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`.

What are the cases in this data set? How many cases are there in our sample?

Answer:

Each case or observation in this data set represents the birth of a child in North Carolina.
The sample contains 1,000 cases.

str(nc)

## 'data.frame':    1000 obs. of  13 variables:
##  $ fage          : int  NA NA 19 21 NA NA 18 17 NA 20 ...
##  $ mage          : int  13 14 15 15 15 15 15 15 16 16 ...
##  $ mature        : Factor w/ 2 levels "mature mom","younger mom": 2 2 2 2 2 2 2 2 2 2 ...
##  $ weeks         : int  39 42 37 41 39 38 37 35 38 37 ...
##  $ premie        : Factor w/ 2 levels "full term","premie": 1 1 1 1 1 1 1 2 1 1 ...
##  $ visits        : int  10 15 11 6 9 19 12 5 9 13 ...
##  $ marital       : Factor w/ 2 levels "married","not married": 1 1 1 1 1 1 1 1 1 1 ...
##  $ gained        : int  38 20 38 34 27 22 76 15 NA 52 ...
##  $ weight        : num  7.63 7.88 6.63 8 6.38 5.38 8.44 4.69 8.81 6.94 ...
##  $ lowbirthweight: Factor w/ 2 levels "low","not low": 2 2 2 2 2 1 2 1 2 2 ...
##  $ gender        : Factor w/ 2 levels "female","male": 2 2 1 2 1 2 2 2 2 1 ...
##  $ habit         : Factor w/ 2 levels "nonsmoker","smoker": 1 1 1 1 1 1 1 1 1 1 ...
##  $ whitemom      : Factor w/ 2 levels "not white","white": 1 1 2 2 1 1 1 1 2 2 ...

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.

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.

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

Answer:

The boxplot suggests that smoking mothers tend to give birth to babies with slightly lower birth weights, on average, than do non-smoking mothers, based on the lower median birthweight. However, the relationship is not totally clear, as the non-smoking mothers have a wider distribution of birthweights, with many outliers on the low side.
```
boxplot(weight ~ habit, data = nc,
        main = "Birth weight vs. smoking status",
        xlab = "Mother's smoking status",
        ylab = "Child's birth weight")
```

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

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:
The conditions necessary for inference are satisfied:
- The observations in the total sample were randomly selected from the overall data set, and the total sample size of 1,000 observations is well under 10% of the population of all births in North Carolina that year. Furthermore, after separating the total sample into the two habit groups, the observations within each group are still independent. While the distributions of the two habit groups are skewed (toward low weights), the sample sizes of 873 and 126 for non-smokers and smokers, respectively, are high enough to mitigate this.
- The two samples are independent, for the same reason above that observations within each group are independent.
```
by(nc$weight, nc$habit, length)
```
```
## nc$habit: nonsmoker
## [1] 873
## -------------------------------------------------------- 
## nc$habit: smoker
## [1] 126
```
Write the hypotheses for testing if the average weights of babies born to smoking and non-smoking mothers are different.

Answer:

$H_0: \mu_{ns} - \mu_s = 0$: average birth weights for non-smoking and smoking mothers are the same.

$H_A: \mu_{ns} - \mu_s \neq 0$: average birth weights for non-smoking and smoking mothers are different.

Note this is a two-tailed test.

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.

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:

The 95% confidence interval for the difference in birth weight between non-smoking and smoking mothers is [0.05, 0.58] lbs (non-smoker - smoker).

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

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.

Answer:

The 95% confidence interval for the average length of pregnancies is [38.15, 38.52] weeks. This means that we are 95% confident that the true population mean length of pregnancies falls within this interval.

It appears that the inference function assumes a normal distribution instead of a t-distribution for inference testing.

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 )

# check numbers
m <- mean(nc$weeks, na.rm = TRUE)
s <- sd(nc$weeks, na.rm = TRUE)
n <- length(nc$weeks[!is.na(nc$weeks)])
se <- s / sqrt(n)
t <- qt(0.975, df = n - 1)
(ci <- m + c(-t, t) * se)

## [1] 38.15257 38.51677

# looks like `inference` assumes normal distribution instead of t-distribution
z <- qnorm(0.975)
(ci2 <- m + c(-z, z) * se)

## [1] 38.15279 38.51655

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.

