19 Crickets chirp faster when it gets warmer. The number of chirps per minute by crickets (x) is given along
with the surrounding temperature (y):
(10 pts)
#chirps/min 20.0 16.0 19.8 18.4 17.1
Temperature (°F) 88.6
71.6 93.3 84.3 80.6
a)
Find the correlation coefficient, r. Would you say there is a significant linear correlation? Why or
why not?
b)
Give the linear regression equation. Round your numbers to three decimal places,
c)
Use your equation to predict the temperature when a cricket chirps 17 times per minute, or state
that it is inappropriate.
d)
Use your equation to predict the temperature when a cricket chirps 30 times per minute, or state
that it is inappropriate.
e)
Find and interpret the coefficient of determination (r2)

Step-by-step explanation:

Step 1.  Enter the data into the lists.  

For basic entry of data, see Basic Commands.

cricket1  

Step 2.  Create a scatter plot of the data.  

cricket4

Obviously, there is some scatter to this data. This variability is the norm, rather than the exception when working with biological data sets.  Real-life data seldom creates a nice straight line.

Step 3.  Choose the Linear Regression Model.

    Press STAT, arrow right to CALC, and arrow down to 4: LinReg (ax+b).  Hit ENTER.  When LinReg appears on the home screen, type the parameters L1, L2, Y1.  The Y1 will put the equation into Y= for you.  

           (Y1 comes from VARS → YVARS, #Function, Y1)

cricket6  

cricket7Older OS form.

cricket newNewer OS form.

cricket8

The linear regression equation is

y = 3.41x + 22.85

(answer to part a)

Step 4.  Graph the Linear Regression Equation from Y1.

    ZOOM #9 ZoomStat to see the graph.

cricket9

(answer to part b)

Step 5.  Is this model a "good fit"?

    The correlation coefficient, r, is .8320424586 which just barely places the correlation into the "strong" category.  (0.8 or greater is a "strong" correlation)

    The coefficient of determination, r 2, is .6922946529 which means that 70% of the total variation in y can be explained by the relationship between x and y.  The other 30% remains unexplained.

    Yes, it is somewhat of a "good fit".

        (answer to part c)

a3

Step 6.  Extrapolate:  (beyond the data set)

     If the ground temperature reached 95º, then at what approximate rate would you expect the crickets to be chirping?

Step 7.  Interpolate: (within the data set)

    With a listening device, you discovered that on a particular morning the crickets were chirping at a rate of 18 chirps per second.  What was the approximate ground temperature that morning?

 The power equation, type 18, hit ENTER, and the answer will appear at the bottom of the screen.

cricket12

(answer to part e --  the ground

the temperature will be approx. 84.23º F)

Step 8.  If the ground temperature should drop to freezing (32º F), what happens to the cricket's chirping?

a4

The TABLE tells us that at 32º F there are 2.68 chirps per second.  So, what does this really mean?  Are the crickets cold?

These findings are a bit deceiving.  At 32º F, the crickets are dead.  The lifespan of cricket in a cold climate is very short.  The crickets spend the winter as eggs laid in the soil.  These eggs hatch in late spring or early summer and tiny immature crickets called nymphs to emerge.  Nymphs develop into adults within approximately 90 days. The adults mate and lay eggs in late summer before succumbing to old age or freezing temperatures in the fall.

hope it

step-by-step explanation:

x^(1/6) · x^(1/6) · x^(1/6)   ·x^(1/6) = x^(4/6)   since we add the exponents

4/6 can be simplified to 2/3

the simplified form is x^(2/3)

is there a picture

step-by-step explanation: