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givi [52]
3 years ago
12

Cricket chirps: Temperature Anyone who has been outdoors on a summer evening has probably heard crickets. Did you know that it i

s possible to use the cricket as a thermometer? Crickets tend to chirp more frequently as temperatures increase. This phenomenon was studied in detail by George W. Pierce, a physics professor at Harvard. In the following data, x is a random variable representing chirps per second and y is the random variable representing temperature (F). These data are also available for download at the online study center.X 20.0 16.0 19.8 18.4 17.1 15.5 14.7 17.1Y 88.6 71.6 93.3 84.3 80.6 75.2 69.7 82.0X 15.4 16.2 15.0 17.2 16.0 17.0 14.4Y 69.4 83.3 79.6 82.6 80.6 83.5 76.3Complete parts (a) through (e), given ?x= 249.8, ?y=1200.6, ?x^2= 4200.56, ?y^2= 96,752.86, ?xy= 20,127.47 and r = 0.835.A) Draw a scatter diagram displaying the data.B) Find X- bar, Y- bar, a and b. Then find the equation of the least squares line y=a+bx.C) Graph the least squares line on your scatter diagram. Be sure to use the point ( X-bar, Y- bar) as one of the points on the line.D) Find the value of the coefficient of determination r ^2. What percentage of the variation in y can be explained by the corresponding variation in x and the least squares line? What percentage is unexplained?E) What is the predicted temperature when x= 19 chirps per second.
Mathematics
1 answer:
nasty-shy [4]3 years ago
7 0

Answer:

Kindly check explanation

Step-by-step explanation:

Using technology on the data above, the least square regression equation obtained is ;

y = 33.4528+2.7975x

Where, slope = 2.7975 ; Intercept = 33.4528

x = chirps per second, independent variable

y = temperature, predicted variable.

The Correlation Coefficient, r = 0.71

The Coefficient of determination, r² = 0.71² = 0.5041

The percentage of variation in y due to regression is the r² value, which is about 50.41%

Hence, the percentage variation in temperature due to the regression line is 50.4, while the rest is due to other factors.

Predicted temperature, y when x = 19 ;

Put x = 19

y = 33.4528+2.7975(19)

y = 86.60°

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