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.
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?
The cases in this data sets are the births.
How many cases are there in our sample?
There are 1,000 cases in this data set.
dim(nc)
## [1] 1000 13
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.
Categrical: mature, premie, marital, lowbirthweight, gender, habit and whitemom.
Numerical: fage, mage, weeks, visits, gained and weight.
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
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.
habit
and weight
. What does the plot highlight about the relationship between these two variables?# Boxplot of habit and weight
boxplot(weight~habit,data=nc, main="Relation Between Mother's Habit and Baby's Weight",
ylab="Baby's Weight", xlab="Mother Smoker/Non-Smoker")
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.
by
command above but replacing mean
with length
.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.
H0: µ{nonsmoker} - µ{smoker} = 0, There is no difference in the mean of the
weight of babies born to smoking and to nonsmoking mothers.
HA: µ{nonsmoker} - µ{smoker} != 0, There is a difference in the mean of the birth weight
of babies born to smoking and to nonsmoking 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")
## Warning: package 'BHH2' was built under R version 3.5.3
## 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.
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.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 )
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$weeks, est = "mean", type = "ci", null = 0,
alternative = "twosided", method = "theoretical")
## Single mean
## Summary statistics:
## mean = 38.3347 ; sd = 2.9316 ; n = 998
## Standard error = 0.0928
## 95 % Confidence interval = ( 38.1528 , 38.5165 )
Based on our computation We are 95% confident that the population mean falls between ( 38.1528 , 38.5165 ) weeks.
conflevel = 0.90
.inference(y = nc$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 )
Based on our computation We are 90% confident that the population mean falls between ( 38.182 , 38.4873 ) weeks.
inference(y = nc$weight, x = nc$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 = 133, mean_mature mom = 7.1256, sd_mature mom = 1.6591
## n_younger mom = 867, mean_younger mom = 7.0972, sd_younger mom = 1.4855
## Observed difference between means (mature mom-younger mom) = 0.0283
##
## H0: mu_mature mom - mu_younger mom = 0
## HA: mu_mature mom - mu_younger mom != 0
## Standard error = 0.152
## Test statistic: Z = 0.186
## p-value = 0.8526
Based on statistical data, we cannot reject the null hypothesis since the p-value of 0.8526 is greater than alpha of 0.05. Thus the data does not provide convincing evidence that there is a difference between the average weight gained by young mothers and the average weight gained by mature mothers.
by(nc$mage, nc$mature, range)
## nc$mature: mature mom
## [1] 35 50
## --------------------------------------------------------
## nc$mature: younger mom
## [1] 13 34
We can use the range to find teh cutoff age between the young mothers and mature mothers. From this function we see that the cutoff is 34 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.
Research question: Does marital status affect weight gained by mothers during pregnancy?
H0: µ{married} - µ{not married} = 0, There is no difference in the mean of the weight gained during pregnancy of between married and single mothers.
HA: µ{married} - µ{not married} != 0, There is a difference in the mean of the weight gained during pregnancy between married and single mothers.
inference(y = nc$gained, x = nc$marital, est = "mean", type = "ht", null = 0,
alternative = "twosided", method = "theoretical")
## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_married = 370, mean_married = 29.873, sd_married = 15.2721
## n_not married = 603, mean_not married = 30.6036, sd_not married = 13.5757
## Observed difference between means (married-not married) = -0.7307
##
## H0: mu_married - mu_not married = 0
## HA: mu_married - mu_not married != 0
## Standard error = 0.967
## Test statistic: Z = -0.755
## p-value = 0.4502
inference(y = nc$gained, x = nc$marital, est = "mean", type = "ci", null = 0,
alternative = "twosided", method = "theoretical")
## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_married = 370, mean_married = 29.873, sd_married = 15.2721
## n_not married = 603, mean_not married = 30.6036, sd_not married = 13.5757
## Observed difference between means (married-not married) = -0.7307
##
## Standard error = 0.9675
## 95 % Confidence interval = ( -2.6269 , 1.1655 )
Based on the hypothesis test the p-value of 0.4502 is greater than alpha of 0.05 and thus we fail to reject the null hypothesis since there is no sufficient evidence that marital status affects the weights of mothers.