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.

library(ggplot2)

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: Individual child births are the cases in this data set. There are 1000 cases in our sample.

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 outlines? If you aren’t sure or want to take a closer look at the data, make a graph.

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?

Answer:

ggplot(aes(y = weight , x = habit, fill = habit), data = na.omit(nc)) + geom_boxplot()+theme_classic()

From the boxplot we can say that there is relationship between habit and birthweight. Nonsmoker mothers child have higher median weight than smoker mothers child.

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:

cd <- na.omit(nc)
ggplot(cd,aes(x=weight)) + 
  geom_histogram() + 
  facet_grid(~habit)+theme_classic()
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.

From the graph we can see that the samples are approximately normally distributed in both group.

by(nc$weight, nc$habit, length)
## nc$habit: nonsmoker
## [1] 873
## -------------------------------------------------------- 
## nc$habit: smoker
## [1] 126
by(nc$weight, nc$habit, var)
## nc$habit: nonsmoker
## [1] 2.306391
## -------------------------------------------------------- 
## nc$habit: smoker
## [1] 1.921494

Since the sample size in each group is > 30 the the samples are large enough. The assumption of equality of variance is not satisfied.

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

Answer: Null Hypothesis: Average weights of babies born to smoking and non-smoking mothers are the same

Alternative Hypothesis: Average weights of babies born to smoking and non-smoking mothers are not the same

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

inference( nc$weight, nc$habit, est = "mean", type = "ht", null = 0, 
          alternative = "twosided", method = "theoretical")
## Please visit openintro.org for free statistics materials
## 
## Attaching package: 'openintro'
## The following object is masked from 'package:ggplot2':
## 
##     diamonds
## The following objects are masked from 'package:datasets':
## 
##     cars, trees
## 
## Attaching package: 'BHH2'
## The following object is masked from 'package:openintro':
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##     dotPlot
## One quantitative and one categorical variable
## Difference between two means
## n_nonsmoker = 873 ; n_smoker = 126 
## Observed difference between means = 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.

Answer:

inference( nc$weight, nc$habit, est = "mean", type = "ci", null = 0, 
          alternative = "twosided", method = "theoretical")
## One quantitative and one categorical variable
## Difference between two means
## n_nonsmoker = 873 ; n_smoker = 126 
## Observed difference between means = 0.3155

## Standard error = 0.1338 
## 95 % Confidence interval = ( 0.05 , 0.58 )

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

inference(nc$weight,nc$habit, est = "mean", type = "ci", null = 0, 
          alternative = "twosided", method = "theoretical", 
          order = c("smoker","nonsmoker"))
## One quantitative and one categorical variable
## Difference between two means
## n_smoker = 126 ; n_nonsmoker = 873 
## Observed difference between means = -0.3155

## Standard error = 0.1338 
## 95 % Confidence interval = ( -0.58 , -0.05 )

On your own

Answer:

inference(nc$weeks, est = "mean", type = "ci", null = 0, 
          alternative = "twosided", method = "theoretical")
## One quantitative variable 
## Single mean 
## Observed mean = 38.3347

## Standard error = 0.0928 
## 95 % Confidence interval = ( 38.15 , 38.52 )

Out of 100 sample drawn from the same population 95% time we will have the mean pregnancy length in weeks of the population between 38.15 weeks and 38.52 weeks.

inference(nc$weeks, est = "mean", type = "ci", null = 0, 
          alternative = "twosided", method = "theoretical",conflevel = 0.90)
## One quantitative variable 
## Single mean 
## Observed mean = 38.3347

## Standard error = 0.0928 
## 90 % Confidence interval = ( 38.18 , 38.49 )

Answer: Null Hypothesis: The average weight gained by younger mothers is not different than the average weight gained by mature mothers.

Alternative Hypothesis: The average weight gained by younger mothers is different than the average weight gained by mature mothers.

inference( nc$gained,  nc$mature, est = "mean", type = "ci", null = 0, 
          alternative = "twosided", method = "theoretical")
## One quantitative and one categorical variable
## Difference between two means
## n_mature mom = 129 ; n_younger mom = 844 
## Observed difference between means = -1.7697

## Standard error = 1.2857 
## 95 % Confidence interval = ( -4.29 , 0.75 )

Since the confidence interval (-4.2896 , 0.7502) contains 0 we accept the Null Hypothesis that there is no difference in mean weight gain of the two populations.

Answer:

by(nc$mage, nc$mature, range)
## nc$mature: mature mom
## [1] 35 50
## -------------------------------------------------------- 
## nc$mature: younger mom
## [1] 13 34

The age cut off value for younger and mature mothers is 35. We can validate this from the range of ages in two groups.

Answer: I’ve choose weight and gender. Research Questions: Are there difference between average weight of male and female child?

Null Hypothesis: There is no difference between average male and female child weight Alternative Hypothesis: There is difference between average male and female child weight.

inference( nc$weight,  nc$gender, est = "mean", type = "ht", null = 0, 
          alternative = "twosided", method = "theoretical")
## One quantitative and one categorical variable
## Difference between two means
## n_female = 503 ; n_male = 497 
## Observed difference between means = -0.3986
## H0: mu_female - mu_male = 0 
## HA: mu_female - mu_male != 0
## Standard error = 0.095 
## Test statistic: Z =  -4.211 
## p-value:  0

The test statistic value is -4.211 . Observed difference between means is 0.3986 Since the p value 0 < 0.05 we reject the null hypothesis and conclude that there is significance difference between male and female child average weight. Males always tends to have more weight than females for this reason there is the observed difference.