The following problems have JMP output I\'ve attached below. The simple linear r
ID: 3218121 • Letter: T
Question
The following problems have JMP output I've attached below.
The simple linear regression model can be summarized in math notation as ya Bo Bixi ei, ei's are iid normal(0, a3) Start with a simple linear regression involving explanatory variable X1 and response variable Y. The results are given in Item B. a) What are the estimates of Bo and Ba (also called bo and bi) for this regression? b) What is the estimate of Hylr (also called y) for this regression? Note that the answer will be in terms of x c) What is the estimate of a (also called stF) for this regression? d Examine the simple linear regression model written above. Note that yi is a random variable, and so Varlyl 303. Also, a does not have the index subscript i in contrast to y, x, and e). What does that say about Varly e) Continuing from before, realize that Elya] Bo Bixi. Hence, the mean of yi is a function of xi. Clearly, xi has the index subscript i. What does that say about E Using values in the output, make two t distributed 95% confidence intervals: one for Bo and one for Bh. Showthe degrees of freedom calculation in solving the t value. g) Complete the sentence: The degrees of freedom in part (f is equal to the formula for simple linear regression. (The generic denominator of the answer is "variance estimate", so instead, say what it is specifically for this model) h) Notice the "t Ratio" for "X1" is 1.86. This is the test statistic for a hypothesis test about BI. State the null and alternative hypotheses for this test, and derive 1.86 using the other values in the output. State the p-value and conclusion of the hypothesis test described in part (h) (Hint: as usual, be careful in phrasing the conclusion: do not conclude that B is 0!Explanation / Answer
a) From the item B SLR:
b0=intercept= 0.927
b1=slope= 0.0119
b) Y_hat = b0+b1*x = .927+.0119*x
c) Sigma^2= Error variance =MSE= (RMSE)^2= 0.411^2=0.169
SSE=df*MSE = (20-1-1)*.169= 3.041
R-sq=0.161 = SSR/SST = SSR/(SSR+SSE) = > (1-R-sq)/R-sq= SSE/SSR =>
SSR = 0.583
SSTotal = SSR+SSE = 3.624 (this is total variability in Y)
d) Var[y]= SST/(n-1) = 3.624/19 = 0.191
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