From the
Exploring Data website - http://curriculum.qed.qld.gov.au/kla/eda/
© Education Queensland, 1997
Using Statistics in Human Movements
One measure of form for a runner is stride rate, defined as the number of steps per second. A runner is considered to be efficient if the stride rate is close to optimum. The stride rate is related to speed; the greater the speed, the greater the stride rate.
In a study of 21 top female runners, researchers measured the stride rate for different speeds. The following table gives the average stride rate of these women versus the speed.
Source: R.C. Nelson, C.M. Brooks, and N.L. Pike, Biomechanical comparison of male and female runners, in P. Milvy (ed.), The Marathon: Physiological, Medical, Epistemiological, and Psychological Studies, New York Academy of Sciences, 1977, pp. 793-807.
| Speed | 15.86 |
16.88 |
17.50 |
18.62 |
19.97 |
21.06 |
22.11 |
| Stride Rate | 3.05 |
3.12 |
3.17 |
3.25 |
3.36 |
3.46 |
3.55 |
1. Plot the data on a scattergram. Decide if the data appears to be linear.
2. Find the equation of the least squares regression line.
3. Interpret the gradient of this function. In other word, what are the units on the gradient?
4. Plot the least squares regression line on the scattergram.
5. Make a prediction of the stride rate if the speed is 19 feet per second.
6. What is the stride rate if the speed is 0? Interpret your result. Do the results make sense? What should be done about this?
Extension
Do a residual plot of the data. Do you notice any patterns? Can you find a better mathematical model for this data?
Using Statistics in Human Movements - A Solution Using the TI-83 Graphical Calculator
1. Enter the following data into a Statistics Lists, say L1 and L2.
| Speed | 15.86 |
16.88 |
17.50 |
18.62 |
19.97 |
21.06 |
22.11 |
| Stride Rate | 3.05 |
3.12 |
3.17 |
3.25 |
3.36 |
3.46 |
3.55 |
| 2.
Construct a scatterplot from this data. |
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| 3.
Use (Zoom, 9:ZoomStat) to resize the scatterplot window
automatically. The data appears to follow a linear relation quite nicely. |
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| 4.
To get the regression equation, press:
(Stat,Calc,4:LinReg(ax+b)), then (L1, L2,
Vars,Y-Vars,1:Function,1:Y1, Enter). This gives the gradient and y-intercept of the regression equation. In this example we are finding the relationship between two rates so the units on the gradient are a bit tricky. The units are (steps per second) per (feet per second). Real data can be messy! 5. Use Graph to see the scatterplot and the graph of the regression line. The regression line fits the data extremely well. |
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| 6.
To use the fitted values to predict the stride rate for a
speed of 19 feet per second, press (Vars,Y-Vars,
Function,Y1,(19),Enter). The predicted stride rate for a speed of 19 feet per second is 3.29. The predicted stride rate for a speed of 0 feet per second is 1.766, which is silly! Obviously, the model doesnt work very effectively for all speeds. A decision needs to be made about the domain for which the model is valid. |
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Extension
| 7.
Extension - Fitting residuals to the scatterplot. The residual for each data value is found by the formula Residual = Data Value - Fitted Value A scatterplot of residuals helps to determine the validity of the linear model. Ideally, the residuals should show no strong patterns. To get a list of the residuals, put the cursor at the top of list L3, and press (List,RESID,Enter). Now do a scatterplot of List L3 verus List L1. The scatterplot shows the data has a strongly curved pattern. So another model, say a quadratic one, may give a slightly better fit. Should a more complicated model be used? In this case there is no single correct answer to this question.. A quadratic function may be slightly more accurate, but with a loss of simplicity. In this case, the linear model is certainly acceptable and would have a correlation coefficient close to 1. My decision would be to use the linear model, and note in a report that the residual plot does have a strong underlying pattern, and that this indicates that the data might better be modelled by a non-linear function. |
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