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.

library(DATA606)
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## Welcome to CUNY DATA606 Statistics and Probability for Data Analytics 
## This package is designed to support this course. The text book used 
## is OpenIntro Statistics, 3rd Edition. You can read this by typing 
## vignette('os3') or visit www.OpenIntro.org. 
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load("C:/Users/chitr/OneDrive/Documents/R/win-library/3.6/DATA606/labs/Lab7/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?

Case corresponds to each child birth and the related information that we have corresponding to child’s mother father etc. In our sample we have a total of 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             premie   
##  Min.   :14.00   Min.   :13   mature mom :133   Min.   :20.00   full term:846  
##  1st Qu.:25.00   1st Qu.:22   younger mom:867   1st Qu.:37.00   premie   :152  
##  Median :30.00   Median :27                     Median :39.00   NA's     :  2  
##  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                      
##      visits            marital        gained          weight      
##  Min.   : 0.0   married    :386   Min.   : 0.00   Min.   : 1.000  
##  1st Qu.:10.0   not married:613   1st Qu.:20.00   1st Qu.: 6.380  
##  Median :12.0   NA's       :  1   Median :30.00   Median : 7.310  
##  Mean   :12.1                     Mean   :30.33   Mean   : 7.101  
##  3rd Qu.:15.0                     3rd Qu.:38.00   3rd Qu.: 8.060  
##  Max.   :30.0                     Max.   :85.00   Max.   :11.750  
##  NA's   :9                        NA's   :27                      
##  lowbirthweight    gender          habit          whitemom  
##  low    :111    female:503   nonsmoker:873   not white:284  
##  not low:889    male  :497   smoker   :126   white    :714  
##                              NA's     :  1   NA's     :  2  
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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.

As can be seen from the graphs below: 1. fage and mage do not seem to have any outliers - the distributions appear quite symmetric also. Although evaluating using the outlier definition of Q3+1.5IQR - there are outliers in each of fage and mage

  1. weeks variable appears left skewed with many observations falling below Q1 confirming the pesence of outliers
  2. visis variable appears left skewed with many observations falling above Q3 - confirming the presence of outliers
  3. gained and weight variables also exhibit outliers with the distributions being left and right skewed respectively
par(mfrow = c(2, 2))
hist(nc$fage)
hist(nc$mage)
hist(nc$weeks)
hist(nc$visits)

hist(nc$gained)
hist(nc$weight)

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.

plot(nc$weight ~ nc$habit, ylab = "Child Weight", xlab = "Smoker/Non-Smoker")

  1. Make a side-by-side boxplot of habit and weight. What does the plot highlight about the relationship between these two variables?

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.

As can be seen from the sample sizes which are well above 30 for both the non-smoker and smoker group.

by(nc$weight, nc$habit, length)
## nc$habit: nonsmoker
## [1] 873
## ------------------------------------------------------------ 
## nc$habit: smoker
## [1] 126
  1. Write the hypotheses for testing if the average weights of babies born to smoking and non-smoking mothers are different.

Ho = The mean weights of the childern having smoker and non-smoker mothers are same Ha = The mean weights are different

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(nc$weeks, est = "mean", type = "ci", 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 )
inference(nc$weeks, est = "mean", type = "ci", method="theoretical", conflevel = 0.9)
## Single mean 
## Summary statistics:

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

since p-value is greater than 0.05 - we do not have sufficient evidence to reject the null hypothesis. hence there is no evidence that the weight gained by mature and young mothers is statistically different.

inference(y = nc$gained, x = nc$mature, est = "mean", type = "ht", null = 0, 
          alternative = "twosided", method = "theoretical")
## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_mature mom = 129, mean_mature mom = 28.7907, sd_mature mom = 13.4824
## n_younger mom = 844, mean_younger mom = 30.5604, sd_younger mom = 14.3469
## Observed difference between means (mature mom-younger mom) = -1.7697
## 
## H0: mu_mature mom - mu_younger mom = 0 
## HA: mu_mature mom - mu_younger mom != 0 
## Standard error = 1.286 
## Test statistic: Z =  -1.376 
## p-value =  0.1686

The cut-off age is 34. Any mother having age more than 34 is categorized as mature. The method groups the sample based on mature variable and then identifies the max and the min ages for the respective groups. Logically there should not be overlap of ages in the two categoties. This is getting verified from the output wherein the max age of younger mom is lower than min age of mature mom. thus, the max age of younger mom serves as the cut-off.

library(RCurl)
library(tidyverse)
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nc%>%group_by(mature)%>%summarize(max=max(mage), min=min(mage))
## # A tibble: 2 x 3
##   mature        max   min
##   <fct>       <int> <int>
## 1 mature mom     50    35
## 2 younger mom    34    13

We can pick weeks=length of pregnancy in weeks and habit=smoker or non-smoker and try to answer the questions that whether smoker moms tend to have shorter pregnancies.

Ho= The mean pregnancy weeks for smoker and non-smoker moms are same Ha= The mean pregnancy weeks are same

As can be inferred from the high p-value that we cannot reject the null and therefore can conclude that the mean pregnancies are same for smoker and non-smoker.

inference(y = nc$weeks, x = nc$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 = 872, mean_nonsmoker = 38.3188, sd_nonsmoker = 2.9936
## n_smoker = 126, mean_smoker = 38.4444, sd_smoker = 2.4676
## Observed difference between means (nonsmoker-smoker) = -0.1256
## 
## H0: mu_nonsmoker - mu_smoker = 0 
## HA: mu_nonsmoker - mu_smoker != 0 
## Standard error = 0.242 
## Test statistic: Z =  -0.519 
## p-value =  0.6038