Line of Best Fit

Exercises

1. Exercise 1 Supplied answer

   Use least-squares to judge if the coin in this experiment is fair.

   flips	8	16	24	32	40
   heads	4	9	13	17	20

   Back to Exercise 1

   Answer. As with the first example discussed above, we are trying to find a best $m$ to “solve” this system.

   $$\begin{array}{rcr}8m& =& 4\\ 16m& =& 9\\ 24m& =& 13\\ 32m& =& 17\\ 40m& =& 20\end{array}$$

   Projecting into the linear subspace gives this

   $$\frac{\left(\begin{array}{c}4\\ 9\\ 13\\ 17\\ 20\end{array}\right)\cdot \left(\begin{array}{c}8\\ 16\\ 24\\ 32\\ 40\end{array}\right)}{\left(\begin{array}{c}8\\ 16\\ 24\\ 32\\ 40\end{array}\right)\cdot \left(\begin{array}{c}8\\ 16\\ 24\\ 32\\ 40\end{array}\right)}\cdot \left(\begin{array}{c}8\\ 16\\ 24\\ 32\\ 40\end{array}\right)=\frac{1832}{3520}\cdot \left(\begin{array}{c}8\\ 16\\ 24\\ 32\\ 40\end{array}\right)$$

   so the slope of the line of best fit is approximately $0.52$.

   

2. Exercise 2 Supplied answer

   For the men’s mile record, rather than give each of the many records and its exact date, we’ve “smoothed” the data somewhat by taking a periodic sample. Do the longer calculation and compare the conclusions.

   Back to Exercise 2

   Answer. With this input

   $$A=\left(\begin{array}{cc}1& 1852.71\\ 1& 1858.88\\ ⋮& ⋮\\ 1& 1985.54\\ 1& 1993.71\end{array}\right)b=\left(\begin{array}{c}292.0\\ 285.0\\ ⋮\\ 226.32\\ 224.39\end{array}\right)$$

   (the dates have been rounded to months, e.g., for a September record, the decimal $.71\approx (8.5/12)$ was used), Maple responded with an intercept of $b=994.8276974$ and a slope of $m=-0.3871993827$.

   

3. Exercise 3 Supplied answer

   Find the line of best fit for the men’s $1500$ meter run. How does the slope compare with that for the men’s mile? (The distances are close; a mile is about $1609$ meters.)

   Back to Exercise 3

   Answer. With this input (the years are zeroed at $1900$)

   $$A:=\left(\begin{array}{cc}1& .38\\ 1& .54\\ ⋮⋮\\ 1& 92.71\\ 1& 95.54\end{array}\right)b=\left(\begin{array}{c}249.0\\ 246.2\\ ⋮\\ 208.86\\ 207.37\end{array}\right)$$

   (the dates have been rounded to months, e.g., for a September record, the decimal $.71\approx (8.5/12)$ was used), Maple gives an intercept of $b=243.1590327$ and a slope of $m=-0.401647703$. The slope given in the body of this Topic for the men’s mile is quite close to this.

   

4. Exercise 4 Supplied answer

   Find the line of best fit for the records for women’s mile.

   Back to Exercise 4

   Answer. With this input (the years are zeroed at $1900$)

   $$A=\left(\begin{array}{cc}1& 21.46\\ 1& 32.63\\ ⋮& ⋮\\ 1& 89.54\\ 1& 96.63\end{array}\right)b=\left(\begin{array}{c}373.2\\ 327.5\\ ⋮\\ 255.61\\ 252.56\end{array}\right)$$

   (the dates have been rounded to months, e.g., for a September record, the decimal $.71\approx (8.5/12)$ was used), MAPLE gave an intercept of $b=378.7114894$ and a slope of $m=-1.445753225$.

   

5. Exercise 5 Supplied answer

   Do the lines of best fit for the men’s and women’s miles cross?

