Sebastian Tello
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RD Graph: Universal Pre-K on Test Scores
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RD Graph: Universal Pre-K on Test Scores

RD Graph.pdf11037.7KB

Universal Pre-Kindergarten is the name of a policy that has provided high-quality free schools for 4-year-olds. If it works, as advocates say, Universal Kindergarten, or Pre-K, will counteract socioeconomic disparities, boost productivity, and increase crime. But does it work? Gormley, Phillips, and Gayer (2008) used RD analysis to evaluate one piece of the puzzle by looking at Universal Pre-K's impact on test scores in Tulsa, Oklahoma. They could do so because children born on or before September 1, 2001, were eligible to enroll in the program for the 2005-2006 school year, while children born after this date had to wait until the following year to enroll. The figure below is a bin plot for this analysis. The dependent variable is test scores from a letter word identification test that measures early writing skills. The children took the test a year after the older kids started Pre-K. The children born after September 1 spend the year doing whatever 4-year-olds do when they are not in Pre-K. The horizontal axis shows age measured in days from the Pre-K cutoff date. The data is binned in groups of 14 days so that each data point shows the average test score for children with ages in the 14-day range. While the actual statistical analysis uses all observations, the bin plot helps us see the relationship between cutoff and test scores better than this cutoff plot of all observations.

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Click the toggle to see how the scatter plot looks
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Relationship between date born and writing test scores

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  1. What is the running variable?
  2. ‣
    Answer
    ✅
    Age in days (or months) away from September 1, 2001
  3. What is the treatment group?
  4. ‣
    Answer
    ✅
    Any four year old born before September 1, 2001
  5. What is the outcome variable?
  6. ‣
    Answer
    ✅
    Scores on a letter word identification test that measures early writing skills.
  7. What is the research question?
  8. ‣
    Answer
    ✅
    The research question is what is the effect of universal pre-K on student’s outcomes
  9. Is this sharp or fuzzy?
  10. ‣
    Answer
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    Since we want the effect of attendance on student’s outcomes, this is a fuzzy design unless we think all kids that are eligible kids attended pre-K.
  11. In order to run a regression to measure the jump, we run the following regression
    1. WritingScores=α0+δ1(Age>=0)+β1(Age−cutoff)WritingScores=\alpha_0+\delta_1(Age>=0)+\beta_1(Age-cutoff)WritingScores=α0​+δ1​(Age>=0)+β1​(Age−cutoff)
    2. Find in the graph, where these coefficients are.
    3. ‣
      Answer
      ✅
      The jump seems to be from 6 to something a little bit higher than 8, say 8.2, so delta will be 2.2. The beta 1 represents the slope, since this regressions treats both of the slopes the same, let’s just see the change in y and x. Change in Y (approximately) 12-4 =8, change in X: 325- (-375)=700, so then the slope is 8/700 or 0.011 approximately. Finally, alpha is the average right before the cutoff, so around 6.
  12. Now imagine that the equation is the one below
    1. WritingScores=α0+δ1(Age>=0)+β1(Age−cutoff)+β2(Age>=0)×(Age−cutoff)WritingScores=\alpha_0+\delta_1(Age>=0)+\beta_1(Age-cutoff)+\beta_2(Age>=0)\times (Age-cutoff)WritingScores=α0​+δ1​(Age>=0)+β1​(Age−cutoff)+β2​(Age>=0)×(Age−cutoff)
    2. Find in the graph, where these coefficients are
    3. ‣
      Answer
      ✅
      Alpha and delta are similar as above, the only changes is beta 1 and beta 2. Similar process as above, just different changes. For beta 1 now, we do change in Y: 6-3, and change in x:0-(-375)=375, so the slope is 3/375=0.008. Then the same process for beta 2, is change in y=12-8-=4, change in x: 325 so 4/325=0.012, so the difference is 0.012-0.008=0.0043 which is the value for beta 2.

More practice

If that was a bit confusing or if you need more practice, for each panel below, indicate whether each β1,β2,β3\beta_1, \beta_2, \beta_3β1​,β2​,β3​ is less than equal to, or greater than zero for the varying slopes of the RD model. But you could do better than “more than” or “less than” zero, find the actual values!

Yi=β0+β1Ti+β2(X1i−C)+β3(X1i−C)×Ti+ϵiY_i=\beta_0+\beta_1T_i+\beta_2(X_{1i}-C)+\beta_3(X_{1i}-C)\times T_i+\epsilon_iYi​=β0​+β1​Ti​+β2​(X1i​−C)+β3​(X1i​−C)×Ti​+ϵi​
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Answer
  • If you find a mistake, let me know! I could not have seen something or missed something.
  • A: β0=4,β1=0,β2=0,β3=−1\beta_0=4,\beta_1=0,\beta_2=0,\beta_3=-1β0​=4,β1​=0,β2​=0,β3​=−1
  • B: β0=8,β1=−4,β2=0,β3=1\beta_0=8,\beta_1=-4,\beta_2=0,\beta_3=1β0​=8,β1​=−4,β2​=0,β3​=1
  • C: β0=1,β1=7,β2=1,β3=0\beta_0=1,\beta_1=7,\beta_2=1,\beta_3=0β0​=1,β1​=7,β2​=1,β3​=0
  • D: β0=10,β1=−5,β2=97,β3=(0.95 or 97−13)\beta_0=10,\beta_1=-5,\beta_2=\frac{9}{7},\beta_3=(0.95\ or\ \frac{9} {7}-\frac{1}{3})β0​=10,β1​=−5,β2​=79​,β3​=(0.95 or 79​−31​)
  • E: β0=6,β1=3,β2=53,β3=−53\beta_0=6,\beta_1=3,\beta_2=\frac{5}{3},\beta_3=-\frac{5}{3}β0​=6,β1​=3,β2​=35​,β3​=−35​
  • F: β0=1,β1=5,β2=−53,β3=53\beta_0=1,\beta_1=5,\beta_2=-\frac{5}{3},\beta_3=\frac{5}{3}β0​=1,β1​=5,β2​=−35​,β3​=35​