Inference for numerical data
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.
load("more/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. |
- What are the cases in this data set? How many cases are there in our sample?
dim(nc)## [1] 1000 13Each row consists of a case (with 13 variables). There is a 1000 cases (rows) in this sample.
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.
- Make a side-by-side boxplot of
habitandweight. What does the plot highlight about the relationship between these two variables?
boxplot(nc$weight ~ nc$habit)
The side by side box plots, on visual glance, appear to have similar medians and fairly close variation as well.
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 .
Inference
- 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
bycommand above but replacingmeanwithlength.
by(nc$weight, nc$habit, length)## nc$habit: nonsmoker
## [1] 873
## --------------------------------------------------------
## nc$habit: smoker
## [1] 126hist(nc$weight[nc$habit == "smoker"], main = "Weight for Smoking Mothers", xlab = "Weight")
hist(nc$weight[nc$habit == "nonsmoker"], main = "Weight for Non-Smoking Mothers", xlab = "Weight")
In order to satisfy the necessary conditions for inference, the data needs to be normal or near normal distribution and independent. Given that this is a sample of all mothers, and presumably they did not know each other, both sample subsets (smoking, nonsmoking) are considered independent. There is some small left sided skew, but given that the sample sizes are quite large, this skew can be ignored. So yes, this satisfies the necessary conditions.
- Write the hypotheses for testing if the average weights of babies born to smoking and non-smoking mothers are different.
Null Hypothesis: The average weights of babies born to smoking mothers == the average weights of babies born to non-smoking mothers. Alternative Hypothesis: The average weights of babies born to smoking mothers != the average weights of babies born to 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.
- Change the
typeargument 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 )On your own
- 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 thexvariable from the function.
inference(y = nc$weeks, est = "mean", type = "ci",
alternative = "twosided", method = "theoretical")## Single mean
## Summary statistics:
## mean = 38.3347 ; sd = 2.9316 ; n = 998
## Standard error = 0.0928
## 95 % Confidence interval = ( 38.1528 , 38.5165 )We are 95% confident that the average length of pregnancy for the population in weeks is within 38.1528 and 38.5165 weeks.
- Calculate a new confidence interval for the same parameter at the 90% confidence level. You can change the confidence level by adding a new argument to the function:
conflevel = 0.90.
inference(y = nc$weeks, est = "mean", type = "ci", confleve = 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 )90% Confidence Interval: ( 38.182 , 38.4873 )
- Conduct a hypothesis test evaluating whether the average weight gained by younger mothers is different than the average weight gained by mature mothers.
inference(y = nc$gained, x = nc$mature, est = "mean", type = "ht", null = 0,
alternative = "twosided", method = "theoretical",
order = c("mature mom", "younger mom"))## 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
The p value is: 0.1686. Therefore, we do NOT reject the null hypothesis.
- 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.
young.mom <- subset(nc, nc$mature == "younger mom")
mature.mom <- subset(nc, nc$mature == "mature mom")
summary(young.mom)## fage mage mature weeks
## Min. :14.00 Min. :13.00 mature mom : 0 Min. :22.00
## 1st Qu.:24.00 1st Qu.:21.00 younger mom:867 1st Qu.:37.00
## Median :29.00 Median :25.00 Median :39.00
## Mean :28.86 Mean :25.44 Mean :38.38
## 3rd Qu.:33.00 3rd Qu.:30.00 3rd Qu.:40.00
## Max. :48.00 Max. :34.00 Max. :45.00
## NA's :160 NA's :1
## premie visits marital gained
## full term:737 Min. : 0.00 married :361 Min. : 0.00
## premie :129 1st Qu.:10.00 not married:506 1st Qu.:21.00
## NA's : 1 Median :12.00 Median :30.00
## Mean :12.03 Mean :30.56
## 3rd Qu.:15.00 3rd Qu.:38.25
## Max. :30.00 Max. :85.00
## NA's :7 NA's :23
## weight lowbirthweight gender habit
## Min. : 1.000 low : 93 female:435 nonsmoker:752
## 1st Qu.: 6.380 not low:774 male :432 smoker :115
## Median : 7.310
## Mean : 7.097
## 3rd Qu.: 8.000
## Max. :11.750
##
## whitemom
## not white:255
## white :611
## NA's : 1
##
##
##
## summary(mature.mom)## fage mage mature weeks
## Min. :26.00 Min. :35.00 mature mom :133 Min. :20.00
## 1st Qu.:35.00 1st Qu.:35.00 younger mom: 0 1st Qu.:38.00
## Median :38.00 Median :37.00 Median :39.00
## Mean :38.36 Mean :37.18 Mean :38.02
## 3rd Qu.:41.00 3rd Qu.:38.00 3rd Qu.:40.00
## Max. :55.00 Max. :50.00 Max. :44.00
## NA's :11 NA's :1
## premie visits marital gained
## full term:109 Min. : 3.00 married : 25 Min. : 0.00
## premie : 23 1st Qu.:10.00 not married:107 1st Qu.:20.00
## NA's : 1 Median :12.00 NA's : 1 Median :28.00
## Mean :12.61 Mean :28.79
## 3rd Qu.:15.00 3rd Qu.:36.00
## Max. :30.00 Max. :70.00
## NA's :2 NA's :4
## weight lowbirthweight gender habit
## Min. : 1.380 low : 18 female:68 nonsmoker:121
## 1st Qu.: 6.380 not low:115 male :65 smoker : 11
## Median : 7.310 NA's : 1
## Mean : 7.126
## 3rd Qu.: 8.190
## Max. :10.250
##
## whitemom
## not white: 29
## white :103
## NA's : 1
##
##
##
## mom <- c(young.mom, mature.mom)
hist(mature.mom$mage, breaks = 20, xlim = c(10,50))
hist(young.mom$mage, breaks = 20, xlim = c(10,50))
paste("Maximum age of a younger mom: ", max(young.mom$mage))## [1] "Maximum age of a younger mom: 34"paste("Minimum age of a mature mom: ", min(mature.mom$mage))## [1] "Minimum age of a mature mom: 35"It appears visually on a historgram as well as looking for the max age of a younger mom and the minimum age of a mature mom that the age cut off for the younger mom and mature mom group is: 34 years old and younger for young mom and 35 years old and older for mature mom.
- 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
inferencefunction, report the statistical results, and also provide an explanation in plain language.
# Let's examine the relationship between a smoker and how many visits she had made to the hospital.
# The question is: do mothers who smoke go to the hospital more frequently than non-smoking mothers?
# Null hypothesis: visits to hospital by moms who smoke == visits to hospital by moms who DO NOT smoke.
# Alternative hypothesis: visits to hospital by moms who smoke != visits to hospital by moms who DO NOT smoke.
inference(y = nc$visits, x = nc$habit, est = "mean", type = "ht", null = 0, alternative = "twosided",
method = "theoretical", order = c("nonsmoker", "smoker"))## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_nonsmoker = 866, mean_nonsmoker = 12.2159, sd_nonsmoker = 3.9139
## n_smoker = 125, mean_smoker = 11.336, sd_smoker = 4.1639## Observed difference between means (nonsmoker-smoker) = 0.8799
##
## H0: mu_nonsmoker - mu_smoker = 0
## HA: mu_nonsmoker - mu_smoker != 0
## Standard error = 0.395
## Test statistic: Z = 2.225
## p-value = 0.026
The p value is: 0.026, thus rejecting the null hypothesis. It appears that mom’s who smoke go to the hospital more frequently than non-smokers.
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.