Example: Using a Scatter Plot to Investigate Cricket Chirps

The table below shows the number of cricket chirps in 15 seconds, for several different air temperatures, in degrees Fahrenheit.[footnote]Selected data from http://classic.globe.gov/fsl/scientistsblog/2007/10/. Retrieved Aug 3, 2010[/footnote] Plot this data, and determine whether the data appears to be linearly related.

Chirps	44	35	20.4	33	31	35	18.5	37	26
Temperature	80.5	70.5	57	66	68	72	52	73.5	53

Answer: Plotting this data suggests that there may be a trend. We can see from the trend in the data that the number of chirps increases as the temperature increases. The trend appears to be roughly linear, though certainly not perfectly so.

Example: Finding a Line of Best Fit

Find a linear function that fits the data in the table below by "eyeballing" a line that seems to fit.

Chirps	44	35	20.4	33	31	35	18.5	37	26
Temperature	80.5	70.5	57	66	68	72	52	73.5	53

Answer: On a graph, we could try sketching a line. Using the starting and ending points of our hand drawn line, points (0, 30) and (50, 90), this graph has a slope of [latex]m=\frac{60}{50}=1.2[/latex] and a y-intercept at 30. This gives an equation of [latex]T\left(c\right)=1.2c+30[/latex] where c is the number of chirps in 15 seconds, and T(c) is the temperature in degrees Fahrenheit. The resulting equation is represented in the graph below.

Analysis of the Solution

This linear equation can then be used to approximate answers to various questions we might ask about the trend.

Example: Understanding Interpolation and Extrapolation

Chirps	44	35	20.4	33	31	35	18.5	37	26
Temperature	80.5	70.5	57	66	68	72	52	73.5	53

Use the cricket data above to answer the following questions:

1. Would predicting the temperature when crickets are chirping 30 times in 15 seconds be interpolation or extrapolation? Make the prediction, and discuss whether it is reasonable.
2. Would predicting the number of chirps crickets will make at 40 degrees be interpolation or extrapolation? Make the prediction, and discuss whether it is reasonable.

Answer:

1. The number of chirps in the data provided varied from 18.5 to 44. A prediction at 30 chirps per 15 seconds is inside the domain of our data, so would be interpolation. Using our model: [latex-display]\begin{array}{l}T\left(30\right)=30+1.2\left(30\right)\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=66\text{ degrees}\hfill \end{array}[/latex-display] Based on the data we have, this value seems reasonable.
2. The temperature values varied from 52 to 80.5. Predicting the number of chirps at 40 degrees is extrapolation because 40 is outside the range of our data. Using our model: [latex]\begin{array}{l}40=30+1.2c\hfill \\ 10=1.2c\hfill \\ c\approx 8.33\hfill \end{array}[/latex]

We can compare the regions of interpolation and extrapolation using the graph below.

Analysis of the Solution

Our model predicts the crickets would chirp 8.33 times in 15 seconds. While this might be possible, we have no reason to believe our model is valid outside the domain and range. In fact, generally crickets stop chirping altogether below around 50 degrees.

Example: Finding a Least Squares Regression Line

Find the least squares regression line using the cricket-chirp data in the table below.

Chirps	44	35	20.4	33	31	35	18.5	37	26
Temperature	80.5	70.5	57	66	68	72	52	73.5	53

Answer:  

1. Click the plus button (add item) in the upper left corner, and select table.
2. Enter chirps data in the x1 column.
3. Enter temperature data in the y1 column.

   x1	44	35	20.4	33	31	35	18.5	37	26
   y1	80.5	70.5	57	66	68	72	52	73.5	53

4. If you can't see the points on the grid, use the plus and minus buttons in the upper right hand corner to zoom in or out on the grid, or click on the wrench and change the upper bound of x1 to 60, and 100 for the y1.
5. In the empty cell below the table you created, enter the expression y1~mx1+b
6. You can add labels to your graph by clicking on the wrench in the upper right hand corner, and typing them into the cells that say "add a label"

Here is an example of how your graph may look: https://www.desmos.com/calculator/ruvzg6iy3o

Analysis of the Solution

Notice that this line is quite similar to the equation we "eyeballed" but should fit the data better. Notice also that using this equation would change our prediction for the temperature when hearing 30 chirps in 15 seconds from 66 degrees to:

[latex]\begin{array}{l}T\left(30\right)=30.281+1.143\left(30\right)\hfill \\ \text{ }=64.571\hfill \\ \text{ }\approx 64.6\text{ degrees}\hfill \end{array}[/latex]

Example: Finding a Correlation Coefficient

Calculate the correlation coefficient for cricket-chirp data in the table below.

Chirps	44	35	20.4	33	31	35	18.5	37	26
Temperature	80.5	70.5	57	66	68	72	52	73.5	53

Answer: Desmos provides you with the correlation coefficient when you use it to calculate a linear regression. The correlation coefficients is labeled as r = 0.951 for this dataset. This value is very close to 1, which suggests a strong increasing linear relationship. https://www.desmos.com/calculator/ruvzg6iy3o

Example: Using a Regression Line to Make Predictions

Gasoline consumption in the United States has been steadily increasing. Consumption data from 1994 to 2004 is shown in the table below.[footnote]http://www.bts.gov/publications/national_transportation_statistics/2005/html/table_04_10.html[/footnote] Determine whether the trend is linear, and if so, find a model for the data. Use the model to predict the consumption in 2008.Is this an interpolation or an extrapolation?

Year	'94	'95	'96	'97	'98	'99	'00	'01	'02	'03	'04
Consumption (billions of gallons)	113	116	118	119	123	125	126	128	131	133	136

Answer: We can introduce new input variable, t, representing years since 1994, this makes entering the data into Desmos easier. Read the value for b, and the value for the slope, m, from Desmos to create the equation for the regression line:

[latex]C\left(t\right)=113.318+2.209t[/latex]

The correlation coefficient was calculated to be 0.997, suggesting a very strong increasing linear trend. Using this to predict consumption in 2008, which is 14 years after 1994,  [latex]\left(t=14\right)[/latex],

[latex]\begin{array}{l}C\left(14\right)=113.318+2.209\left(14\right)\hfill \\ =144.244\hfill \end{array}[/latex]

The model predicts 144.244 billion gallons of gasoline consumption in 2008. This is an extrapolation because there is not a datapoint whose x1 value is 2008. The scatter plot of the data, including the least squares regression line, is shown below in Desmos. Note how we changed the viewing window for the y-axis to 100 < y < 150. https://www.desmos.com/calculator/lv27pmtdbh