In this lab we will explore the data using the dplyr
package and visualize it using the ggplot2
package for data visualization. The data can be found in the companion package for this course, statsr
.
Let’s load the packages.
library(statsr)
library(dplyr)
library(ggplot2)
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.
data(nc)
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 . |
As a first step in the analysis, we should take a look at the variables in the dataset. This can be done using the str
command:
str(nc)
## Classes 'tbl_df', 'tbl' and 'data.frame': 1000 obs. of 13 variables:
## $ fage : int NA NA 19 21 NA NA 18 17 NA 20 ...
## $ mage : int 13 14 15 15 15 15 15 15 16 16 ...
## $ mature : Factor w/ 2 levels "mature mom","younger mom": 2 2 2 2 2 2 2 2 2 2 ...
## $ weeks : int 39 42 37 41 39 38 37 35 38 37 ...
## $ premie : Factor w/ 2 levels "full term","premie": 1 1 1 1 1 1 1 2 1 1 ...
## $ visits : int 10 15 11 6 9 19 12 5 9 13 ...
## $ marital : Factor w/ 2 levels "married","not married": 1 1 1 1 1 1 1 1 1 1 ...
## $ gained : int 38 20 38 34 27 22 76 15 NA 52 ...
## $ weight : num 7.63 7.88 6.63 8 6.38 5.38 8.44 4.69 8.81 6.94 ...
## $ lowbirthweight: Factor w/ 2 levels "low","not low": 2 2 2 2 2 1 2 1 2 2 ...
## $ gender : Factor w/ 2 levels "female","male": 2 2 1 2 1 2 2 2 2 1 ...
## $ habit : Factor w/ 2 levels "nonsmoker","smoker": 1 1 1 1 1 1 1 1 1 1 ...
## $ whitemom : Factor w/ 2 levels "not white","white": 1 1 2 2 1 1 1 1 2 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.
We will first start with analyzing the weight gained by mothers throughout the pregnancy: gained
.
Using visualization and summary statistics, describe the distribution of weight gained by mothers during pregnancy. The summary
function can also be useful.
summary(nc$gained)
## Min. 1st Qu. Median Mean 3rd Qu. Max. NA's
## 0.00 20.00 30.00 30.33 38.00 85.00 27
Next, 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
. Which of the following is false about the relationship between habit and weight?
# type your code for the Question 3 here, and Knit
ggplot(data = nc, aes(x = habit, y = weight)) +
geom_boxplot()
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 to first group the data by the habit
variable, and then calculate the mean weight
in these groups using the mean
function.
nc %>%
group_by(habit) %>%
summarise(mean_weight = mean(weight))
## # A tibble: 3 x 2
## habit mean_weight
## <fct> <dbl>
## 1 nonsmoker 7.14
## 2 smoker 6.83
## 3 <NA> 3.63
There is an observed difference, but is this difference statistically significant? In order to answer this question we will conduct a hypothesis test.
Exercise: Are all conditions necessary for inference satisfied? Comment on each. You can compute the group sizes using the same by
command above but replacing mean(weight)
with n()
.
Next, we introduce a new function, inference
, that we will use for conducting hypothesis tests and constructing confidence intervals.
Then, run the following:
inference(y = weight, x = habit, data = nc, statistic = "mean", type = "ht", null = 0,
alternative = "twosided", method = "theoretical")
## Response variable: numerical
## Explanatory variable: categorical (2 levels)
## n_nonsmoker = 873, y_bar_nonsmoker = 7.1443, s_nonsmoker = 1.5187
## n_smoker = 126, y_bar_smoker = 6.8287, s_smoker = 1.3862
## H0: mu_nonsmoker = mu_smoker
## HA: mu_nonsmoker != mu_smoker
## t = 2.359, df = 125
## p_value = 0.0199
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: weight
. The second argument is the explanatory variable, x
, which is the variable that splits the data into two groups, smokers and non-smokers: habit
. The third argument, data
, is the data frame these variables are stored in. Next is statistic
, which is the sample statistic we’re using, or similarly, the population parameter we’re estimating. In future labs we can also work with “median” and “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.
For more information on the inference function see the help file with ?inference
.
Exercise: What is the conclusion of the hypothesis test?
type
argument to "ci"
to construct and record a confidence interval for the difference between the weights of babies born to nonsmoking and smoking mothers, and interpret this interval in context of the data. Note that by default you’ll get a 95% confidence interval. If you want to change the confidence level, add a new argument (conf_level
) which takes on a value between 0 and 1. Also note that when doing a confidence interval arguments like null
and alternative
are not useful, so make sure to remove them.