Answer:

The 90% confidence interval is [38.18, 38.49] weeks.

inference(y = nc$weeks, est = "mean", type = "ci", null = 0, 
      alternative = "twosided", 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 )

# check numbers
z1 <- qnorm(0.95)
(ci3 <- m + c(-z1, z1) * se)

## [1] 38.18203 38.48731

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

Answer:

$H_0: \mu_{young} - \mu_{mature} = 0$: average weight gain by younger and mature mothers is the same.

$H_A: \mu_{young} - \mu_{mature} \neq 0$: average weight gain by younger and mature mothers is different.

The p-value is 16.9%, which is greater than the significance level of $\alpha = 0.05$, so we fail to reject $H_0$.

inference(y = nc$gained, x = nc$mature, est = "mean", type = "ht", null = 0, 
      alternative = "twosided", method = "theoretical", order = c("younger mom","mature mom"))

## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_younger mom = 844, mean_younger mom = 30.5604, sd_younger mom = 14.3469
## n_mature mom = 129, mean_mature mom = 28.7907, sd_mature mom = 13.4824

## Observed difference between means (younger mom-mature mom) = 1.7697
## 
## H0: mu_younger mom - mu_mature mom = 0 
## HA: mu_younger mom - mu_mature mom != 0 
## Standard error = 1.286 
## Test statistic: Z =  1.376 
## p-value =  0.1686

# check numbers
se <- sqrt(14.347^2 / 844 + 13.482^2 / 129)
z <- 1.77 / se
(p <- pnorm(-z) * 2)

## [1] 0.1685956

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.

Answer:

The age cutoff is 35 years: younger mom is up to 34 years old, while mature mom is 35 years and older.
See the boxplot below, and the age distribution of younger and mature moms.

# boxplot
boxplot(mage ~ mature, data = nc,
        main = "Mother's age vs. maturity status",
        xlab = "Mother's maturity status",
        ylab = "Mother's age")

# age distribution
summary(nc$mage[nc$mature == "younger mom"])

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   13.00   21.00   25.00   25.44   30.00   34.00

summary(nc$mage[nc$mature == "mature mom"])

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   35.00   35.00   37.00   37.18   38.00   50.00

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.

Answer:
- Is the average weight gained during pregnancy different for white and non-white mothers? The numerical variable is gained and the categorical variable is whitemom.
- Hypothesis test: two-tailed test
  $H_0: \mu_{white} - \mu_{non} = 0$: average weight gain by white and non-white mothers is the same.
  $H_A: \mu_{white} - \mu_{non} \neq 0$: average weight gain by white and non-white mothers is different.
- Result: p-value = 2.9%, which is less than the significance level of $\alpha = 0.05$, so we reject $H_0$ in favor of $H_A$ and conclude that the average weight gain is indeed different for white and non-white mothers. In other words, if the null hypothesis is true that the weight gain is the same for white and non-white mothers, then there would be only a 2.9% chance that the 2.3 lbs observed difference of sample means can be true. Since this likelihood is below our 5% threshold, we reject the null hypothesis and conclude that the alternative hypothesis must be true.
```
inference(y = nc$gained, x = nc$whitemom, est = "mean", type = "ht", null = 0, 
      alternative = "twosided", method = "theoretical", order = c("white","not white")) 
```
```
## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_white = 694, mean_white = 30.9798, sd_white = 13.8234
## n_not white = 277, mean_not white = 28.6751, sd_not white = 15.1845
```
```
## Observed difference between means (white-not white) = 2.3047
## 
## H0: mu_white - mu_not white = 0 
## HA: mu_white - mu_not white != 0 
## Standard error = 1.052 
## Test statistic: Z =  2.19 
## p-value =  0.0286
```
```
# check numbers
se <- sqrt(13.82^2 / 694 + 15.18^2 / 277)
z <- 2.305 / se
(p <- pnorm(-z) * 2)
```
```
## [1] 0.02847485
```

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.