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The data show the bug chirps per minute at different temperatures. Find the regression equation, letting the first variable be the independent (x) variable. Find the best predicted temperature for a time when a bug is chirping at the rate of 3000 chirps per minute. Use a significance level of 0.05. What is wrong with this predicted value? Chirps in 1 min Temperature (F) 894 965 83 856 949 1233 69.9 81 74.3 77 75 88.3 What is the regression equation? (Round the x-coefficient to four decimal places as needed. Round the constant to two decimal places as needed.)

math   Statistics-And-Probability   
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Answer #1
X Y XY X2
894 69.9 62490.6 799236
965 81 78165 931225
893 74.3 66349.9 797449
856 77 65912 732736
949 75 71175 900601
1233 88.3 108873.9 1520289
Total = 5790 465.5 452966.4 5681536

a=rac{sum Ysum X^{2}-sum Xsum XY}{nsum X^{2}-(sum X)^{2}}

(465.5 × 568 1536)-(5790 × 452966.4) (6 × 568 1536)-57902 39.071

b=rac{nsum XY-sum Xsum Y}{nsum X^{2}-(sum X)^{2}}

=rac{(6 imes 452966.4)-(5790 imes 465.5)}{(6 imes 5681536)-5790^{2}}=0.0399

So,

the Regression Equation is:

y = 39.0T+ 0.03992.

For x = 3000:

hat{y}=39.07+(0.0399 imes 3000)=358.7980

The wrong in the predicted value is that x = 3000 is an extrapolation i.e., beyond the given range of values of x and so the predicted value is not reliable.

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