library(ggplot2)
library(dplyr)
## 
## Attaching package: 'dplyr'
## The following objects are masked from 'package:stats':
## 
##     filter, lag
## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union

North Carolina births

In 2004, the state of North Carolina released a large data set containing information on births recorded in this state. This data set is useful to researchers studying the relation between habits and practices of expectant mothers and the birth of their children. We will work with a random sample of observations from this data set.

Exploratory analysis

Load the nc data set into our workspace.

load("more/nc.RData")

We have observations on 13 different variables, some categorical and some numerical. The meaning of each variable is as follows.

variable description
fage father’s age in years.
mage mother’s age in years.
mature maturity status of mother.
weeks length of pregnancy in weeks.
premie whether the birth was classified as premature (premie) or full-term.
visits number of hospital visits during pregnancy.
marital whether mother is married or not married at birth.
gained weight gained by mother during pregnancy in pounds.
weight weight of the baby at birth in pounds.
lowbirthweight whether baby was classified as low birthweight (low) or not (not low).
gender gender of the baby, female or male.
habit status of the mother as a nonsmoker or a smoker.
whitemom whether mom is white or not white.
  1. What are the cases in this data set? How many cases are there in our sample?

Answer - The cases are for all the births population in North Carolina in 2004. Each case has the relevant information about the parents of the child, and some child’s information as well. The sample has 1000 cases.

As a first step in the analysis, we should consider summaries of the data. This can be done using the summary command:

summary(nc)
##       fage            mage            mature        weeks      
##  Min.   :14.00   Min.   :13   mature mom :133   Min.   :20.00  
##  1st Qu.:25.00   1st Qu.:22   younger mom:867   1st Qu.:37.00  
##  Median :30.00   Median :27                     Median :39.00  
##  Mean   :30.26   Mean   :27                     Mean   :38.33  
##  3rd Qu.:35.00   3rd Qu.:32                     3rd Qu.:40.00  
##  Max.   :55.00   Max.   :50                     Max.   :45.00  
##  NA's   :171                                    NA's   :2      
##        premie        visits            marital        gained     
##  full term:846   Min.   : 0.0   married    :386   Min.   : 0.00  
##  premie   :152   1st Qu.:10.0   not married:613   1st Qu.:20.00  
##  NA's     :  2   Median :12.0   NA's       :  1   Median :30.00  
##                  Mean   :12.1                     Mean   :30.33  
##                  3rd Qu.:15.0                     3rd Qu.:38.00  
##                  Max.   :30.0                     Max.   :85.00  
##                  NA's   :9                        NA's   :27     
##      weight       lowbirthweight    gender          habit    
##  Min.   : 1.000   low    :111    female:503   nonsmoker:873  
##  1st Qu.: 6.380   not low:889    male  :497   smoker   :126  
##  Median : 7.310                               NA's     :  1  
##  Mean   : 7.101                                              
##  3rd Qu.: 8.060                                              
##  Max.   :11.750                                              
##                                                              
##       whitemom  
##  not white:284  
##  white    :714  
##  NA's     :  2  
##                 
##                 
##                 
##

As you review the variable summaries, consider which variables are categorical and which are numerical. For numerical variables, are there outliers? If you aren’t sure or want to take a closer look at the data, make a graph.

Answer: Categorical variables are: mature premie marital lowbirthweight gender habit whitemom

Numerical variables are: fage mage weeks visits gained weight

So out of total 13 variables, 7 are categorical and 6 are numerical.

Checking for the distribution of the numeric variables, and any outliers: We are creating histograms for all the numeric variables below.

hist(nc$fage)

hist(nc$mage)

hist(nc$weeks)

hist(nc$visits)

hist(nc$gained)

hist(nc$weight)

From the histogram plots, we will deduce the following: 1) Father’s age - no outliers as per the histogram. 2) Mother’s age - no outliers as per the histogram. 3) Lengths of pregeneancy is highly shewed on the lefyt. That clearly shows that there are outliers in this case. 4) Number of hospital visits is having outliers on the right, that means more number of visits. 5) Weight gained also has a strong right skew, and there are outliers on the right side. 6) Weight of the baby has strong skew on the left. It means there are some observations where new born babies have very less weights.

