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: The cases in this data set are birth details for baby born in North Carolina. There are 100 cases in this sample.

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.

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: Median birth weight of babies born to non-smoker mothers is slightly higher than that of babies born to smoker mothers. The range of birth weights of babies born to non-smoker mothers is greater than that of babies born to smoker mothers. But the IQRs of the distributions are roughly equal.

# Boxplot for smoker and nonsmoker mothers for their babies' weight.
boxplot(weight~habit,data=nc, main="Relationship Between Mother's Smoking Habit & Baby's Weight", ylab="Weight", xlab="Mother Habit")

# Compare the two groups by summary.
by(nc$weight, nc$habit, summary)

## nc$habit: nonsmoker
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   1.000   6.440   7.310   7.144   8.060  11.750 
## -------------------------------------------------------- 
## nc$habit: smoker
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   1.690   6.077   7.060   6.829   7.735   9.190

# Compre the two groups by IQR.
by(nc$weight, nc$habit, IQR)

## nc$habit: nonsmoker
## [1] 1.62
## -------------------------------------------------------- 
## nc$habit: smoker
## [1] 1.6575

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 sample observations within the groups are randomly selected and independent as well as across the sample groups. The samples sizes are less than 10% of the population size. Since the sample size is large enough, the distribution will be approximately normal.

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: Null Hypothesis, Ho: There is no difference in the mean of the birth weight between smoking and non-smoking mothers. Alternative Hypothesis, Ha: There is a difference in the mean of the birth weight between smoking and non-smoking mothers.

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 between the weights of babies born to smoking and non-smoking mothers is (0.0534 , 0.5777).

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 (38.1528 , 38.5165). This interval will capture the true population mean 95% of the time.

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.

Answer: The 90% confidence interval for the average length of pregnancies is (38.182 , 38.4873). This interval will capture the true population mean 90% of the time. Also, as you can see the 90% confidence interval is narrower than the 95% confidence interval because its confidence level is lower.

inference(y = nc$weeks, est = "mean", type = "ci", null = 0, alternative = "twosided", conflevel = 0.90, method = "theoretical")

## 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.

Answer: Null Hypothesis, Ho: The average weight gained by

younger mothers have NO different than the average weight gained by mature mothers. Alternative Hypothesis, Ha: The average weight gained by younger mothers is different than the average weight gained by mature mothers. As the p-value 0.1686 is greater than the significant level 0.05, the Null Hypothesis cannot be rejected. Therefore, we can say that the average weight gained by younger mothers have NO different than the average weight gained by mature mothers.

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.

Answer: From below summary and boxplot, we can see that the maximum age of younger moms is 34 and minimum age of mature moms is 35. So, the age cutoff for younger moms is 34 and the age cutoff for mature moms is 35. I use the by function to split the mother ager `mage` variable into the maturiry status of mother `mature` groups, then using the `summary` function to summarize the statistic for each. I also do a boxplot to visualize the age cutoff between mature moms and yourger moms.

by(nc$mage, nc$mature, summary)

## nc$mature: mature mom
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   35.00   35.00   37.00   37.18   38.00   50.00 
## -------------------------------------------------------- 
## nc$mature: younger mom
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   13.00   21.00   25.00   25.44   30.00   34.00

boxplot(nc$mage ~ nc$mature, ylab = "Years")

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: As the p-value is very small, equal to 0, we can reject the Null Hypothesis. Therefore, we can conclude that there is statistical significant evidence that the average number of hospital visits is greater for full term births than the premature births.

- The pair of numerical and categorical variables that I pick is visits (numerical) and premie (categorical). The research question I interest to know is whether there is a significant evidence that the average number of hospital visits is greater for full term births than the premature births.

- Null Hypothesis, Ho: The average number of hospital visits is equal between full term births and the premature births. Alternative Hypothesis, Ha: The average number of hospital visits is greater for full term births than the premature births.

inference(y = nc$visits, x = nc$premie, est = "mean", type = "ht", null = 0, alternative = "greater", method = "theoretical")

## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_full term = 840, mean_full term = 12.3524, sd_full term = 3.7515
## n_premie = 150, mean_premie = 10.74, sd_premie = 4.7323

## Observed difference between means (full term-premie) = 1.6124
## 
## H0: mu_full term - mu_premie = 0 
## HA: mu_full term - mu_premie > 0 
## Standard error = 0.407 
## Test statistic: Z =  3.957 
## p-value =  0