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Getting Started

Load packages

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)

The data

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.
  1. There are 1,000 cases in this data set, what do the cases represent?
    1. The hospitals where the births took place
    2. The fathers of the children
    3. The days of the births
    4. The births

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)
## tibble [1,000 × 13] (S3: tbl_df/tbl/data.frame)
##  $ fage          : int [1:1000] NA NA 19 21 NA NA 18 17 NA 20 ...
##  $ mage          : int [1:1000] 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 [1:1000] 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 [1:1000] 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 [1:1000] 38 20 38 34 27 22 76 15 NA 52 ...
##  $ weight        : num [1:1000] 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.

Exploratory data analysis

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
  1. How many mothers are we missing weight gain data from?
    1. 0
    2. 13
    3. 27
    4. 31

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.

  1. Make side-by-side boxplots of habit and weight. Which of the following is false about the relationship between habit and weight?
    1. Median birth weight of babies born to non-smoker mothers is slightly higher than that of babies born to smoker mothers.
    2. Range of birth weights of babies born to non-smoker mothers is greater than that of babies born to smoker mothers.
    3. Both distributions are extremely right skewed.
    4. The IQRs of the distributions are roughly equal. 
# type your code for the Question 3 here, and Knit
ggplot(na.omit(nc), aes(x=habit, y=weight)) +
  geom_boxplot() +
  ggtitle("Distribution of baby weights accross smoking habits") + 
  ylab("Smoking habit") + 
  xlab("Baby weight")

# We can compare the two groups using the by() function together with the summary() function.
by(nc$weight, nc$habit, summary)
## nc$habit: nonsmoker
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   1.000   6.440   7.310   7.144   8.060  11.750 
## ------------------------------------------------------------ 
## nc$habit: smoker
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   1.690   6.077   7.060   6.829   7.735   9.190
# Similarly, we can use the IQR()
by(nc$weight, nc$habit, IQR)
## nc$habit: nonsmoker
## [1] 1.62
## ------------------------------------------------------------ 
## nc$habit: smoker
## [1] 1.6575

The IQRs for the two distributions seem very similar. The range of birth weights of babies born to non-smoker mothers is greater than that of babies born to smoker mothers.

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.

na.omit(nc) %>%
  group_by(habit) %>%
  summarise(mean_weight = mean(weight))
## `summarise()` ungrouping output (override with `.groups` argument)
## # A tibble: 2 x 2
##   habit     mean_weight
##   <fct>           <dbl>
## 1 nonsmoker        7.25
## 2 smoker           6.89

There is an observed difference, but is this difference statistically significant? In order to answer this question we will conduct a hypothesis test.

Inference

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().

by(nc$weight, nc$habit, mean)
## nc$habit: nonsmoker
## [1] 7.144273
## ------------------------------------------------------------ 
## nc$habit: smoker
## [1] 6.82873
na.omit(nc) %>%
  group_by(habit) %>%
  summarise(n = n())
## `summarise()` ungrouping output (override with `.groups` argument)
## # A tibble: 2 x 2
##   habit         n
##   <fct>     <int>
## 1 nonsmoker   716
## 2 smoker       84

The two means compared are not dependent. This is because for each observation in the smoker group, there isn’t a single specially-corresponding observation in the non-smoker group; and vice-versa. The conditions for inference for comparing two independent means:

  1. What are the hypotheses for testing if the average weights of babies born to smoking and non-smoking mothers are different?
    1. \(H_0: \mu_{smoking} = \mu_{non-smoking}\); \(H_A: \mu_{smoking} > \mu_{non-smoking}\)
    2. Ans: \(H_0: \mu_{smoking} = \mu_{non-smoking}\); \(H_A: \mu_{smoking} \ne \mu_{non-smoking}\)
    3. \(H_0: \bar{x}_{smoking} = \bar{x}_{non-smoking}\); \(H_A: \bar{x}_{smoking} > \bar{x}_{non-smoking}\)
    4. \(H_0: \bar{x}_{smoking} = \bar{x}_{non-smoking}\); \(H_A: \bar{x}_{smoking} > \bar{x}_{non-smoking}\)
    5. \(H_0: \mu_{smoking} \ne \mu_{non-smoking}\); \(H_A: \mu_{smoking} = \mu_{non-smoking}\)

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.

?inference

Exercise: What is the conclusion of the hypothesis test?

  1. Change the 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.
    1. We are 95% confident that babies born to nonsmoker mothers are on average 0.05 to 0.58 pounds lighter at birth than babies born to smoker mothers.
    2. We are 95% confident that the difference in average weights of babies whose moms are smokers and nonsmokers is between 0.05 to 0.58 pounds.
    3. We are 95% confident that the difference in average weights of babies in this sample whose moms are smokers and nonsmokers is between 0.05 to 0.58 pounds.
    4. We are 95% confident that babies born to nonsmoker mothers are on average 0.05 to 0.58 pounds heavier at birth than babies born to smoker mothers. 
# type your code for the Question 5 here, and Knit
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)

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)

  1. Calculate a 99% confidence interval for the average length of pregnancies (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?
    1. (38.1526 , 38.5168)
    2. (38.0892 , 38.5661)
    3. (6.9779 , 7.2241)
    4. (38.0952 , 38.5742) 
# type your code for Question 6 here, and Knit
inference(y = weeks, data = nc, statistic = "mean", type = "ci", conf_level = 0.99,
          method = "theoretical")
## 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", conf_level = 0.90,
          method = "theoretical")
## Single numerical variable
## n = 998, y-bar = 38.3347, s = 2.9316
## 90% CI: (38.1819 , 38.4874)

As the confidence level increases, the width of the confidence interval also increases. We expect this to happen with the same parameter.