# type your code for the Question 5 here, and Knit
inference(y = weight, x = habit, data = nc, statistic = "mean", type = "ci", method = "theoretical")
## Response variable: numerical, Explanatory variable: categorical (2 levels)
## n_nonsmoker = 873, y_bar_nonsmoker = 7.1443, s_nonsmoker = 1.5187
## n_smoker = 126, y_bar_smoker = 6.8287, s_smoker = 1.3862
## 95% CI (nonsmoker - smoker): (0.0508 , 0.5803)
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 = weight, x = habit, data = nc, statistic = "mean", type = "ci",
method = "theoretical", order = c("smoker","nonsmoker"))
## Response variable: numerical, Explanatory variable: categorical (2 levels)
## n_smoker = 126, y_bar_smoker = 6.8287, s_smoker = 1.3862
## n_nonsmoker = 873, y_bar_nonsmoker = 7.1443, s_nonsmoker = 1.5187
## 95% CI (smoker - nonsmoker): (-0.5803 , -0.0508)
weeks
). 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. Which of the following is the correct interpretation of this interval?
# type your code for Question 6 here, and Knit
inference(y = weeks, data = nc, statistic = "mean", type = "ci",
method = "theoretical", conf_level = 0.99)
## Single numerical variable
## n = 998, y-bar = 38.3347, s = 2.9316
## 99% CI: (38.0952 , 38.5742)
Exercise: Calculate a new confidence interval for the same parameter at the 90% confidence level. Comment on the width of this interval versus the one obtained in the the previous exercise.
# type your code for the Exercise here, and Knit
inference(y = weeks, data = nc, statistic = "mean", type = "ci",
method = "theoretical", conf_level = 0.9)
## Single numerical variable
## n = 998, y-bar = 38.3347, s = 2.9316
## 90% CI: (38.1819 , 38.4874)
The width of the distribution at the 90% confidence level is narrower than that the distribution at the 99% confidence level.
Exercise: Conduct a hypothesis test evaluating whether the average weight gained by younger mothers is different than the average weight gained by mature mothers.
# type your code for the Exercise here, and Knit
inference(y = weight, x = mature, data = nc, statistic = "mean", type = "ht", null = 0, alternative = "twosided", method = "theoretical")
## Response variable: numerical
## Explanatory variable: categorical (2 levels)
## n_mature mom = 133, y_bar_mature mom = 7.1256, s_mature mom = 1.6591
## n_younger mom = 867, y_bar_younger mom = 7.0972, s_younger mom = 1.4855
## H0: mu_mature mom = mu_younger mom
## HA: mu_mature mom != mu_younger mom
## t = 0.1858, df = 132
## p_value = 0.8529
# type your code for Question 7 here, and Knit
nc %>%
group_by(mature) %>%
summarise(max(mage),min(mage))
## # A tibble: 2 x 3
## mature `max(mage)` `min(mage)`
## <fct> <dbl> <dbl>
## 1 mature mom 50.0 35.0
## 2 younger mom 34.0 13.0
Exercise: Pick a pair of variables: one numerical (response) and one categorical (explanatory). 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. Be sure to check all assumptions,state your \(\alpha\) level, and conclude in context. (Note: Picking your own variables, coming up with a research question, and analyzing the data to answer this question is basically what you’ll need to do for your project as well.)
# type your code for the Exercise here, and Knit
sum(is.na(nc$weeks))
## [1] 2
nc_new <- nc %>%filter(!is.na(weeks))
nc_new %>%
group_by(mature) %>%
summarise(mean = mean(weeks))
## # A tibble: 2 x 2
## mature mean
## <fct> <dbl>
## 1 mature mom 38.0
## 2 younger mom 38.4
inference(y = weeks, x = mature, data = nc_new, statistic = "mean", type = "ht", null = 0, alternative = "twosided", method = "theoretical", conf_level = 0.95)
## Response variable: numerical
## Explanatory variable: categorical (2 levels)
## n_mature mom = 132, y_bar_mature mom = 38.0227, s_mature mom = 3.2184
## n_younger mom = 866, y_bar_younger mom = 38.3822, s_younger mom = 2.8844
## H0: mu_mature mom = mu_younger mom
## HA: mu_mature mom != mu_younger mom
## t = -1.2113, df = 131
## p_value = 0.228
Variables picked: mature
| maturity status of mother. weeks
| length of pregnancy in weeks.
Question:At a 95% confidence level,is the difference in the average length of pregnancy in weeks born to younger and mature mothers statistical significant?
Answer& Explanation: The p_value is greater than 5%, thus the observation fail to reject the null hypothesis. There is no obvious diffrence in lenght of pregnancy between younger and mature mothers.
This is a product of OpenIntro that is released under a Creative Commons Attribution-ShareAlike 3.0 Unported. This lab was written for OpenIntro by Andrew Bray and Mine Çetinkaya-Rundel.