1. We return to the teaching effectiveness data that was considered during the course. A sample of 23 student teachers w

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1. We return to the teaching effectiveness data that was considered during the course. A sample of 23 student teachers w

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1 We Return To The Teaching Effectiveness Data That Was Considered During The Course A Sample Of 23 Student Teachers W 1
1 We Return To The Teaching Effectiveness Data That Was Considered During The Course A Sample Of 23 Student Teachers W 1 (129.76 KiB) Viewed 113 times
1 We Return To The Teaching Effectiveness Data That Was Considered During The Course A Sample Of 23 Student Teachers W 2
1 We Return To The Teaching Effectiveness Data That Was Considered During The Course A Sample Of 23 Student Teachers W 2 (143.47 KiB) Viewed 113 times
1. We return to the teaching effectiveness data that was considered during the course. A sample of 23 student teachers was evaluated and given an overall teaching effectiveness score, the response variable teff. In- dividuals i = 1, ..., 12 were male and individuals i = 13,...,23 were female. In addition, each student teacher was given four standardised tests and the results of these tests were recorded in the explanatory variables x1, x2, x3 and 24. Here we just focus on the second stan- dardised test, corresponding to the covariate x2. In the Routput below, Gender is a factor with two levels, M (male) and F (female), and x2 is treated as a numerical covariate. > out1=lm(teff“gender*x2) > summary (out1) Call: lm (formula = teff gender * x2) Residuals: Min 1Q -195.507 -51.305 Median 3.493 3Q Max 23.728 161.387 Coefficients: Estimate Std. Error t value Pr(>ltl) (Intercept) 107.6944 280.2802 0.384 0.705 gender -0.6509 372.9534 -0.002 0.999 x2 2.2895 1.9822 1.155 0.262 genderM:x2 0.1259 2.6027 0.048 0.962 Residual standard error: 84.09 on 19 degrees of freedom Multiple R-squared: 0.174, Adjusted R-squared: 0.04356 F-statistic: 1.334 on 3 and 19 DF, p-value: 0.2929 > vcov (out1) (Intercept) gender x2 genderM:x2 (Intercept) 78556.9770 -78556.9770 -553.296127 553.296127 gender -78556.9770 139094.2143 553.296127 -966.255830 x2 -553.2961 553.2961 3.929153 -3.929153 genderM:x2 553.2961 -966.2558 -3.929153 6.773881

(a) What is the name of the type of analysis being performed here. Be as specific as possible. [2 marks] (b) Write down, in mathematical form, the model under consideration, using the “reference” parametrisation (i.e. the same parametrisa- tion that is used by R). Briefly state the modelling assumptions that are being made. (3 Marks] (c) Briefly explain how, after fitting the above model in R, you would use R to check for (i) constant variance and (ii) normal errors. [4 Marks) (d) Give a brief summary of what you conclude from the R output that has been presented. [3 marks] (e) An equivalent way to write the model, putting y=teff, is Yi = am + Bmxi + Ei, i = 1,...,12; af + Biti te i = 13,... 23. What estimates of the parameters am, BM, Qf and Bf are implied by the reference parameter estimates produced by R? (3 Marks] (f) Determine the standard error of each estimate âm,âf, BM, Bf of, respectively, am, QF, BM and Bf. [4 Marks) (g) What are the degrees of freedom associated with the standard errors you have calculated? Briefly explain your reasoning. [2 Marks (h) Calculate 95% confidence intervals for the parameters am, af, BM and Br. Note: these confidence intervals should be based on the relevant t-distribution, not the normal distribution. Briefly discuss the extent to which these confidence intervals confirm or contradict your findingss in part (d). [4 Marks] HINT: If X and Y are two random variables, both with finite variance, and a and b are constants, then VarſaX +bY] = aʼVar[X] + bạVar[Y] + 2abCov[X, Y). =
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