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A regression of length of eruption ( y ) for Old Faithful on the waiting time between eruptions ( x ) yields y ^ =?1.874+0.0753

A regression of length of eruption (y) for Old Faithful on the waiting time between eruptions (x) yields

y^ =?1.874+0.0753x. The output, at right, from R gives the 95% confidence interval for the mean response whenx0= 80 s.

  1. a)(1) From the output, what is the value ofse(?^y)?
  2. b)(1) From the output, what is the value of MSE?
  3. c)(2) Determine the standard error for predicting a neweruption whenx0= 80 s, that isse(y^).
  4. d)(3) Determine the 95%predictioninterval for a new eruption withx0= 80 s. Hint: the degrees of freedom is the same as from the R output.

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A regression of length of eruption (y) for Old Faithful on the waiting time between eruptions (x) yields 3}: l.874+0.0753 x . The output, at right, from R gives the 95% condence interval for the mean response when xo = 80 s. a) (1) From the output, what is the value of se(@) .7 b) (1) From the output, what is the value of MSE? c) (2) Determine the standard error for predicting a new eruption when xo = 80 s, that is 86(1)) . d) (3) Determine the 95% prediction interval for a new eruption with xo = 80 s. Hint: the degrees of freedom is the same as from the R output. $fit fit lwr upr 1 4.17622 4.104848 4.247592 $se.fit [1] 0.0362517? $df [1] 27a $residual. scale [1] 0.4965129

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