Need answers for your academic questions?

The table needs to be inputed into R in order to answer the following questions. Thank you



ANSWER

We can easily calculate the expected cell counts Eij using the Minitab command chisq, as shown below.

MTB > read c1-c4

DATA> 12 8 31 41

DATA> 307 246 439 245

DATA> end

2 rows read.

MTB > chisq c1-c4

Expected counts are as below observed counts

C1     C2     C3     C4     Total

1       12    8      31    41        92

22.08   17.58     32.54    19.80

2      307 246 439   245 1237

296.92 236.42 437.46 266.20

Total          319   254 470 286 1329

ChiSq = 4.604 + 5.223 + 0.072 + 22.704 + 0.342 + 0.388 + 0.005 + 1.689 = 35.028

df = 3

MTB > cdf 35.028;

SUBC> chisq 3.

35.0280 1.0000

The p-value is essentially zero, so the evidence of a relationship is very strong. The same computation is shown below in S-PLUS, using the function chisq.test().

> x_c(12,8,31,41,307,246,439,245)

> x_matrix(x,4,2,byrow=T)

> chisq.test(x)

Pearson’s chi-square test without Yates’ continuity correction

data: x

X-squared = 35.0285, df = 3, p-value = 0

Through the X2 test for independence, we have demonstrated beyond a reasonable doubt that a relationship exists between cholesterol and CHD.

It would make sense to estimate the conditional probabilities of CHD within the four cholesterol groups. To do this, we estimate P(Y = i|Z = j).

P(Y = i|Z = j) = P(Y = i, Z = j) / P(Z = j)

[(n_ij/n₊₊) / (n_+j/n₊₊)] = n_ij / n_+j

12/319 = .038

8/254 = .031

31/470 = .066

41/286 = .143

307/319 = .962

246/254 = .969

439/470 = .934

245/286 = .857

The risk of CHD appears to be essentially constant for the 0–199 and 200–219 groups.

a test of independence for the 2 × 2 table

12	8
307	246

yields X ² = 0.157, p-value = .69.

For an I × J table, the usual X ² or G ² test for independence has

(IJ − 1) − (I − 1) − (J − 1) = (I − 1)(J − 1)