Exercise 4: Solutions fourth part

Load the data in R.

dat <- read.csv("NHANES1.csv")

# make factors
integer_info <- sapply(dat, is.integer)
integer_info[which(names(integer_info) == "age")] <- FALSE  # age should stay an integer
dat[integer_info] <- lapply(dat[integer_info], as.factor)

Task 4.1

Is the variance the same for males and females? Perform an appropriate test. What is the null hypothesis?

var.test(dat$height[dat$male], dat$height[!dat$male])

    F test to compare two variances

data:  dat$height[dat$male] and dat$height[!dat$male]
F = 1.1147, num df = 2352, denom df = 2367, p-value = 0.00838
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
 1.028257 1.208407
sample estimates:
ratio of variances 
          1.114693

Task 4.2

Which test can we use, if, instead, we want to check if males are on average taller than females? Set an adequate alternative hypothesis.

t.test(dat$height[dat$male], dat$height[!dat$male], alternative='greater')

    Welch Two Sample t-test

data:  dat$height[dat$male] and dat$height[!dat$male]
t = 61.102, df = 4701.7, p-value < 2.2e-16
alternative hypothesis: true difference in means is greater than 0
95 percent confidence interval:
 12.96329      Inf
sample estimates:
mean of x mean of y 
 173.9717  160.6497

Task 4.3

Analyze the confidence interval obtained in the previous point. Why doesn’t it have an upper bound?

ci <- t.test(dat$height[dat$male], dat$height[!dat$male], alternative='greater')
ci$conf.int
[1] 12.96329      Inf
attr(,"conf.level")
[1] 0.95

Task 4.4

Now look at the distribution of the variable weight: can we graphically state its normality? Perform a transformation in order to recover it.

hist(dat$weight,breaks=50)

qqnorm(dat$weight)
qqline(dat$weight)

hist(log(dat$weight),breaks=50)

qqnorm(log(dat$weight))
qqline(log(dat$weight))

Task 4.5

Test if the mean of the variable weight is 80 kg, testing \(H_0: log(weight) = log(80)\) versus \(H_1: log(weight) \neq log(80)\). Can you state an interval estimate with a level of 0.99?

t.test(log(dat$weight),mu=log(80))

    One Sample t-test

data:  log(dat$weight)
t = -8.1594, df = 4721, p-value = 4.277e-16
alternative hypothesis: true mean is not equal to 4.382027
95 percent confidence interval:
 4.344784 4.359213
sample estimates:
mean of x 
 4.351998

Task 4.6

Provide a punctual and an interval estimate for the prevalence of heart diseases and lung pathology.

prop.test(table(!dat$heartdis_ever))

    1-sample proportions test with continuity correction

data:  table(!dat$heartdis_ever), null probability 0.5
X-squared = 3359.6, df = 1, p-value < 2.2e-16
alternative hypothesis: true p is not equal to 0.5
95 percent confidence interval:
 0.07042734 0.08594290
sample estimates:
         p 
0.07783669 
prop.test(table(!dat$lungpath_ever))

    1-sample proportions test with continuity correction

data:  table(!dat$lungpath_ever), null probability 0.5
X-squared = 3562.7, df = 1, p-value < 2.2e-16
alternative hypothesis: true p is not equal to 0.5
95 percent confidence interval:
 0.05882014 0.07316893
sample estimates:
         p 
0.06563625

Task 4.7

Test if the prevalence is statistically different between men and women.

table(dat$male,dat$heartdis_ever)
       
        FALSE TRUE
  FALSE  2212  170
  TRUE   2136  197
prop.test(table(dat$male,!dat$heartdis_ever))

    2-sample test for equality of proportions with continuity correction

data:  table(dat$male, !dat$heartdis_ever)
X-squared = 2.6267, df = 1, p-value = 0.1051
alternative hypothesis: two.sided
95 percent confidence interval:
 -0.028799184  0.002655111
sample estimates:
    prop 1     prop 2 
0.07136860 0.08444063 
chisq.test(table(dat$male, dat$heartdis_ever))

    Pearson's Chi-squared test with Yates' continuity correction

data:  table(dat$male, dat$heartdis_ever)
X-squared = 2.6267, df = 1, p-value = 0.1051

Task 4.8

Turn the variable smokstat into a binary variable (put the non-smokers and the people who almost never smoked into one group).

table(dat$smokstat)

   1    2    3 
1089  162  788 
smoke_bin <- NULL
smoke_bin[dat$smokstat == 1 | dat$smokstat == 2] <- 0
smoke_bin[dat$smokstat == 3] <- 1
smoke_bin <- as.factor(smoke_bin)
dat$smoke_bin <- smoke_bin
table(smoke_bin)
smoke_bin
   0    1 
1251  788

Task 4.9

Do you expect smokers to have a higher cancer prevalence than non-smokers? Test if there is a significant difference in cancer prevalence between smokers and non-smokers (using the variable from before). If so, which group has a higher prevalence? Did you get the result you expected?

(tab <- table(dat$smoke_bin, dat$cancer_ever))
   
    FALSE TRUE
  0  1080  171
  1   740   46
prop.table(tab, margin = 1)  # normalize by first margins to 1
   
         FALSE       TRUE
  0 0.86330935 0.13669065
  1 0.94147583 0.05852417
prop.test(table(dat$smoke_bin, !dat$cancer_ever))

    2-sample test for equality of proportions with continuity correction

data:  table(dat$smoke_bin, !dat$cancer_ever)
X-squared = 30.171, df = 1, p-value = 3.955e-08
alternative hypothesis: two.sided
95 percent confidence interval:
 0.05199798 0.10433497
sample estimates:
    prop 1     prop 2 
0.13669065 0.05852417

Task 4.10

Do the same test in the subgroup of people who are between 20 and 49 years old. What do you see now? How can you explain the results?

table(dat$smoke_bin[dat$age < 50], !dat$cancer_ever[dat$age < 50])
   
    FALSE TRUE
  0     9  417
  1    14  443
prop.test(table(dat$smoke_bin[dat$age < 50], !dat$cancer_ever[dat$age < 50]))

    2-sample test for equality of proportions with continuity correction

data:  table(dat$smoke_bin[dat$age < 50], !dat$cancer_ever[dat$age < 50])
X-squared = 0.45555, df = 1, p-value = 0.4997
alternative hypothesis: two.sided
95 percent confidence interval:
 -0.03265876  0.01364314
sample estimates:
    prop 1     prop 2 
0.02112676 0.03063457