RLab 7 - Inference for Numerical Data
R. Lee
October 28, 2019
North Carolina births
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.
Exploratory analysis 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.
Exercise 1
What are the cases in this data set? How many cases are there in our sample?
The cases are births in North Carolina in 2004. There are 1000 cases.
As a first step in the analysis, we should consider summaries of the data. This can be done using the summary command:
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.
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
##
##
##
## str(nc)## '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 ...Exercise 2
Make a side-by-side boxplot of habit and weight. What does the plot highlight about the relationship between these two variables?
The plots have a similar shape. Both are left-skewed. Their medians are about the same. The smoker distribution has a smaller spread.
boxplot (nc$weight ~ nc$habit)
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] 6.82873There is an observed difference, but is this difference statistically significant? In order to answer this question we will conduct a hypothesis test .
Exercise 3
Check if the conditions necessary for inference are satisfied. Note that you will need to obtain sample sizes to check the conditions. You can compute the group size using the same by command above but replacing mean with length.
by(nc$weight, nc$habit, length)## nc$habit: nonsmoker
## [1] 873
## --------------------------------------------------------
## nc$habit: smoker
## [1] 126CONDITIONS FOR INFERENCE: 1. random samples 2. observations within sample are independent 3. the 2 samples are independent of each other 4. sampling distributions are approximately normal
Condition 4 does not hold (distributions are right skewed). But both sample sizes are sufficiently large and there is only one outlier in the nonsmoker distribution. So sampling distributions meet the condition.
Exercise 4
Write the hypotheses for testing if the average weights of babies born to smoking and non-smoking mothers are different.
Null Hypothesis: There is no difference in the average birth weight of babies born to smoking mothers and babies born to non-smoking mothers.
Alternative Hypothesis: There is a difference in the average birth weight of babies born to smoking mothers and babies born to 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")## 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.
Exercise 6
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.
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 (μnonsmoker−μsmoker) . 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 )#ON YOUR OWN
1. 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.
95 % Confidence interval = ( 38.1528 , 38.5165 )
We are 95% confident that the true population mean length of pregnancy for babies born in North Carolina is between 38.15 and 38.52 weeks.
inference(y = nc$weeks, est = "mean", type = "ci", null = 0,
alternative = "twosided", method = "theoretical")## Single mean
## Summary statistics:
## mean = 38.3347 ; sd = 2.9316 ; n = 998
## Standard error = 0.0928
## 95 % Confidence interval = ( 38.1528 , 38.5165 )2. Calculate a new confidence interval for the same parameter at the 90% confidence level. You can change the confidence level by adding a new argument to the function: conflevel = 0.90.
90 % Confidence interval = ( 38.182 , 38.4873 )
inference(y = nc$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 )3. Conduct a hypothesis test evaluating whether the average weight gained by younger mothers is different than the average weight gained by mature mothers.
Null Hypothesis: There is no difference in the average weight gained by younger mothers and the average weight gained by nature mothers.
Alternative Hypothesis: There is a difference in the average weight gained by younger mothers and the average weight gained by nature mothers.
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
4. 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.
Plot a histogram of mother’s age (mage) to see distribution of ages. This provides a clue as to whether there is a natural cutoff - a noticeable drop-off in pregnant women at a certain age.
From the histogram, it looks like there is a big drop-off at age 38.
We could also pick a particular percentile to determine the cutoff. For example, if we choose the 75th percentile, the age is 32 y/o.
The 95th percentile is 39.4 y/o. The mean is 27 y/o. The SD is 6.2 years. 2 SD’s from the mean is 39.4 y/o.
hist(nc$mage,breaks=30)
summary(nc$mage)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 13 22 27 27 32 50sd(nc$mage)## [1] 6.2135835. Pick a pair of numerical and categorical variables and come up with a research question evaluating the relationship between these variables. Formulate the question in a way that it can be answered using a hypothesis test and/or a confidence interval. Answer your question using the inference function, report the statistical results, and also provide an explanation in plain language.
numerical: weight (baby’s birth weight) categorical: gender (baby’s gender)
Research Question: Does a baby’s gender have an effect on its birth weight?
H0: The baby’s gender has no effect on its birth weight. HA: The baby’s gender has an effect on its birth weight.
# Hypothesis Test
inference(y = nc$weight, x = nc$gender, est = "mean", type = "ht", null = 0,
alternative = "twosided", method = "theoretical")## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_female = 503, mean_female = 6.9029, sd_female = 1.4759
## n_male = 497, mean_male = 7.3015, sd_male = 1.5168## Observed difference between means (female-male) = -0.3986
##
## H0: mu_female - mu_male = 0
## HA: mu_female - mu_male != 0
## Standard error = 0.095
## Test statistic: Z = -4.211
## p-value = 0
The mean birth weight is 6.9 lbs for females and 7.3 lbs for males.
With a p-value of 0 at the 0.05 significance level, there is very strong evidence for the alternative hypothesis, so the null is rejected.
The baby’s gender has an effect on its birth weight.