5}(y_{i}-\bar{y})^{2}=(41 - 66)^{2}+(63 - 66)^{2}+(70 - 66)^{2}+(72 - 66)^{2}+(84 - 66)^{2}$
$=(-25)^{2}+(-3)^{2}+4^{2}+6^{2}+18^{2}$
$=625+9+16+36+324$
$=1010$
$r=\frac{834}{\sqrt{706.8\times1010}}\approx\frac{834}{\sqrt{713868}}\approx\frac{834}{844.91}\approx0.99$
A value of $r\approx0.99$ indicates a very strong positive linear relationship, which is consistent with the scatter - plot.

Step3: Assessing new specimen

For the new specimen with $x = 70$ and $y = 40$, if it is from the same species, it should follow the existing linear relationship. But the existing relationship has a strong positive trend. Plotting this point $(70,40)$ on the scatter - plot, it is far below the trend line of the previous points. So, it is likely not from the same species.

Step4: Effect on correlation

The new point $(70,40)$ is an outlier. Since it does not follow the positive linear trend of the other points, it will decrease the value of the correlation coefficient $r$. The outlier will pull the regression line in its direction and weaken the overall linear relationship.

Answer:

1. After making the scatter - plot, the points seem to follow a positive linear trend, suggesting they may be from the same species.
2. $r\approx0.99$, which indicates a very strong positive linear relationship, consistent with the scatter - plot.
3. No, because the point $(70,40)$ does not follow the positive linear trend of the existing points.
4. It will decrease the correlation coefficient $r$ as it is an outlier that weakens the overall linear relationship.