Anyone can help solve this statistics. 1. a medical research team obtained the index of adiposity,?

basal insulin, and basal glucose values of 10 normal subjects. these results are shown in the following table. the researchers wished to investigate the strength of association among these variables. index of Adiposity (X):90,112,127,137,103,140


,105,92,92,96. basal insulin (uU/ml) X1: 12,10,15,11,10, 38,9,6,8,6. for basal glucose ( mg/100 m10) X2: 98,103,101,90,108,100,101,92,91.


2. the following are the data for the average number of cigarettes smoked daily and the number of free radicals found in the lungs of eigth individuals. no. of cigarettes: 0,10,14,5,18,20,30,40. free radicals: 94,144,182,120,240,234,321,400.


a. determine the degree of relationship between the average number of cigarettes smoked daily and number of free radicals found in the lungs. b. represent the data in a scatter diagram. c. determine the regression line. d. predict the amount of free radicals ins someone who smokes an average of 26 cigarettes daily.

Anyone can help solve this statistics. 1. a medical research team obtained the index of adiposity,?
1. you only listed 9 values for the glucose, I added the value 98 to the list so that the lengths are the same.





From what I read here I would use a linear model and Analysis of Variance to find out what is important and what is not.





in R, (you can get R free from www.r-project.org) use the following code:





adiposity %26lt;- c(90, 112, 127, 137, 103, 140, 105, 92, 92, 96);


insulin %26lt;- c(12, 10, 15, 11, 10, 38, 9, 6, 8, 6);


glucose %26lt;- c(98, 103, 101, 90, 108, 100, 101, 92, 91, 98);


model %26lt;- lm(adiposity ~ glucose + insulin)


model


anova(model)





the output is that the model is:





adiposity = 117.1957 + 1.4207 * insulin - 0.2602 * glucose





Analysis of Variance Table





Response: adiposity


. . . . . . . . . . . Df . Sum Sq . . .Mean Sq . .F value . Pr(%26gt;F)


glucose. . . . .1 . . 10.31 . . . .10.31 . . . 0.0426 . . . . 0.84235


insulin. . . . . . .1 . .1532.39 .. 1532.39 . 6.3333 . . . .0.04001 *


Residuals. . . 7 . . 1693.71 . . 241.96





this indicates that insulin is statistically significant in the model but glucose is not.





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2.





cigarettes %26lt;- c(0, 10, 14, 5, 18, 20, 30, 40);


freeRadicals %26lt;- c(94, 144, 182, 120, 240, 234, 321, 400);





plot(freeRadicals ~ cigarettes)


fit %26lt;- lm(freeRadicals ~ cigarettes)


abline(fit)





fit





the relationship is direct,





the above code will give you a scatter plot with the regression line.





the regression line is:


freeRadicals = 81.087 + 7.929 * cigarettes





if cigarettes = 26 then we estimate freeRadicals = 287.241





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Reply:Cheaters never prosper.