Inference for numerical data

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")
head(nc)

##   fage mage      mature weeks    premie visits marital gained weight
## 1   NA   13 younger mom    39 full term     10 married     38   7.63
## 2   NA   14 younger mom    42 full term     15 married     20   7.88
## 3   19   15 younger mom    37 full term     11 married     38   6.63
## 4   21   15 younger mom    41 full term      6 married     34   8.00
## 5   NA   15 younger mom    39 full term      9 married     27   6.38
## 6   NA   15 younger mom    38 full term     19 married     22   5.38
##   lowbirthweight gender     habit  whitemom
## 1        not low   male nonsmoker not white
## 2        not low   male nonsmoker not white
## 3        not low female nonsmoker     white
## 4        not low   male nonsmoker     white
## 5        not low female nonsmoker not white
## 6            low   male nonsmoker not white

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?

jmb answer: each row or case of the data represents a baby born and mother pair. There are 1,000 cases in this data set.

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

library(ggplot2)
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.

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.

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

jmb answer: It appears that mothers who smoke have babies with lower birth weight.

boxplot(nc$weight~nc$habit)

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

3. 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.

jmb answer: Independence: The sample is random and each and the sample size is at least 30 and the sample makes up less than 10% of all baby-mother pairs. No information is provided about the skew.

by(nc$weight, nc$habit, length)

## nc$habit: nonsmoker
## [1] 873
## -------------------------------------------------------- 
## nc$habit: smoker
## [1] 126

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

jmb answer:

\(H_{0}: \mu_{diff} = 0\) i.e. there is no difference in means
\(H_{A}: \mu_{diff} \neq 0\) i.e. there is a difference in the means

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")

## Warning: package 'BHH2' was built under R version 3.3.3

## 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.

5. 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.

jmb answer:

inference(y = nc$weight, x = nc$habit, est = "mean", type = "ci", 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
## 
## Standard error = 0.1338 
## 95 % Confidence interval = ( 0.0534 , 0.5777 )

# 95 % Confidence interval = ( 0.0534 , 0.5777 )

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

• Calculate a 95% confidence interval for the average length of pregnancies (weeks) and interpret it in context. Note that since you’re doing inference on a single population parameter, there is no explanatory variable, so you can omit the x variable from the function.

jmb answer using the inference function, I’ve calculated the mean weeks of pregnancy to 38.3 weeks. The 95% confidence interval is: ( 38.1528 , 38.5165 )

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 )

• Calculate a new confidence interval for the same parameter at the 90% confidence level. You can change the confidence level by adding a new argument to the function: conflevel = 0.90.

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 )

• Conduct a hypothesis test evaluating whether the average weight gained by younger mothers is different than the average weight gained by mature mothers.

\(H_{0}: \mu_{diff} = 0\) i.e. there is no difference in weight gains
\(H_{A}: \mu_{diff} \neq 0\) i.e. there is a difference in the avg weight gains

Given the below, I’ve failed to reject \(H_{0}\).

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

• Now, a non-inference task: Determine the age cutoff for younger and mature mothers. Use a method of your choice, and explain how your method works.

jmb answer using the by function, I’ll select mage and mature to get the max in each group which tells me the max age for mature, younger moms is 34.

#qplot(x = mage, y = weight, data = nc, geom = "point", color = mature)
by(nc$mage, nc$mature, max)

## nc$mature: mature mom
## [1] 50
## -------------------------------------------------------- 
## nc$mature: younger mom
## [1] 34

• Pick a pair of numerical and categorical variables and come up with a research question evaluating the relationship between these variables. Formulate the question in a way that it can be answered using a hypothesis test and/or a confidence interval. Answer your question using the inference function, report the statistical results, and also provide an explanation in plain language.

jmb answer:

Average weight gained between smokers and non smokers is equal:

\(H_{0}: \mu_{diff} = 0\) i.e. there is no difference.
\(H_{A}: \mu_{diff} \neq 0\) i.e. there is a difference

\(p>\alpha\): fail to reject the null hypothesis.

inference(y = nc$gained, x = nc$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 = 851, mean_nonsmoker = 30.0964, sd_nonsmoker = 14.0226
## n_smoker = 122, mean_smoker = 31.9262, sd_smoker = 15.6512

## Observed difference between means (nonsmoker-smoker) = -1.8299
## 
## H0: mu_nonsmoker - mu_smoker = 0 
## HA: mu_nonsmoker - mu_smoker != 0 
## Standard error = 1.496 
## Test statistic: Z =  -1.223 
## p-value =  0.2214

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.