STA 302 / 1001 H - Fall 2003Test 1October 6, 2003LAST NAME:FIRST NAME:STUDENT NUMBER:ENROLLED IN: (circle one) STA 302 STA 1001INSTRUCTIONS:• Time: 50 minutes• Aids allowed: calculator.• A table of values from the t distribution is on the last page.• Total points: 30Some formulae:b1=P(Xi−X)(Yi−Y )P(Xi−X )2b0=Y − b1XVar(b1) =σ2P(Xi−X)2Var(b0) = σ21n+X2P(Xi−X)2Cov(b0, b1) = −σ2XP(Xi−X)2SSTO =P(Yi− Y )2SSE =P(Yi−ˆYi)2SSR = b21P(Xi−X)2=P(ˆYi− Y )2σ2{ˆY∗} = Var(ˆY∗) = σ21n+(X∗−X)2P(Xi−X)2σ2{pred} = Var(Y∗−ˆY∗) = σ21 +1n+(X∗−X)2P(Xi−X)21 (a) (b) 1 (c) (d) (e) 2 (a) (b) (c) 2 (d) 311. (a) (4 points) State the simple linear regression model for dependent variable Yand independent variable X and the Gauss-Markov conditions.(b) (5 points) The least squares estimate of the Y intercept for the model in (a)is b0as given on the first page. Under the model you specified in (a), derivethe formula for the variance of b0. You may not asssume that any of theformulae for variance or covariance are known.2(c) (3 points) In order to do inference about the slope (such as testing whetheror not the slope is 0) we need to make one more assumption about the modelin (a). What is the usual assumption and why is it necessary?(d) (2 points) An estimate is more precise than another if it has smaller variance.The estimate of the expected value of Y varies with the value of X. At whatvalue of X will there be the most precise estimate of the expected value ofY ? Justify your answer.(e) (2 points) Suppose the dependent variable Y could be measured with lesserror. Why would this lead to more precise estimation of the intercept of theregression line?32. Some engineers are interested in examining the relationship between load, inpounds, and the corresponding deformation, in inches, on a mild steel bar. Basedon the physical properties of the steel, the engineers believe there will be a linearrelationship between the natural logarithm of load (logL) and the natural loga-rithm of deformation (logD). Data were collected for 24 loads ranging from 1000to 9900 pounds and a regression of logD on logL was run. Some output from SASis given below. (Some numbers have been purposely removed from the output.)Descriptive StatisticsUncorrected StandardVariable Sum Mean SS Variance DeviationIntercept 24.00000 1.00000 24.00000 0 0logL 212.18446 8.84102 1883.83553 0.34386 0.58639logD 174.88129 7.28672 1304.29718 1.30375 1.14182Dependent Variable: logDAnalysis of VarianceSum of MeanSource DF Squares Square F Value Pr > FModel 1 20.57353 20.57353 48.09 <.0001Error 22 9.41263 0.42785Corrected Total 23 29.98616Parameter EstimatesParameter StandardVariable DF Estimate Error t Value Pr > |t|Intercept 1 -6.97275 2.06066 -3.38 0.0027logL 1 1.61288 0.23259 6.93 <.00014The following questions (labelled (a) through (d)) relate to the SAS output on theprevious page.(a) (3 points) Construct a 95% confidence interval for the slope. What can youconclude from this confidence interval about a test of H0: β1= 0?(b) (2 points) What is the value of R2? What does this value mean?(c) (4 points) What is the predicted value of logD when the load is 9600 pounds?Construct an appropriate 90% interval for the expected value of logD at thisload.5(d) (2 points) Do you trust your prediction in (c)? Explain.3. (3 points) A simple linear regression is performed to study the relationship betweena child’s intelligence at age 5 (the dependent variable) and the length of time thechild was breastfed as an infant (the independent variable). The value of r2was0.56 and for the two-sided test of whether the slope was 0, the p-value was 0.013.A newspaper headline reporting on the study said “Study shows that to make asmarter child, mothers should breastfeed longer”. Comment on the validity of