10. The data show the bug chir
10. The data show the bug chirps per minute at differenttemperatures. Find the regression equation, letting the firstvariable be the independent (x) variable. Find the best predictedtemperature for a time when a bug is chirping at the rate of 3000chirps per minute. Use a significance level of 0.05. What is wrongwith this predicted value?
Chirps in 1 min Temperature (°F)
985 83.8
1087 85.2
867 74.7
1060 83.2
764 64.9
807 65.5
What is the regression equation?
Ŷ=____+____x
(Round the x-coefficient to four decimal places as needed.Round the constant to two decimal places as needed.)
What is the best predicted temperature for a time when a bug ischirping at the rate of 3000 chirps per minute?
The best predicted temperature when a bug is chirping at 3000chirps per minute is ____F.
(Round to one decimal place as needed.)
What is wrong with this predicted value? Choose the correctanswer below.
A.It is unrealistically high. The value 3000 is far outside ofthe range of observed values.
B.The first variable should have been the dependentvariable.
C.It is only an approximation. An unrounded value would beconsidered accurate.
D.Nothing is wrong with this value. It can be treated as anaccurate prediction
Answer:
This is a simple problem related to formulation of a regressionequation based on the trend of the independent and dependent data.
|
Chirps/min (x) |
Temp(°F) (y) |
| 985 | 83.8 |
| 1087 | 85.2 |
| 867 | 74.7 |
| 1060 | 83.2 |
| 764 | 64.9 |
| 807 | 65.5 |
Thus the best fit linear regression equation is given by
where x denotes chirps/min and y represents the temperature.
Now think !
What can we do with this regression equation ?
We can easily predict the value of y (temperature) based on theactual value of x (chirps/min)
Thus if we have a chirps/min value of 3000 /min then thecorresponding pricited temperature in fahrenheit will be givenby
y =14.8691+0.0661*3000
or, y =213.2 °F
Now think again !
We had predicted the value for a given chirp value . But look atthe value !
It is 3000 chirps/min
and look at the range of the value of x over which we havemodelled our regression equation .
It is (807-1087)
Now observe the chirp rate of 3000
This is way ahead of our range of regression. Hence we may notbe absolutely correct with the predicted value at such a far waypoint from the range.
This will thus give us only an vague insight into the potentialtemperature .
But this cant be a very close value in comparison to actualtemperature value at 3000 chirps/min.
The error (Actual -observed) might be very large.
Hence Option (A).
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