1. Research performed at NASA and led by Dr. Emily R. Morey-Holton measured the lengths of the right humerus and right tibia in 11 rats that were sent into space on Spacelab Life Sciences 2. Treating the length of the right humerus as the explanatory variable and the length of the right tibia as the response variable the researchers are interested to t a simple linear regression to the data. The following data and statistics were collected. The total sum of squares (SST) is given as 23.1533. The sample mean and sample variance of the length of right humerus are 25.34 and 1.08424 respectively. The sample mean of the length of right tibia is 35.34091. The Adjusted Rsquared is 0.8944. AN OVA TABLE Source Degrees ereeedem P-velue Regression 20.9538 Residuals (a) Complete the ANOVA table. (b) Write down the basic model appropriately and explain each component of the model. (c) Compute the least square estimates of the two coefcients in your model and write down the regression equation involving estimates. Show the formulae you are using and show your work. ((1) Interpret the estimated value of the slope appropriately in the context of the given data. (e) What percentage of variation in the length of the right tibia is explained by the regression equation. (f) Plot the data and the regression line on the same graph. 1 (g Show clearly using appropriate formula that the standard error of the estimate of the slope is 0.15. 1What is the standard error of the estimate of the intercept. (h) Test if linear relation exists between length of the right humerus and the length of the right tibia, at or = 0.01. State your hypotheses with appropriate symbol. Write expression of the test statistic and its value. Show your work. Determine the critical region. Compute the P-value. State the conclusion of the test. (i) Predict the mean length of right tibia for all rats having right humerus length as 25.83 mm. (j) Predict the length of the right tibia for a randomly selected rat whose right humerus is 25.83 mm. (1:) Construct a 95% prediction interval for the length of the right tibia for a randomly selected rat whose right humerus is 25.83 mm. 6+2x5+5+5+1+1+3=mn