   Back to Exercise 5

   Answer. These are the equations of the lines for men’s and women’s mile (the vertical intercept term of the equation for the women’s mile has been adjusted from the answer above, to zero it at the year $0$, because that’s how the men’s mile equation was done).

   $$\begin{array}{rl}y& =994.8276974-0.3871993827x\\ y& =3125.6426-1.445753225x\end{array}$$

   Obviously the lines cross. A computer program is the easiest way to do the arithmetic: MuPAD gives $x=2012.949004$ and $y=215.4150856$ ($215$ seconds is $3$ minutes and $35$ seconds). Remark. Of course all of this projection is highly dubious —for one thing, the equation for the women is influenced by the quite slow early times —but it is nonetheless fun.

   

6. Exercise 6 Supplied answer

   (This illustrates that there are data sets for which a linear model is not right, and that the line of best fit doesn’t in that case have any predictive value.) In a highway restaurant a trucker told me that his boss often sends him by a roundabout route, using more gas but paying lower bridge tolls. He said that New York State calibrates the toll for each bridge across the Hudson, playing off the extra gas to get there from New York City against a lower crossing cost, to encourage people to go upstate. This table, from [Cost Of Tolls] and [Google Maps], lists for each toll crossing of the Hudson River, the distance to drive from Times Square in miles and the cost in US dollars for a passenger car (if a crossings has a one-way toll then it shows half that number).

   Crossing	Distance	Toll
   Lincoln Tunnel Holland Tunnel George Washington Bridge Verrazano-Narrows Bridge Tappan Zee Bridge Bear Mountain Bridge Newburgh-Beacon Bridge Mid-Hudson Bridge Kingston-Rhinecliff Bridge Rip Van Winkle Bridge	$2$ $7$ $8$ $16$ $27$ $47$ $67$ $82$ $102$ $120$	$6.00$ $6.00$ $6.00$ $6.50$ $2.50$ $1.00$ $1.00$ $1.00$ $1.00$ $1.00$

   Find the line of best fit and graph the data to show that the driver was practicing on my credulity.

   Back to Exercise 6

   Answer. Sage gives the line of best fit as $\text{toll}=-0.05\cdot \text{dist}+5.63$.

   sage: dist = [2, 7, 8, 16, 27, 47, 67, 82, 102, 120]
   sage: toll = [6, 6, 6, 6.5, 2.5, 1, 1, 1, 1, 1]
   sage: var('a,b,t')
   (a, b, t)
   sage: model(t) = a*t+b
   sage: data = zip(dist,toll)
   sage: fit = find_fit(data, model, solution_dict=True)
   sage: model.subs(fit)
   t |--> -0.0508568169130319*t + 5.630955848442933
   sage: p = plot(model.subs(fit), (t,0,120))+points(data,size=25,color='red')
   sage: p.save('bridges.pdf')        

   But the graph shows that the equation has little predictive value.

   

   Apparently a better model is that (with only one intermediate exception) crossings in the city cost roughly the same as each other, and crossings upstate cost the same as each other.

7. Exercise 7 Supplied answer

   When the space shuttle Challenger exploded in 1986, one of the criticisms made of NASA’s decision to launch was in the way they did the analysis of number of O-ring failures versus temperature (O-ring failure caused the explosion). Four O-ring failures would be fatal. NASA had data from $24$ previous flights.

   temp ${}^{\circ }$F	53	75	57	58	63	70	70	66	67	67	67
   failures	3	2	1	1	1	1	1	0	0	0	0

   
   

   68	69	70	70	72	73	75	76	76	78	79	80	81
   0	0	0	0	0	0	0	0	0	0	0	0	0

   The temperature that day was forecast to be ${31}^{\circ }\text{F}$.

   1. NASA based the decision to launch partially on a chart showing only the flights that had at least one O-ring failure. Find the line that best fits these seven flights. On the basis of this data, predict the number of O-ring failures when the temperature is $31$, and when the number of failures will exceed four.

   2. Find the line that best fits all 24 flights. On the basis of this extra data, predict the number of O-ring failures when the temperature is $31$, and when the number of failures will exceed four.

   Which do you think is the more accurate method of predicting? (An excellent discussion is in [Dalal, et. al.].)

   Back to Exercise 7

   Answer.