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 = gained, x = mature, data = nc, statistic = "mean", type = "ht", null=0,
          method = "theoretical", alternative = "twosided")
## Response variable: numerical
## Explanatory variable: categorical (2 levels) 
## n_mature mom = 129, y_bar_mature mom = 28.7907, s_mature mom = 13.4824
## n_younger mom = 844, y_bar_younger mom = 30.5604, s_younger mom = 14.3469
## H0: mu_mature mom =  mu_younger mom
## HA: mu_mature mom != mu_younger mom
## t = -1.3765, df = 128
## p_value = 0.1711

  1. 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.
# type your code for Question 7 here, and Knit
na.omit(nc) %>%
  group_by(mature) %>%
  summarise(min = min(mage), max = max(mage))
## `summarise()` ungrouping output (override with `.groups` argument)
## # A tibble: 2 x 3
##   mature        min   max
##   <fct>       <int> <int>
## 1 mature mom     35    50
## 2 younger mom    15    34

Using the maxmimum and minimum age for each group, the mature mom age ranges from 35-50 years while the younger mom age ranges from 15-34.

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.)

The two variables we will be exploring are:

variable description
premie whether the birth was classified as premature (premie) or full-term.
weight weight of the baby at birth in pounds.

We will try and explore whether a premature babies tend to have a lower weight on average than full-term babies and whether this difference is statistically significant. Our exploratory variable in this case would be premie while our response variable is weight

Conditions for inference for comparing two independent means.

The two means compared are not dependent. This is because for each observation in the premature group, there isn’t a single specially-corresponding observation in the full-term group; and vice-versa. The conditions for inference for comparing two independent means:

# type your code for the Exercise here, and Knit
na.omit(nc) %>%
  group_by(premie) %>%
  summarise(n = n())
## `summarise()` ungrouping output (override with `.groups` argument)
## # A tibble: 2 x 2
##   premie        n
##   <fct>     <int>
## 1 full term   694
## 2 premie      106
  • Independence:
    • Within groups: since we’re sampling without replacement, each group needs to be less than 10% of the given population. This seems to be the case with both the premature and full-term groups.
    • Between groups: as mentioned above, there is no reason to believe that the premature and full-term groups are dependent in any way.
mdat <- na.omit(nc) %>%
  group_by(premie) %>%
  summarise(mean = mean(weight))
## `summarise()` ungrouping output (override with `.groups` argument)
mdat
## # A tibble: 2 x 2
##   premie     mean
##   <fct>     <dbl>
## 1 full term  7.51
## 2 premie     5.24
ggplot(na.omit(nc), aes(x=weight, fill=premie)) +
  geom_density(alpha=.3) + 
  geom_vline(data=mdat, aes(xintercept=mean,  colour=premie),
               linetype="dashed", size=1)

  • Sample size/skew
    • The full term group is heavily left skewed but the large sample size of n = 694, mitigates this. The premie group appears to be nearly normal and symmetric.

Inference

\(H_0: \mu_{full} = \mu_{premie}\) The observed difference between the means of full and premie is due to chance and due to sampling variability. \(H_A: \mu_{full} \ne \mu_{premie}\) The observed difference is statistically significant and cannot be attributed to sampling variability.

We’ll be using a signficance level of 0.05 and a null value of 0. The null value represents no difference in the means of the two groups.

inference(y = weight, x = premie, data = nc, statistic = "mean", type = "ht", null=0,
          method = "theoretical", alternative = "twosided")
## Response variable: numerical
## Explanatory variable: categorical (2 levels) 
## n_full term = 846, y_bar_full term = 7.4594, s_full term = 1.075
## n_premie = 152, y_bar_premie = 5.1284, s_premie = 1.9696
## H0: mu_full term =  mu_premie
## HA: mu_full term != mu_premie
## t = 14.2156, df = 151
## p_value = < 0.0001

t.test(weight ~ premie, data = nc, conf.level=0.95)
## 
##  Welch Two Sample t-test
## 
## data:  weight by premie
## t = 14.216, df = 167.51, p-value < 2.2e-16
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  2.007267 2.654709
## sample estimates:
## mean in group full term    mean in group premie 
##                7.459409                5.128421

Since our p-value is well below 0.05, we can reject the null hypothesis and the observed difference between the two means is not due to chance. The difference is statistically significant and a premature birth can be attributed to a lower birth weight.

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.