Consider the possible relationship between a mother’s smoking habit and the weight of her baby. Plotting the data is a useful first step because it helps us quickly visualize trends, identify strong associations, and develop research questions.

  1. Make a side-by-side boxplot of habit and weight. What does the plot highlight about the relationship between these two variables?
qplot(nc$habit, nc$weight, geom = "boxplot", na.rm = TRUE)

The median weight of the new brons from the mothers who smoke is less than the same fromt eh mother who does not smoke. Even though there are many outliers on the lower side in case of the non-smoking mothers, but the general range of weight for the non-smoker mothers is slightly more than the same for the smoker mothers.

The box plots show how the medians of the two distributions compare, but we can also compare the means of the distributions using the following function to split the weight variable into the habit groups, then take the mean of each using the mean function.

by(nc$weight, nc$habit, mean)
## nc$habit: nonsmoker
## [1] 7.144273
## -------------------------------------------------------- 
## nc$habit: smoker
## [1] 6.82873

There is an observed difference, but is this difference statistically significant? In order to answer this question we will conduct a hypothesis test .

Inference

  1. Check if the conditions necessary for inference are satisfied. Note that you will need to obtain sample sizes to check the conditions. You can compute the group size using the same by command above but replacing mean with length.

Answer: So, here we are dealing with the weight of the new borns under the 2 categories - smoking mothers and non-smoking mothers. We will reate the histograms of the 2 categories separately:

nc_smoking_mothers <- nc[nc$habit == "nonsmoker",]
nrow(nc_smoking_mothers)
## [1] 874
hist(nc_smoking_mothers$weight)

nc_non_smoking_mothers <- nc[nc$habit == "smoker",]
nrow(nc_non_smoking_mothers)
## [1] 127
hist(nc_non_smoking_mothers$weight)

To confirm the conditions necessary for inference: 1) Both the data sets - smomking and non-smking mothers are individually independent 2) The distributions are skewed for both but the counts are large enough to ignore the skewness, and assume that this follows normal distribution.

  1. Write the hypotheses for testing if the average weights of babies born to smoking and non-smoking mothers are different.

Answer: Null Hypothesis H0 : The difference between the average weight of babies born to smoking and no-smoking mothers is Zero ??-non-smoking - ??-smoking = 0

Alternate Hypothesis H0 : The difference between the average weight of babies born to smoking and no-smoking mothers is not Zero ??-non-smoking - ??-smoking <> 0

Next, we introduce a new function, inference, that we will use for conducting hypothesis tests and constructing confidence intervals.

inference(y = nc$weight, x = nc$habit, est = "mean", type = "ht", null = 0, 
          alternative = "twosided", method = "theoretical")
## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_nonsmoker = 873, mean_nonsmoker = 7.1443, sd_nonsmoker = 1.5187
## n_smoker = 126, mean_smoker = 6.8287, sd_smoker = 1.3862
## Observed difference between means (nonsmoker-smoker) = 0.3155
## 
## H0: mu_nonsmoker - mu_smoker = 0 
## HA: mu_nonsmoker - mu_smoker != 0 
## Standard error = 0.134 
## Test statistic: Z =  2.359 
## p-value =  0.0184

Let’s pause for a moment to go through the arguments of this custom function. The first argument is y, which is the response variable that we are interested in: nc$weight. The second argument is the explanatory variable, x, which is the variable that splits the data into two groups, smokers and non-smokers: nc$habit. The third argument, est, is the parameter we’re interested in: "mean" (other options are "median", or "proportion".) Next we decide on the type of inference we want: a hypothesis test ("ht") or a confidence interval ("ci"). When performing a hypothesis test, we also need to supply the null value, which in this case is 0, since the null hypothesis sets the two population means equal to each other. The alternative hypothesis can be "less", "greater", or "twosided". Lastly, the method of inference can be "theoretical" or "simulation" based.

  1. Change the type argument to "ci" to construct and record a confidence interval for the difference between the weights of babies born to smoking and non-smoking mothers.

