Question

1. Contin 1. Crickets make a chirping noise by sliding their wings rapidly over each other. Perhaps you have noticed that the number of chirps seems to increase with the temperature. The following data list the temperature (in degrees Fahrenheit) and the number of chirps per second for the striped ground cricket. Temperature, Chirps per Temperature, Chirps per Second, Second, 88.6 20.0 16.0 933 716 79.6 80. 6 16. 01 153 17.0 76. 3 144 Source: George W. Pierce. The Songs of Insects. Cambridge MA: Harvard University Press, 1949. pp. 12-21 (a) What is the likely explanatory variable in these data? Why? 11 (b) Draw a scatter diagram of the data. 13C) Compute the linear correlation coefficient between temperature and chirps per seconds 14 (d) Does a linear relation exist between temperature and chirps per second? 21 (e) Find the least-squares regression line treating temperature as the explanatory variable and chirps per second as the response variablea 22 (1). Interpret the slope and y-intercept, if appropriate. (g) Predict the chirps per second if it is 83.3°F (h) A cricket chirps 15 times per second when the temperature is 82°F Is this rate of chirping above or below average at this temperature?

Answer 1

(a)

Temperature is explanatory variable because researcher wants to find the effect of temperature on the number of Chirps.

(b)

Following is the scatter plot:

20 19 18 16 14 65 70 75 80 85 90 95 х 60 21 17 15

(c)

Following table shows the calculations

	X	Y	X^2	Y^2	XY
	88.6	20	7849.96	400	1772
	93.3	19.8	8704.89	392.04	1847.34
	80.6	17.1	6496.36	292.41	1378.26
	69.7	14.7	4858.09	216.09	1024.59
	69.4	15.4	4816.36	237.16	1068.76
	79.6	15	6336.16	225	1194
	80.6	16	6496.36	256	1289.6
	76.3	14.4	5821.69	207.36	1098.72
	71.6	16	5126.56	256	1145.6
	84.3	18.4	7106.49	338.56	1551.12
	75.2	15.5	5655.04	240.25	1165.6
	82	17.1	6724	292.41	1402.2
	83.3	16.2	6938.89	262.44	1349.46
	82.6	17.2	6822.76	295.84	1420.72
	83.5	17	6972.25	289	1419.5
Total	1200.6	249.8	96725.86	4200.56	20127.47

Sample size: n = 15 From table we have: (Σ3 96725.86- ((1200.6)^2/15) = = 629.836 (Σν)" SST Sry Σν. 4200.56- ((249.8)^2/15) = 40.55733 Σ-/Σ) Συ- SP SyT = 20127.47- ((249.8 * 1200.6) /15) п = 133.478 The correlation coefficient is: 133.478/ sqrt( 629.836 * 40.55733) r 0.835

(d)

The correlation coefficient between the variables is strong and positive. Scatter plot also shows a strong , positive linear relationship between the variables. So there exists a relationship between the variables.

(e)

The slope is: 133.478/629.836 b0.212 The intercept is 1 bo(y 2x)=(249.8- (0.212 * 1200.6)/ 15 п bo-0.3151 The regression equation is: y' 0.212* x -0.3151

(f)

Slope: It is 0.212. It shows that for each unit increase in temperature chirps increased by 0.212 units.

Intercept: It is -0.3151. When temperature is zero the number of chirps is -0.3151. It is not possible so it is meaningless in this case.

(g)

The predicted value for x 83.3 is: y' 0.212 * 83.3 = 17.3445 0.3151

(h)

The predicted value for x 82 is y' -0.3151 0.212 * 82 = 17.0689

The actual value 15 is lower than predicted value so it is below average.