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.
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?
Each row in the dataset represents a case. There are 1000 different cases.
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.
boxplot(nc$fage, nc$mage, nc$weeks, nc$visits, nc$gained, nc$weight, names =c("age(dad)", "age(mom)", "weeks", "visits", "gained", "wgt(baby)"))
There are outliers for all numerical variable but not too extreme. The outliers for weight gained among mothers are more defined than the others.
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?ggplot(nc, aes(habit, weight)) + geom_boxplot()
On average, both groups give birth to babes at an averge of 7 lbs. It is not very common to have babies weighing more tha 9lbs with smoking mothers. However, a lot more babies are born pre-mature with non-smokers. There may be other factors that contribute to this finding due to health, physical ability, nutrition etc.
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
Independence - Both samples are random and less than 10% of the population size
The sample sizes are greater than 30 hence we can be lenient on any skewness in the distribution.
\(H_0: \mu_n = \mu_s\)   There is no difference in average birth for newborns from smoking and non-smoking mothers.
\(H_A: \mu_n \neq \mu_s\)   There is a difference in the average birth of newborns from 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")
## Warning: package 'openintro' was built under R version 3.5.2
## 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$weeks, est = "mean", type = "ci", null = 0,
alternative = "twosided", method = "theoretical")
## Single mean
## Summary statistics:
## mean = 38.3347 ; sd = 2.9316 ; n = 998
## Standard error = 0.0928
## 95 % Confidence interval = ( 38.1528 , 38.5165 )
We are 95% confident that the average length of pregnancies is from 38.1528 to 38.5165 among expentant monthers.
conflevel = 0.90
.inference(y = nc$weeks, est = "mean", type = "ci", null = 0, conflevel = 0.90,
alternative = "twosided", 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 )
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
We failed to reject the null hypthesis since the p-value is greater than 0.05. Therefore there is no difference in the weight gained between mature and young mothers.
Option 1:
mature_mom <- filter(nc, mature == "mature mom")
## Warning: package 'bindrcpp' was built under R version 3.5.2
young_mom <- filter(nc, mature =="younger mom")
#find the mid point between the average of ages for both groups
summary(mature_mom$mage)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 35.00 35.00 37.00 37.18 38.00 50.00
summary(young_mom$mage)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 13.00 21.00 25.00 25.44 30.00 34.00
The min age for mature is 35 and the max age for young is 34 which means the cut-off age is 35.
- 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.
Does the amount of weight gained stay the same or varies among races? Significance level: \(\alpha = 0.05\)
\(H_0:\) Weight is the same \(H_A:\) Weight varies among races
inference(y = nc$gained, x = nc$whitemom, est = "mean", type = "ht", null = 0,
alternative = "twosided", method = "theoretical")
## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_not white = 277, mean_not white = 28.6751, sd_not white = 15.1845
## n_white = 694, mean_white = 30.9798, sd_white = 13.8234
## Observed difference between means (not white-white) = -2.3047
##
## H0: mu_not white - mu_white = 0
## HA: mu_not white - mu_white != 0
## Standard error = 1.052
## Test statistic: Z = -2.19
## p-value = 0.0286
The p-value is less than 0.05 there we reject the null hypothesis that the weight stays the same among races.
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.