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?

Statistically, cases are called as observations. There are around 1000 observations in this dataset.

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?

plot_ly(x=~nc$habit,y=~nc$weight,type = "box")

## Warning: Ignoring 1 observations

The above box plot higlights that the median weight is almost equal. Both nonsmoker and smoker has many outliers.

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, sum)

## nc$habit: nonsmoker
## [1] 6236.95
## -------------------------------------------------------- 
## nc$habit: smoker
## [1] 860.42

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.

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

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

E2 1. Independent: In this case study, one birth does not depend on another case. Also it is assumed that the simple random sample is less than 10% of the population. So the independence is assumed.

E2 2. Although, there are little bit of skew, as the sample size is more than 30, we assume it is a normal distribution.

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

H0 : Average weights of babies born to smoking mothers are same compared to the non-smoking mothers.

HA: Average weights of babies born to smoking and non-smoking mothers are different.

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.

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 is 0.0534 to 0.5777. It means the on average, the weight difference of of non-smoker vs smoker mother is between 0.0534 to 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.

inference(y = nc$weeks,est = "mean",conflevel = 95,type = "ci",alternative = "twosided",method = "theoretical")

## Warning: Confidence level converted to 0.95.

## 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",conflevel = 0.90,alternative = "twosided",type = "ci",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.

$H_{0}$: Weight gained by younger mom and mature mom is same or equal

$H_A$: Weight gained by younger mom and mature mom is different.

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.

#Calculating confidence interval

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

## nc$mature: mature mom
## [1] 133
## -------------------------------------------------------- 
## nc$mature: younger mom
## [1] 867

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

## nc$mature: mature mom
## [1] 37.18045
## -------------------------------------------------------- 
## nc$mature: younger mom
## [1] 25.43829

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

## nc$mature: mature mom
## [1] 2.430347
## -------------------------------------------------------- 
## nc$mature: younger mom
## [1] 5.027804

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

#95% confidence interval for mature moms
37.18 + (2.34*(2.43/sqrt(133)))

## [1] 37.67306

37.18 - (2.34*(2.43/sqrt(133)))

## [1] 36.68694

#95% confidence interval for younger moms
25.43 + (1.96*(5.02/sqrt(867)))

## [1] 25.76416

25.43 - (1.96*(5.02/sqrt(867)))

## [1] 25.09584

Above code fetches the mean, sd and length of mature and younger mom’s age and it calculates the z-score for 99% confidence interval.

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.

H0: Visits by married mom and non-married mom are equal HA: Visits by married mom and non-married mom are different

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

## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_married = 380, mean_married = 10.9553, sd_married = 4.2408
## n_not married = 611, mean_not married = 12.82, sd_not married = 3.5883

## Observed difference between means (married-not married) = -1.8647
## 
## H0: mu_married - mu_not married = 0 
## HA: mu_married - mu_not married != 0 
## Standard error = 0.262 
## Test statistic: Z =  -7.13 
## p-value =  0

inference(y = nc$visits,x=nc$marital,est = "mean",null = 0,alternative = "twosided",type = "ci",conflevel = 0.95,method = "theoretical",order = c("not married","married"))

## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_not married = 611, mean_not married = 12.82, sd_not married = 3.5883
## n_married = 380, mean_married = 10.9553, sd_married = 4.2408

## Observed difference between means (not married-married) = 1.8647
## 
## Standard error = 0.2615 
## 95 % Confidence interval = ( 1.3521 , 2.3773 )

This clearly mentions that the not married mom visits are more than married mom. p value is almost 0. Hence we reject null hypothesis.