Data606 Lab5
Ritesh Lohiya
March 23, 2018
library(ggplot2)
library('DATA606') # Load the package##
## Welcome to CUNY DATA606 Statistics and Probability for Data Analytics
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## Attaching package: 'DATA606'## The following object is masked from 'package:utils':
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## demolibrary(knitr)
library(dplyr)##
## Attaching package: 'dplyr'## The following objects are masked from 'package:stats':
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## filter, lag## The following objects are masked from 'package:base':
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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("nc.RData")
head(nc)## fage mage mature weeks premie visits marital gained weight
## 1 NA 13 younger mom 39 full term 10 married 38 7.63
## 2 NA 14 younger mom 42 full term 15 married 20 7.88
## 3 19 15 younger mom 37 full term 11 married 38 6.63
## 4 21 15 younger mom 41 full term 6 married 34 8.00
## 5 NA 15 younger mom 39 full term 9 married 27 6.38
## 6 NA 15 younger mom 38 full term 19 married 22 5.38
## lowbirthweight gender habit whitemom
## 1 not low male nonsmoker not white
## 2 not low male nonsmoker not white
## 3 not low female nonsmoker white
## 4 not low male nonsmoker white
## 5 not low female nonsmoker not white
## 6 low male nonsmoker not whiteWe have observations on 13 different variables, some categorical and some numerical. The meaning of each variable is as follows.
Exercise 1: What are the cases in this data set? How many cases are there in our sample?
Answer: The cases in this data set are individual pregnancies which includes data on the mother, babies and father.
nrow(nc)## [1] 1000There 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:
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.
boxplot(nc$fage,nc$mage,nc$weeks,nc$visits, nc$gained,nc$weight)
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.
Exercise 2: Make a side-by-side boxplot of habit and weight. What does the plot highlight about the relationship between these two variables?
Answer;
boxplot(nc$weight~nc$habit, ylab="Weight", xlab="Habit")
nsmokewgt<- subset(nc, habit == "nonsmoker")
smokewgt<- subset(nc, habit == "smoker")
summary(nsmokewgt$weight)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 1.000 6.440 7.310 7.144 8.060 11.750summary(smokewgt$weight)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 1.690 6.077 7.060 6.829 7.735 9.190From the plot we see that babies born to non smokers on average have higher weights than those born to smokers.The Median weight is lower for smokers, so habits affect birth weights.
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
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] 126Exercise 4: Write the hypotheses for testing if the average weights of babies born to smoking and non-smoking mothers are different.
Answer:
Ho is that there is no difference in the mean of the weights for the two populations. HA is that there is a difference in the mean for the weights of the two populations.
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.
Exercise 5: 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.
Answer:
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 )Since the confidence interval of (0.0534, 0.5777) pounds does not span 0, there is a statistically significance in the weight of the two populations. We reject Ho and accept HA.
By default the function reports an interval for (??nonsmoker?????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 the x variable from the function.
Answer:
inference(y = nc$weeks, est = "mean", type = "ci", null = 0,
alternative = "twosided", method = "theoretical")## Single mean
## Summary statistics:
## mean = 38.3347 ; sd = 2.9316 ; n = 998
## Standard error = 0.0928
## 95 % Confidence interval = ( 38.1528 , 38.5165 )We are 95% confident that we have captured the mean pregnancy length in weeks of the population between 38.1528 weeks and 38.5165 weeks.
- Calculate a new confidence interval for the same parameter at the 90% confidence level. You can change the confidence level by adding a new argument to the function: conflevel = 0.90.
Answer:
inference(y = nc$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 )We are 90% confident that we have captured the mean pregnancy length in weeks of the population between 38.182 weeks and 38.4873 weeks.
- Conduct a hypothesis test evaluating whether the average weight gained by younger mothers is different than the average weight gained by mature mothers.
Answer:
inference(y = nc$gained, x = nc$mature, est = "mean", type = "ci", 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
##
## Standard error = 1.2857
## 95 % Confidence interval = ( -4.2896 , 0.7502 )Because confidence interval (-4.2896 , 0.7502) pounds spans 0 we accept the Null Hypothesis that there is no difference in mean weight gain of the two populations.
- 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.
Answer:
by(nc$mage, nc$mature, range)## nc$mature: mature mom
## [1] 35 50
## --------------------------------------------------------
## nc$mature: younger mom
## [1] 13 34summary(nc$mage)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 13 22 27 27 32 50Younger mom and mature mom cutoff point can be calculated by range of the 2 groups.Using the by function with the range argument we can see that the range for younger mom ends at 34 (MAX) and mature mom begins at 35 (MIN). So the cut off age for the 2 groups is 34.
- Pick a pair of numerical and categorical variables and come up with a research question evaluating the relationship between these variables. Formulate the question in a way that it can be answered using a hypothesis test and/or a confidence interval. Answer your question using the inference function, report the statistical results, and also provide an explanation in plain language.
Answer:
Research Question:
Is there a difference between the average number of hospital visits during pregnancy for a younger mom and a mature mom?
inference(y = nc$visits, 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 = 131, mean_mature mom = 12.6107, sd_mature mom = 4.3793
## n_younger mom = 860, mean_younger mom = 12.0279, sd_younger mom = 3.8832## Observed difference between means (mature mom-younger mom) = 0.5828
##
## H0: mu_mature mom - mu_younger mom = 0
## HA: mu_mature mom - mu_younger mom != 0
## Standard error = 0.405
## Test statistic: Z = 1.439
## p-value = 0.15
Because the p-value is higher than aplha 0.05, we do not reject the Ho.
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.