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.
download.file("http://www.openintro.org/stat/data/nc.RData", destfile = "nc.RData")
load("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?
Answer:
The cases are each new born babies. There are 100 cases in our sample.
## [1] 1000
As a first step in the analysis, we should consider summaries of the data. This can be done using the summary
command:
## 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
##
##
##
##
boxplot(nc$fage, nc$mage, nc$weeks, nc$visits, nc$gained, nc$weight,
col = 'cyan1',
names =c('fage', 'mage', 'weeks', 'visits', 'gained', 'weight'),
horizontal = TRUE,
las=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.
2. Make a side-by-side boxplot of habit
and weight
. What does the plot highlight about the relationship between these two variables?
Answer:
Babies whose mother are non-smokers have higher weight than those whose mother are smokers.
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.
## 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
.## nc$habit: nonsmoker
## [1] 873
## --------------------------------------------------------
## nc$habit: smoker
## [1] 126
4. Write the hypotheses for testing if the average weights of babies born to smoking and non-smoking mothers are different.
Answer:
The hypotheses are as below:
H0: \(\mu_{(baby-weight | Mother = Non-Smoker)} = \mu_{(baby-weight | Mother = Smoker)}\)
H1: \(\mu_{(baby-weight | Mother = Non-Smoker)} \neq \mu_{(baby-weight | Mother = Smoker)}\)
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.
5. 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 )
- 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 is (38.1528, 38.5165).
inference(y=nc$weeks,
est = 'mean',
type = 'ci',
null = 0,
alternative = 'twosided',
method = 'theoretical')
## Single mean
## Summary statistics:
## mean = 38.3347 ; sd = 2.9316 ; n = 998
## Standard error = 0.0928
## 95 % Confidence interval = ( 38.1528 , 38.5165 )
- Calculate a new confidence interval for the same parameter at the 90% confidence level. You can change the confidence level by adding a new argument to the function: conflevel = 0.90
.
Answer:
The 95% confidence interval is (38.182 , 38.4873).
inference(y=nc$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 )
- Conduct a hypothesis test evaluating whether the average weight gained by younger mothers is different than the average weight gained by mature mothers.
Answer:
H0: \(\mu_{(weight-gained | younger-mothers)} = \mu_{(weight-gained | mature-mothers)}\)
H1: \(\mu_{(weight-gained | younger-mothers)} \neq \mu_{(weight-gained | mature-mothers)}\)
As p-value = 0.1686 > 0.05, using 95% confidence interval, we do not reject the null hypothesis. The Average weight gained by younger mothers is similar to the average weight gained by mature mothers.
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
- 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:
As shown in the histogram, the data cut off point for younger and mature mothers is at 35.
data <-nc$mage[nc$mature=='younger mom']
hist(nc$mage[nc$mature=='younger mom'],
breaks = 30,
xlim = c(0,60),
xlab = 'Age',
col = 'cyan',
main = 'Age Cutoff: Younger and Mature Mothers')
hist(nc$mage[nc$mature=='mature mom'],
breaks = 30,
add= TRUE,
col='deeppink')
abline(v = 35, col="black", lwd=3, lty=2)
legend("topright", c("Younger Mothers", "Mature Mothers"), fill=c("cyan", "deeppink"))
- 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 testand/or a confidence interval. Answer your question using the inference
function, report the statistical results, and also provide an explanation in plain language.
Answer:
Research question: whether the average weight of full term babies is different than the average weight of premie babies.
Hypothesis:
H0: the average weight of full term babies and that of premie babies are the same.
H1: the average weight of full term babies and that of premie babies are not the same.
Result: the p-value is 0 <- 0.05. Using 95% confidence interval, we reject the null hypothesis that the average weight of full term babies and that of premie babies are the same.
inference(y = nc$weight,
x= nc$premie,
est = 'mean',
type = 'ht',
null = 0,
alternative = 'twosided',
method = 'theoretical')
## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_full term = 846, mean_full term = 7.4594, sd_full term = 1.075
## n_premie = 152, mean_premie = 5.1284, sd_premie = 1.9696
## Observed difference between means (full term-premie) = 2.331
##
## H0: mu_full term - mu_premie = 0
## HA: mu_full term - mu_premie != 0
## Standard error = 0.164
## Test statistic: Z = 14.216
## p-value = 0