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")
libs <- c("tidyverse", "magrittr")
lapply(libs, require, character.only=TRUE)
## Loading required package: tidyverse
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## v readr 1.1.1 v forcats 0.2.0
## -- Conflicts -------------------------------------------------------------------------------------------------------------------------------------------------------------- tidyverse_conflicts() --
## x dplyr::filter() masks stats::filter()
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## Loading required package: magrittr
##
## Attaching package: 'magrittr'
## The following object is masked from 'package:purrr':
##
## set_names
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## extract
## [[1]]
## [1] TRUE
##
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## [1] TRUE
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 . |
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.
habit
and weight
. What does the plot highlight about the relationship between these two variables?ggplot(nc, aes(x = habit, y = weight)) +
geom_boxplot()
* The relationship seems to indicate that there are more heavy non-smokers than there are heavy smokers, however the habit does not seem to have a causational relationship on weight.
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
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 3.4.4
## 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.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 )
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$weeks, est = "mean", type = "ci", null = 0,
alternative = "twosided", method = "theoretical")
## Single mean
## Summary statistics:
## mean = 38.3347 ; sd = 2.9316 ; n = 998
## Standard error = 0.0928
## 95 % Confidence interval = ( 38.1528 , 38.5165 )
conflevel = 0.90
.inference(y = nc$weeks, est = "mean", type = "ci", null = 0,
alternative = "twosided", method = "theoretical", conflevel = .9)
## 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 = "less", 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 = 0.0843
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
ggplot(nc, aes(x = mature, y = mage, fill = mature)) +
geom_boxplot()
* Younger Moms: 13-34 * Mature Moms: 35-50 * The method uses the summary function to get the range of values for the mother’s age categorized by mature or young.
inference
function, report the statistical results, and also provide an explanation in plain language.inference(y = nc$marital, x = nc$mature, est = "proportion", type = "ht", null = 0,
alternative = "greater", method = "theoretical", success = "married")
## Response variable: categorical, Explanatory variable: categorical
## Two categorical variables
## Difference between two proportions -- success: married
## Summary statistics:
## x
## y mature mom younger mom Sum
## married 25 361 386
## not married 107 506 613
## Sum 132 867 999
## Observed difference between proportions (mature mom-younger mom) = -0.227
##
## H0: p_mature mom - p_younger mom = 0
## HA: p_mature mom - p_younger mom > 0
## Pooled proportion = 0.3864
## Check conditions:
## mature mom : number of expected successes = 51 ; number of expected failures = 81
## younger mom : number of expected successes = 335 ; number of expected failures = 532
## Standard error = 0.045
## Test statistic: Z = -4.989
## p-value = 1
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.