\[ Z_{asian}^2 = (49 - 15.2)^2/15.2 = 75.16 \\ Z_{black}^2 = (10 - 7.08)^2/7.08 = 1.20 \\ Z_{hispanic}^2 = (34 - 44.25)^2/44.25 = 2.37 \\ Z_{white}^2 = (206 - 272.58)^2/272.58 = 16.26 \\ Z_{other}^2 = (55 - 14.87)^2/14.87 = 108.3 \]

\[ Z_{asian}^2 + Z_{black}^2 + Z_{hispanic}^2 + Z_{white}^2 + Z_{other}^2 = 203.29 = \chi^2_{obs} \]

Which of the following is an appropriate null hypothesis?

1. Ethnic diversity and choice of college are independent of one another.
2. Ethnic diversity and home-state are independent of one another.
3. \(p_{asian} = p_{black} = p_{hispanic} = p_{white} = p_{other}\).
4. The first-year class at Reed is sampled from a population that shares the same ethnic distribution as Oregon.

1. Simulation: simulate many first-year Oregonian Reedie classes and calculate the \(\chi^2\) for each.
2. Probability Theory: multinomial distribution. \(X \sim \textrm{multinom}(n, p_1, p_2, p_3, p_4, p_5)\).
3. Chi-Squared Approximation: if conditions are reasonable, the \(\chi^2\) will follow a known distribution under the null hypothesis.

1. Independent observations
2. Each expected cell count is \(\ge\) 5
3. \(k \ge 3\)

then our statistic can be well-approximated by the \(\chi^2\) distribution with \(k - 1\) degrees of freedom.

Table

R

1 - pchisq(203.29, df = 4)

## [1] 0

We reject the hypothesis that the Reed first-year class represents a random sample from Oregon w.r.t ethnicity.

The number of parameters that are free to vary, without violating any constraint imposed on it.

Parameters

\[ p_{asian}, p_{black}, p_{hispanic}, p_{white}, p_{other} \]

Since \(\sum_{i = 1}^k p_i = 1\), one of our parameters is contrained, leaving \(k-1\) that are free to vary.

These make no sense. Why not?

• We don't really care what the true \(\chi^2\) parameter value is.
• A two-sided interval wouldn't make sense
  • Distribution is bounded on the left by zero
  • Only "extreme" values are in the right tail

treatment <- rep(c("acu", "sham", "trad"), c(387, 387, 388))
pain <- c(rep(c("reduc", "noreduc"), c(184, 203)),
          rep(c("reduc", "noreduc"), c(171, 216)),
          rep(c("reduc", "noreduc"), c(106, 282)))
table(pain, treatment)

##          treatment
## pain      acu sham trad
##   noreduc 203  216  282
##   reduc   184  171  106

\(H_0\) implies that the associations between these two vectors are just due to chance, so we mirror that by randomizing the vectors to get another possible data set under the \(H_0\).

Randomization method

library(infer)
null_dist <- acu %>%
  specify(response = pain, explanatory = treatment) %>%
  hypothesize(null = "independence") %>%
  generate(reps = 1000, type = "permute") %>%
  calculate(stat = "Chisq")

The proportion of simulated \(\chi^2\) statistics under \(H_0\) that are greater than \(\chi^2_{obs}\) is 0 \(\rightarrow\) we reject the idea that pain is independent of treatment mode.

The mathematical approximation is good enough when

1. Independent observations
2. At least 5 expected counts in each cell
3. Degrees of freedom \(\ge 2\)

\[ df = (R - 1) \times (C - 1) \]

• \(R\): number of rows in table
• \(C\): number of columns in table

• Pain reduction does appear to be effected by the treatment method.
• We don't yet know which pairwise comparisons are significant \(\rightarrow\) need to follow up with CIs on the differences in proportions.