   1. A computer algebra system like MAPLE or MuPAD will give an intercept of $b=4259/1398\approx 3.239628$ and a slope of $m=-71/2796\approx -0.025393419$ Plugging $x=31$ into the equation yields a predicted number of O-ring failures of $y=2.45$ (rounded to two places). Plugging in $y=4$ and solving gives a temperature of $x=-{29.94}^{\circ }$F.

   2. On the basis of this information

      $$A=\left(\begin{array}{cc}1& 53\\ 1& 75\\ ⋮\\ 1& 80\\ 1& 81\end{array}\right)b=\left(\begin{array}{c}3\\ 2\\ ⋮\\ 0\\ 0\end{array}\right)$$

      MAPLE gives the intercept $b=187/40=4.675$ and the slope $m=-73/1200\approx -0.060833$. Here, plugging $x=31$ into the equation predicts $y=2.79$ O-ring failures (rounded to two places). Plugging in $y=4$ failures gives a temperature of $x={11}^{\circ }$F.

      

8. Exercise 8 Supplied answer

   This table lists the average distance from the sun to each of the first seven planets, using Earth’s average as a unit.

   Mercury	Venus	Earth	Mars	Jupiter	Saturn	Uranus
   $0.39$	$0.72$	$1.00$	$1.52$	$5.20$	$9.54$	$19.2$

   1. Plot the number of the planet (Mercury is $1$, etc.) versus the distance. Note that it does not look like a line, and so finding the line of best fit is not fruitful.

   2. It does, however look like an exponential curve. Therefore, plot the number of the planet versus the logarithm of the distance. Does this look like a line?

   3. The asteroid belt between Mars and Jupiter is what is left of a planet that broke apart. Renumber so that Jupiter is $6$, Saturn is $7$, and Uranus is $8$, and plot against the log again. Does this look better?

   4. Use least squares on that data to predict the location of Neptune.

   5. Repeat to predict where Pluto is.

   6. Is the formula accurate for Neptune and Pluto?

   This method was used to help discover Neptune (although the second item is misleading about the history; actually, the discovery of Neptune in position $9$ prompted people to look for the “missing planet” in position $5$). See [Gardner, 1970]

   Back to Exercise 8

   Answer.

   1. The plot is nonlinear.

      

   2. Here is the plot.

      

      There is perhaps a jog up between planet $4$ and planet $5$.

   3. This plot seems even more linear.

      

   4. With this input

      $$A=\left(\begin{array}{cc}1& 1\\ 1& 2\\ 1& 3\\ 1& 4\\ 1& 6\\ 1& 7\\ 1& 8\end{array}\right)b=\left(\begin{array}{c}-0.40893539\\ -0.1426675\\ 0\\ 0.18184359\\ 0.71600334\\ 0.97954837\\ 1.2833012\end{array}\right)$$

      MuPAD gives that the intercept is $b=-0.6780677466$ and the slope is $m=0.2372763818$.

   5. Plugging $x=9$ into the equation $y=-0.6780677466+0.2372763818x$ from the prior item gives that the log of the distance is $1.4574197$, so the expected distance is $28.669472$. The actual distance is about $30.003$.

   6. Plugging $x=10$ into the same equation gives that the log of the distance is $1.6946961$, so the expected distance is $49.510362$. The actual distance is about $39.503$.

9. Exercise 9 Supplied answer

   Suppose that $W$ is a subspace of ${\mathrm{ℝ}}^n$ for some $n$ and suppose that $\overrightarrow{v}$ is not an element of $W$. Let the orthogonal projection of $\overrightarrow{v}$ into $W$ be the vector ${\text{proj}}_W(\overrightarrow{v})=\overrightarrow{p}$. Show that $\overrightarrow{p}$ is the element of $W$ that is closest to $\overrightarrow{v}$.

   Back to Exercise 9

   Answer. For any $\overrightarrow{w}\in W$, the vectors $\overrightarrow{v}-\overrightarrow{p}$ and $\overrightarrow{p}-\overrightarrow{w}$ are orthogonal. So the Triangle Inequality applies to the triangle with those vectors as sides, and with $\overrightarrow{v}-\overrightarrow{w}$ as hypoteneuse. Therefore $\overrightarrow{v}-\overrightarrow{w}$ is at least as long as $\overrightarrow{v}-\overrightarrow{p}$.