# Statistical analysis of chirps and temperature

1. Apr 8, 2009

### xiaobai5883

1. The problem statement, all variables and given/known data

One class application of correlation and regression involves the association between the temperature and the number of times cricket chirps in a minute. Listed below are numbers of chirps in 1 minute and the corresponding temperature in degrees Fahrenheit.

Chirps in 1 min 882 1188 1104 864 1200 1032 960 900
temperature 69.7 93.3 84.3 76.3 88.6 82.6 71.6 79.6

i. Is there sufficient evidence to conclude that there is an association between the number of chirps in a minute and the temperature?
ii. Find the equation of the straight line that best fits the sample data.
iii. If a cricket chirps 1234 times in 1 minute, what is the best predicted value for the temperature?
iv. What percentage of the variation in temperature can be explained by the variation in the cricket chirp rate?
v. Can the correlation/regression results be used to show that changes in temperature cause changes in the cricket chirp rate?

2. Relevant equations

3. The attempt at a solution

2. Apr 8, 2009

### lanedance

Re: Statistic...

the idea is for you to attempt the question first - to get you started try plotting your results for i) then google "2 variable statistics" and "linear regression" or "least squares regression"

Last edited: Apr 8, 2009
3. Apr 8, 2009

### xiaobai5883

Re: Statistic...

i have already plot those results...
is there any functions or equation or formula that i can use??

4. Apr 8, 2009

### lanedance

Re: Statistic...

so do you see any correlation in your plot?

the topics in the last post should give you the equations
for i) look at correlation coefficient r in 2 variable statistics

5. Apr 8, 2009

### xiaobai5883

Re: Statistic...

you mean that r should be put like...
(Temperature)=r(Chirps in 1 min)??
i think should be
(Temperature)=r(Chirps in 1 min)+c
but in almost every case the r and c is different...
so how how??

6. Apr 8, 2009

### lanedance

Re: Statistic...

you want to find the line that best fits the data
if
y = chirps
x = temp

then find m and c such that
y = mx + c
to minimise the sum of squared errror between you data points and the line, this is a mathematical way of "eyeballing" the best line through the data

Any scientific calculator or spreadsheet package should be able to give you m and c

the correlation coefficient r, ranging form 0 to 1 is a measure of how well correlated your data, ie how close to a line

7. Apr 8, 2009

### xiaobai5883

Re: Statistic...

LOL??
actually can you give me one example??
because i don't know what do you mean actually...
and the correlation coefficient r is what formula??
is there any formula??