Question

\(\ln\left(\frac{\hat{P}_i}{1-\hat{P}_i}\right) = -13.021 + 2.8261 \text{GPA}_i + 0.095 \text{TUCE}_i + 2.3787 \text{PSI}_i,\)\n\(\text{se} = (4.9310) \quad (1.2629) \quad (0.142155) \quad (1.0645),\)\n\(t = (-2.6407) \quad (2.2378) \quad (0.7226) \quad (2.2345),\)\n\text{where}\n\text{P}_i \text{ is the probability of obtaining an A grade by i}^{\text{th}} \text{ student, } (1-\text{P}_i) \text{ is the probability of not}\ngetting it, } \hat{P}_i \text{ is the estimated value of } \text{P}_i,\n\text{GPA is the explanatory variable denoting entering grade point average,}\n\text{TUCE is the score on an examination given at the beginning of the term to test}\n\text{macroeconomics knowledge of entering students,}\n\text{PSI is the explanatory variable denoting personalized system of instructions;}\n\text{PSI = 1 if the new method of the personalized system of instructions is used and is 0}\notherwise,\nthe figures in the first set of parentheses corresponding to \text{"se"} are the estimated standard\nerror of the estimated regression coefficients, and those in the second set of parentheses are\nthe estimated t values computed under the null hypothesis that the true population value of\neach regression coefficient individually is zero.\n(i) Interpret the logistic regression.\n(ii) Find out the logit value of a student with GPA=4., TUCE=23, PSI=1, and that of a\nstudent with GPA=2.00, TUCE=20, PSI=0.

          \(\ln\left(\frac{\hat{P}_i}{1-\hat{P}_i}\right) = -13.021 + 2.8261 \text{GPA}_i + 0.095 \text{TUCE}_i + 2.3787 \text{PSI}_i,\)\n\(\text{se} = (4.9310) \quad (1.2629) \quad (0.142155) \quad (1.0645),\)\n\(t = (-2.6407) \quad (2.2378) \quad (0.7226) \quad (2.2345),\)\n\text{where}\n\text{P}_i \text{ is the probability of obtaining an A grade by i}^{\text{th}} \text{ student, } (1-\text{P}_i) \text{ is the probability of not}\ngetting it, } \hat{P}_i \text{ is the estimated value of } \text{P}_i,\n\text{GPA is the explanatory variable denoting entering grade point average,}\n\text{TUCE is the score on an examination given at the beginning of the term to test}\n\text{macroeconomics knowledge of entering students,}\n\text{PSI is the explanatory variable denoting personalized system of instructions;}\n\text{PSI = 1 if the new method of the personalized system of instructions is used and is 0}\notherwise,\nthe figures in the first set of parentheses corresponding to \text{"se"} are the estimated standard\nerror of the estimated regression coefficients, and those in the second set of parentheses are\nthe estimated t values computed under the null hypothesis that the true population value of\neach regression coefficient individually is zero.\n(i) Interpret the logistic regression.\n(ii) Find out the logit value of a student with GPA=4., TUCE=23, PSI=1, and that of a\nstudent with GPA=2.00, TUCE=20, PSI=0.

ln((P̂i)/(1-P̂i)) = -13.021 + 2.8261 GPAi + 0.095 TUCEi + 2.3787 PSIi,se = (4.9310) (1.2629) (0.142155) (1.0645),t = (-2.6407) (2.2378) (0.7226) (2.2345),wherePi is the probability of obtaining an A grade by i^th student, (1-Pi) is the probability of notit, P̂i is the estimated value of Pi,GPA is the explanatory variable denoting entering grade point average,TUCE is the score on an examination given at the beginning of the term to testmacroeconomics knowledge of entering students,PSI is the explanatory variable denoting personalized system of instructions;PSI = 1 if the new method of the personalized system of instructions is used and is 0,figures in the first set of parentheses corresponding to "se" are the estimated standardof the estimated regression coefficients, and those in the second set of parentheses areestimated t values computed under the null hypothesis that the true population value ofregression coefficient individually is zero.(i) Interpret the logistic regression.(ii) Find out the logit value of a student with GPA=4., TUCE=23, PSI=1, and that of awith GPA=2.00, TUCE=20, PSI=0.

Added by Gabriel S.

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