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In a randomized experiment, researchers assigned 407 volunteers to receive 1,000 mg of Vitamin C daily throughout the cold season and 411 to receive a placebo. A physician interviewed the volunteers at the end of the study to determine whether or not they had suffered any colds during the study period.
1 |
A data frame with 2 observations on the following 3 variables.
a factor with levels "Placebo"
and
"VitC"
the number of who got colds
the number that did not get any colds
Ramsey, F.L. and Schafer, D.W. (2013). The Statistical Sleuth: A Course in Methods of Data Analysis (3rd ed), Cenage Learning.
Anderson, T.W., Reid, D.B.W. and Beaton, G. H. (1972). Vitamin C and the Common Cold, Canadian Medial Association Journal 107: 503–508.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 | str(case1802)
attach(case1802)
library(MASS)
## INFERENCE (4 methods)
myTable <- cbind(Cold,NoCold)
row.names(myTable) <- c("Placebo","Vitamin C")
myTable
prop.test(myTable, alternative="greater") # Compare 2 binomial proportions
# Alternative: pop prop. of first column (cold) in larger in first row (placebo)
prop.test(myTable, alternative="greater", correct=TRUE)
prop.test(myTable,correct=TRUE) # Use 2-sided alternative to get CI
chisq.test(myTable) # Chi-square test
fisher.test(myTable, alternative="greater")
fisher.test(myTable) # 2-sided alternative to get CI for odds ratio
myGlm1 <- glm(myTable ~ Treatment, family=binomial) # logistic reg (Ch 21)
summary(myGlm1)
beta <- myGlm1$coef
1 - exp(beta[2]) # 0.3474911
1 - exp(confint(myGlm1,2)) # 0.53365918 0.09042098
# Interpretation: The odds of getting a cold are 35% less on Vitamin C than
# Placebo (95% confidence interval: 9% to 53% less).
detach(case1802)
|
'data.frame': 2 obs. of 3 variables:
$ Treatment: Factor w/ 2 levels "Placebo","VitC": 1 2
$ Cold : int 335 302
$ NoCold : int 76 105
Cold NoCold
Placebo 335 76
Vitamin C 302 105
2-sample test for equality of proportions with continuity correction
data: myTable
X-squared = 5.9196, df = 1, p-value = 0.007487
alternative hypothesis: greater
95 percent confidence interval:
0.02303649 1.00000000
sample estimates:
prop 1 prop 2
0.8150852 0.7420147
2-sample test for equality of proportions with continuity correction
data: myTable
X-squared = 5.9196, df = 1, p-value = 0.007487
alternative hypothesis: greater
95 percent confidence interval:
0.02303649 1.00000000
sample estimates:
prop 1 prop 2
0.8150852 0.7420147
2-sample test for equality of proportions with continuity correction
data: myTable
X-squared = 5.9196, df = 1, p-value = 0.01497
alternative hypothesis: two.sided
95 percent confidence interval:
0.01391972 0.13222111
sample estimates:
prop 1 prop 2
0.8150852 0.7420147
Pearson's Chi-squared test with Yates' continuity correction
data: myTable
X-squared = 5.9196, df = 1, p-value = 0.01497
Fisher's Exact Test for Count Data
data: myTable
p-value = 0.007424
alternative hypothesis: true odds ratio is greater than 1
95 percent confidence interval:
1.14257 Inf
sample estimates:
odds ratio
1.531722
Fisher's Exact Test for Count Data
data: myTable
p-value = 0.01444
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
1.083492 2.172490
sample estimates:
odds ratio
1.531722
Call:
glm(formula = myTable ~ Treatment, family = binomial)
Deviance Residuals:
[1] 0 0
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) 1.4834 0.1271 11.675 <2e-16 ***
TreatmentVitC -0.4269 0.1702 -2.508 0.0121 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 6.3573 on 1 degrees of freedom
Residual deviance: 0.0000 on 0 degrees of freedom
AIC: 16.162
Number of Fisher Scoring iterations: 3
TreatmentVitC
0.3474911
Waiting for profiling to be done...
2.5 % 97.5 %
0.53365918 0.09042098
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