Question

10. 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)

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 is chirping at the rate of 3000 chirps per​ minute?

The best predicted temperature when a bug is chirping at 3000 chirps per minute is ____F.

​(Round to one decimal place as​ needed.)

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

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

B.The first variable should have been the dependent variable.

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

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

Answer 1

This is a simple problem related to formulation of a regression equation 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 the actual value of x (chirps/min)

Thus if we have a chirps/min value of 3000 /min then the corresponding pricited temperature in fahrenheit will be given by

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 at the value !

It is 3000 chirps/min

and look at the range of the value of x over which we have modelled 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 not be absolutely correct with the predicted value at such a far way point from the range.

This will thus give us only an vague insight into the potential temperature .

But this cant be a very close value in comparison to actual temperature value at 3000 chirps/min.

The error (Actual -observed) might be very large.

Hence Option (A).