By default the function reports an interval for (\(\mu_{nonsmoker} - \mu_{smoker}\)) . We can easily change this order by using the order argument:

inference(y = nc$weight, x = nc$habit, est = "mean", type = "ci", null = 0, 
          alternative = "twosided", method = "theoretical", 
          order = c("smoker","nonsmoker"))
## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_smoker = 126, mean_smoker = 6.8287, sd_smoker = 1.3862
## n_nonsmoker = 873, mean_nonsmoker = 7.1443, sd_nonsmoker = 1.5187

## Observed difference between means (smoker-nonsmoker) = -0.3155
## 
## Standard error = 0.1338 
## 95 % Confidence interval = ( -0.5777 , -0.0534 )

On your own

inference(y = nc$weeks, est = "mean", type = "ci", null = 0, 
          alternative = "twosided", method = "theoretical")
## Single mean 
## Summary statistics:

## mean = 38.3347 ;  sd = 2.9316 ;  n = 998 
## Standard error = 0.0928 
## 95 % Confidence interval = ( 38.1528 , 38.5165 )

Hence 95 % confidence level is: ( 38.1528 , 38.5165 ) That means that 95% of the women will have the average length falling in this range.

inference(y = nc$weeks, est = "mean", type = "ci", null = 0, 
          alternative = "twosided", method = "theoretical", conflevel = 0.90)
## Single mean 
## Summary statistics:

## mean = 38.3347 ;  sd = 2.9316 ;  n = 998 
## Standard error = 0.0928 
## 90 % Confidence interval = ( 38.182 , 38.4873 )

H0: average weight gained by younger mothers and average weight gained by mature mothers is same ??-weight-gained-younger-mothers - ??-weight-gained-mature-mothers = 0 H0: average weight gained by younger mothers and average weight gained by mature mothers is not same ??-weight-gained-younger-mothers - ??-weight-gained-mature-mothers <> 0

Now we have to find the p-value of the point estimate.

xbar.weight.gained.younger.mothers <- mean(nc$gained[nc$mature == "younger mom"], na.rm = TRUE)

xbar.weight.gained.mature.mothers <- mean(nc$gained[nc$mature == "mature mom"], na.rm = TRUE)
point.mean.difference <- xbar.weight.gained.younger.mothers - xbar.weight.gained.mature.mothers
mean.null <- 0
s_young <- sd(nc$gained[nc$mature == "younger mom"], na.rm = TRUE)
n_young <- nrow(nc[nc$mature == "younger mom",])
s_mature <- sd(nc$gained[nc$mature == "mature mom"], na.rm = TRUE)
n_mature <- nrow(nc[nc$mature == "mature mom",])

SE.young.minus.mature <- sqrt(s_young ^2/n_young + s_mature ^2/n_mature)
z.value <- (point.mean.difference - mean.null) / SE.young.minus.mature
p.value <- 2 * pnorm(-abs(z.value))

Now since p-value > significance value which is normally taken as 0.05, H0 (null hypothesis) cannot be rejected. Hence there is not much evidence to reject that average weight gained by younger mothers and average weight gained by mature mothers is same.

younger.mom.min.age <- nc$mage[nc$mature == "younger mom"] %>% min()
younger.mom.max.age <- nc$mage[nc$mature == "younger mom"] %>% max()
sprintf("Younger moms min age is: %s", younger.mom.min.age)
## [1] "Younger moms min age is: 13"
sprintf("Younger moms max age is: %s", younger.mom.max.age)
## [1] "Younger moms max age is: 34"
mature.mom.min.age <- nc$mage[nc$mature == "mature mom"] %>% min()
mature.mom.max.age <- nc$mage[nc$mature == "mature mom"] %>% max()
sprintf("Mature moms min age is: %s", mature.mom.min.age)
## [1] "Mature moms min age is: 35"
sprintf("Mature moms max age is: %s", mature.mom.max.age)
## [1] "Mature moms max age is: 50"

This is a product of OpenIntro that is released under a Creative Commons Attribution-ShareAlike 3.0 Unported. This lab was adapted for OpenIntro by Mine Çetinkaya-Rundel from a lab written by the faculty and TAs of UCLA Statistics.