Question

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: 955 1190 1206 804 1230 770

Temp F : 82.5 86.3 87.4 74.7 90.9 70.2

A) What is the regression​ equation?

Y= ?

B) What is the best predicted temperature for a time when a bug is chirping at the rate of

3000 chirps per​ minute?

C) What is wrong with this predicted​ value? Choose the correct answer below.

a) It is only an approximation. An unrounded value would be considered accurate.

b) The first variable should have been the dependent variable.

c) It is unrealistically high. The value 3000 is far outside of the range of observed values.

d) Nothing is wrong with this value. It can be treated as an accurate prediction.

Answer 1

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	6155.00	492.00	220772.83	319.64	8098.70
mean	1025.83	82.00	SSxx	SSyy	SSxy

a)

Sample size,   n =   6      
here, x̅ = Σx / n=   1025.833          
ȳ = Σy/n =   82.000          
SSxx =    Σ(x-x̅)² =    220772.8333      
SSxy=   Σ(x-x̅)(y-ȳ) =   8098.7      
              
estimated slope , ß1 = SSxy/SSxx =   8098.7/220772.8333=   0.0367      
intercept,ß0 = y̅-ß1* x̄ =   82- (0.0367 )*1025.8333=   44.3689      
              
Regression line is, Ŷ=   44.369   + (   0.037   )*x

b)

Predicted Y at X=   3000   is          
Ŷ=   44.36894   +   0.03668   *3000=   154.419

c)

c) It is unrealistically high. The value 3000 is far outside of the range of observed values.