Analysis of NJ Standardized Test Scores: Regression Models and Area Calculations
This project analyzes data from the NJ Department of Education regarding standardized test scores in Mathematics over a nine-year period. It employs various regression techniques including cubic, quartic, and sine regressions to model the average rate of change and instantaneous rates of change. Additionally, it explores the area under the curve using the Fundamental Theorem of Calculus, yielding estimations of student proficiency. Key insights include the identification of maxima, minima, and points of inflection, informing educational assessment strategies.
Analysis of NJ Standardized Test Scores: Regression Models and Area Calculations
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Presentation Transcript
Calculus 151 Regression Project Data collected from the NJ Department of Education Website
NJ Standardized Test Scores 76.8 – 75.2 1.6 Average Rate of Change = 02 - 11 = -9 = - .778
Sine Regression Instantaneous Rate of Change at 2003 = -5.174
Quartic RegressionR2 =.555 Instantaneous Rate of Change at 2003 = -1.826
Split Regressions Limit x 6.5- 75.25657 Limit x 6.5+ 71.669602
Continuous Split Regressions Limit x 6.5 73.463086 Limit x - ∞ ∞ Limit x ∞ DNE
Derivatives of exponential, logarithmic, and sine regressions Y’= 74.56303051 *.9993957215^x *ln(.9993957215) Y’= -.6345494264 x Y’=-6.784189065* cos(-2.304469566 x + 1.333706904)
Newton’s Methodfinding zeros of the cubic regression X0=23.74251964 X0=23.74251964
Mean Value Theorem f’(c) = 75.682- 76.565 11-2 f’(c) = - .883 9 f’(c) = -.098 c = X f(c) = Y4 f’(c) = Y5 Y= -.098(x – 3.4931) + 71.661 Y= -.098(x – 6.9124) + 75.325 Y= -.098(x – 9.67854) +72.782
Max and Min of Cubic Regression The Regression has a minimum at 5.4093854 and a maximum at 10.033224. It is increasing between [5.4093854, 10.033224] ,and is decreasing between (- ∞ , 5.4093854) U (10.033224, ∞).
Second derivative of cubic regression Second Derivative Zero Inflection Point Concave up Concave down First Derivative Maximum
Approximating area under a curve using left endpoints Estimate Area is 668.504 72.432 73.684 77.426 74.644 76.352 71.138 76.517 74.42 71.891
Approximating area under a curve using right endpoints Estimate Area is 670.836 76.352 71.138 76.517 74.42 71.891 77.426 72.432 73.684 76.976
Finding Area under the curve using the Fundamental Theorem of Calculus 11 Area=∫02 2.943926518sin(-2.304469566x+1.333706904) +74.26459702dx F(x)= 1.277485527cos(-2.304469566x +1.333706904)+74.26459702x F(11)- F(02)≈ 817.47-147.26≈ 670.21 Area ≈ 670.21
Average Value Area= the sum of the % of students proficient in Mathematics over the past 9 years Average % of students 670.69193 proficient in Mathematics = 9 ≈ 74.55% for each year
