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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)
[(nij/n++) / (n+j/n++)] = nij / 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)

Comments

  1. UnknownMay 22, 2019 at 10:09 PM

    https://www.chegg.com/homework-help/questions-and-answers/2-survey-find-main-causes-lateness-factory-s-work-force-random-sample-200-employees-late-w-q38226745

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