A study on educational aspirations of high school students (Crysdale, Int. J. Comp. Sociol., 16, 19-36, 1975) measured aspirations using the sale (some high school, high school graduate, some college, college graduate). For students whose family income was low, the counts in these categories were (9, 44, 13, 10); when the family income was middle, the counts were (11, 52, 23, 22); when the family income was high, the counts were (9, 41, 12, 27). a. Use SAS/R to test whether the aspirations and family income are independent, reporting both the X2 and G2 statistics. b. No matter your answer in part a, do the standardized residuals suggest any interesting patterns? c. Using SAS/R, conduct a more powerful test than those in part a

I code an example question with answer.

Step-by-step explanation:

A study on educational aspirations of High School Students ( S. Crysdale, International Journal of Comparative Sociology, Vol 16, 1975, pp 19-36) measured aspirations using the scale (some high school, high school graduate, some college, college graduate). For students whose family income was low, the counts in these categories were (9, 44, 13, 10); when family income was middle, the counts were (11, 52, 23, 22); when family income was high, the counts were (9, 41, 12, 27)

A. Construct a suitable contingency table for the above data.

B. Find the conditional distribution on aspirations for those whose family income was high.

C. Conduct a Chi-square test of Independence between educational aspirations and income levels.

D. Explain what further analyses you could do that would be more informative than a chi-squared test.

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Chi-Squared Test for Homogeneity of Several Categorical Populations

Null Hypothesis: populations of people are homogeneous with respect to the four levels of education (low, med, high income groups educated same)

Alternative Hypothesis: populations not homogeneous.

Chi-Square Test: Some High School, Grad High School, Some College, Grad College

Expected counts are printed below observed counts

Chi-Square contributions are printed below expected counts

Chi-Square Test: Some High School, Grad High School, Some College, Grad College

Expected counts are printed below observed counts

Chi-Square contributions are printed below expected counts

Some Grad

High High Some Grad

School School College College Total

Low Income

9 44 13 10 76 [observed counts]

8.07 38.14 13.36 16.42 [expected counts]

0.106 0.901 0.010 2.513 [Chi-Square contr]

Medium Income

11 52 23 22 108

11.47 54.20 18.99 23.34

0.019 0.089 0.847 0.077

High Income

9 41 12 27 89

9.45 44.66 15.65 19.23

0.022 0.300 0.851 3.135

Total 29 137 48 59 273

Chi-Sq = 8.871, DF = 6, P-Value = 0.181

"P-Value" = 0.181 > 0.10 [90% confidence interval]; thus Null Hypothesis of homogeneity should be rejected (low, med, high income groups educated same). Alternative Hypothesis should be accepted (education dependent upon income level)

correct answer is c. just did it

step-by-step explanation:

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.