Understanding Lifespan Predictions: Correlation and Regression Analysis

Assignment 6a Ryan pollard 3661835 Question 1a To predict the lifespan of an animal with a certain length of gestation life span Y is used as a variable and for explanatory variable then gestation X is used as a variable Question 1b X y x2 y2 xy 8 30 64 900 240 2.1 12 4.41 144 25.2 1.3 6 1.69 36 7.8 1 3 1 9 3 11.5 25 132.25 625 287.5 5.3 12 28.09 144 63.6 3.8 10 14.44 100 38 24.3 40 590.49 1600 972 X57.3 y138 x2)2836.67 y2)3558 xy 1637.1 Question 1b Compute the correlation coefficient between the gestation period and the lifespan. Interpret the value. [4 marks] R= 8 x 1637.1 – 57.3 x 138 / √[ 8 x 836.37 – 57.3^2] x [ 8 x 3558 – 138 ^2] = 0.9159 Question1c c. At α = 0.05, can you conclude that there is a significant correlation between the two variables? [6 marks] ( Hint: Show the key steps of test of Hypothesis.)

H0: p =0 no correlation and Ho: p≠ 0 significant correlation Degree of freedom : 8-2 = 6 A=0.05 R= 0.9159 T = 0.9159 / √1 -(0.9159)^2/ 8-2 = 5.8591 Since t > critical value We reject the null hypothesis and conclude that there is significant correlation between the two Question 1d R^2 = 0.9159^2 = 0.83887 This means that 83.9% of the variability in y is explained by the model Question 1e There are other factors to consider in determining lifespan, such as genetics and Environment Question 1f Slope = M = 8 x 1637.1 – 57.3 x 138 / 8 x 836.37 – 57.3^2 = 1.52285 B = Σy – m Σx / n Intercept = B = 138 – 1.52285 x 57.3 / 8 = 6.3425 Question 1g Interpret the slope of the regression equation with reference to context. [2 marks] The slope is 1.523, this implies that an increase in gestation period by 1 unit increases the life span of the animal by 1.523 unit Question 1h =1.52285872752 (8) + 6.3425 = 18.5254 Residual 30 – 18.5254 = 11.47 Question 2a

N = 9 9(4329) – (60 x 553) / 9(582) – 60^2 = 3.5293 It y-intercept b: 553/9 - 3.5293 x 60/9 = 37.9158 Regression equation = 3.5293x+37.9158 Question 2b SEE = √Σ(y - ŷ)^2 / n -2 =√ 3407.233/7 = √486.7475 = 22.0623 It measures how well a given linear regression model captures the relationship between the dependent variable and the independent variable. Lower the SEE the better fit but here we have a little bit high of SEE so it doesn’t exactly captures the relationship. Question 3a Null, Ho : blood pressure independent of a regular walk Alternative, H1: blood pressure- dependent for a regular walk Question 3b A=0.05 DF = (r-1)(c-1) = (2-1)(2-1) = 2 Ho= Σ( Col1* -Col2*)^2/Col2* = 2.934 =5.991 Question 3d Since Ho< Ha, we fail to reject the null hypothesis. It is a right-tailed value when we are doing a right-tailed test so it has to be rejected Question 4

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Question5 H0 = the square footage has no effect on the city Ha= the square footage does have an effect on the city SSB=8.75 x 3= 26.25 SSW = 71.06 – 26.25 = 44.81 MSW = 44.81 / 7 = 2.49 p-value = 0.036 a=0.01 p-value> a , do not reject null hypothesis conclusion : there is enough evidence that mean square is different among four

assignment 6a

Related Documents