(From Chapter 4) If you have a cricket, the frequency of its chirps can be used

Question

(From Chapter 4) If you have a cricket, the frequency of its chirps can be used to estimate the temperature. Experiments have been run to test this. The table gives the results of one small study. The first column labeled Chirps is the number of chirps by the cricket in 15 seconds. The second column labeled Temperature is the temperature in Fahrenheit degrees at the time of the cricket chirps. Answer the questions below. (2 points each) Chirps Temperature 49 The sample dataset consists of 10 ordered pairs. What is the 56 61 12 critical value (CV) for the linear correlation coefficient to be significant (that is to say, indicate a relation between the variables "Chirps" and "Temperature")? 17 23 30 37 21 29 37 73 60 68 What is the value of the linear correlation coefficient? 80 52 15 Does it indicate a relation between the two variables? Assuming the relation between the variables is significant, the linear regression estimates the relation. What is the linear regression line for this equation? The coefficients of the equation should show 2 digits no matter where the decimal point may be located. If the cricket chirps 35 times, what temperature does the regression line estimate? if the temperature is 60 degrees, how many chirps do you expect out of the cricket? How does this compare with the actual count of chirps at 60 degrees?

Explanation / Answer

Here independent variable is chips and dependent variable is temperature.

We have given 10 data pairs.

Now we have to test the hypothesis that,

H0 : There is no relationship between chips and temperature.

H1 : There is relationship between chips and temperature.

Assume alpha = level of significance = 0.05

The test statistic is,

t = r * sqrt(n-2) / sqrt(1-r2)

where r is sample correlation between chips and temperature.

For this data r = 0.9922

n = no. of data pairs = 10

t = 0.9922*sqrt(10-2) / sqrt(1 - 0.99222)

t = 22.51

Now we have to find p-value for taking the decision.

P-value we can find in excel.

syntax :

=TDIST(x, deg_freedoms,tails)

where x is absolute value of test statistic

deg_freedoms = 10-2 =8

tails = 2

P-value = 1.6082E-08 = 0.000

P-value < alpha

Reject H0 at 5% level of significance.

Conclusion : There is relationship between chips and temperature.

Now we have to fit regression.

We can fit regression in excel

steps :

ENTER data into excel sheet --> Data --> Data analysis --> Regression --> ok --> Input Y range : select temp --> Input X range : select chips --> Labels --> Output range : select one empty cell --> ok --> Output range : select one empty cell --> ok

SO from the output :

intercept = 38.61

slope = 0.96

Therefore the regression equation is,

temp = 38.61 + 0.96*chips

Now we have to find temp when chips = 35

temp = 38.61 + 0.96*35 = 72.21

Now again we have to find chips when temp = 60

60 = 38.61 + 0.96*chips

60 - 38.61 = 0.96*chips

21.39 = 0.96*chips

chips = 21.39 / 0.96 = 22.28

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