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")

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?

Each case is an observation of dad’s age together with other mother and baby characteristics. There is a total of 1000 cases.

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
nrow(nc)
## [1] 1000

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.

Categorical variables: mature, premie, maritial, lowbirthrate, gender, habit, whitemom

Numerical: (outliners defined as having a min or max more than 2 times the standard deviations away from the mean) fage:

testOutliers<-function(data) {
  m<-mean(data,na.rm=TRUE)
  m
  s<-sd(data,na.rm=TRUE)
  s
  min<-min(data,na.rm=TRUE)
  min
  max<-max(data,na.rm=TRUE)
  max
  if (min<m-2*s | max>m+2*s) {
    print("there are outliers")
  } else {
    print("no outliers")
  }
}

testOutliers(nc$fage)
## [1] "there are outliers"

mage:

testOutliers(nc$mage)
## [1] "there are outliers"

weeks:

testOutliers(nc$weeks)
## [1] "there are outliers"

visits:

testOutliers(nc$visits)
## [1] "there are outliers"

gained:

testOutliers(nc$gained)
## [1] "there are outliers"

weight:

testOutliers(nc$weight)
## [1] "there are outliers"

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?

The side by side box plot lets us compare the difference between the means and also the difference between the spread of values in both data sets.

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

  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 below both distributions for smoking and non-smoking seem normal, although with some skew. This can be seen in both the distribution and probability plots. But the sample sizes are large enough, much greater than 30, so some skew is acceptable. Another condition is that the samples need to be random and independent. Because they have been taking from a large population, all births in the state, we can assume they do not represent a large proportion of the population, more than 10%, so they are independent - according to this site the number of births in North Carolina is consistently over 150k https://www.marchofdimes.org/peristats/ViewSubtopic.aspx?reg=37&top=2&stop=1&lev=1&slev=4&obj=1. We have also no reason to assume there was any relationship between the subjects in the samples, so they are taken at random and are independent.

by(nc$weight, nc$habit, length)
## nc$habit: nonsmoker
## [1] 873
## -------------------------------------------------------- 
## nc$habit: smoker
## [1] 126
h<-hist(nc$weight[nc$habit == "smoker"], main = "Weight for Smoking Mothers", xlab = "Weight",freq = FALSE)
x <- seq(0, 12, by = 0.01)
y <- dnorm(x,mean(nc$weight[nc$habit == "smoker"],na.rm=TRUE),sd(nc$weight[nc$habit == "smoker"],na.rm=TRUE))
lines(x = x, y = y, col = "blue")

qqnorm(nc$weight[nc$habit == "smoker"])
qqline(nc$weight[nc$habit == "smoker"])

h<-hist(nc$weight[nc$habit == "nonsmoker"], main = "Weight for Non-Smoking Mothers", xlab = "Weight",freq = FALSE)
x <- seq(0, 12, by = 0.01)
y <- dnorm(x,mean(nc$weight[nc$habit == "smoker"],na.rm=TRUE),sd(nc$weight[nc$habit == "smoker"],na.rm=TRUE))
lines(x = x, y = y, col = "blue")

qqnorm(nc$weight[nc$habit == "nonsmoker"])
qqline(nc$weight[nc$habit == "nonsmoker"])

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

Ho: the average weight of babies born to smoking and non-smoking mothers are not different.
\(\mu_{smoking}\) = \(\mu_{non\_smoking}\)

Ha: the average weight of babies born to smoking and non-smoking mothers are different.
\(\mu_{smoking}\) \(\neq\) \(\mu_{non\_smoking}\)

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.
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 )

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

1- 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.

As can be seen below the mean for the number of pregnancy weeks is 38.3347, and we are 95% confident the real population mean for pregnancy weeks is between 38.1528 and 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 )

2- 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.

As can be seen below the mean for the number of pregnancy weeks is 38.3347, and we are 90% confident the real population mean for pregnancy weeks is between 38.182 and 38.4873

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 )

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

We are using a 95% confidence level. As can be seen below, the p-value for this test is greated than 0.05, which means we do not reject the hull. We have no indication (statistical evidence) that there is a weight gain difference between these two groups. We get a similar result calculating the 95% confidence interval, where the null or a difference of zero is within the interval.

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

inference(y = nc$gained, x = nc$mature, est = "mean", type = "ci", 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
## 
## Standard error = 1.2857 
## 95 % Confidence interval = ( -4.2896 , 0.7502 )

4- 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.

We look for the max of young mons and the min for mature mons to find the cutoff. We that young mons ahve up to 34 years, and mature mons stant at 35 years.

max(subset(nc,nc$mature == "younger mom")$mage)
## [1] 34
min(subset(nc,nc$mature == "mature mom")$mage)
## [1] 35

5- 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.

We ask ourselves if white mons and non white mons have babies are born with different average weights. We state our hypothesis as:

Ho: white mon babies are the same average weight as non white mons
\(\mu_{weight\_white\_mon}\) = \(\mu_{weight\_non\_white\_mon}\)
Ha: white mon babies are not the same average weight as non white mons
\(\mu_{weight\_white\_mon}\) \(\neq\) \(\mu_{weight\_non\_white\_mon}\)

We run both p-value and confidence interval.

For p-value we find it to be very small, almost zero: p-val is 7.142018e-07
Because the p-value is very small, we reject the null and conclude that we are confident that there is a difference between the average weight of white and non-weight moms.

Looking at the confidence interval analysis we arrive at the same conclusion. Because zero difference between the means of the weights of white and non-white mons is not in the interval, we reject the null and thus also conclude there is a difference between the average weights a white and non white mons.

inference(y = nc$weight, x = nc$whitemom, est = "mean", type = "ht", null = 0, 
          alternative = "twosided", method = "theoretical")
## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_not white = 284, mean_not white = 6.7195, sd_not white = 1.6207
## n_white = 714, mean_white = 7.2505, sd_white = 1.4333
## Observed difference between means (not white-white) = -0.5309
## 
## H0: mu_not white - mu_white = 0 
## HA: mu_not white - mu_white != 0 
## Standard error = 0.11 
## Test statistic: Z =  -4.821 
## p-value =  0

inference(y = nc$weight, x = nc$whitemom, est = "mean", type = "ci", null = 0, 
          alternative = "twosided", method = "theoretical")
## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_not white = 284, mean_not white = 6.7195, sd_not white = 1.6207
## n_white = 714, mean_white = 7.2505, sd_white = 1.4333

## Observed difference between means (not white-white) = -0.5309
## 
## Standard error = 0.1101 
## 95 % Confidence interval = ( -0.7467 , -0.3151 )

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.