Lab5: Inference for numerical data

library(plotly)

## Warning: package 'plotly' was built under R version 3.4.2

## Loading required package: ggplot2

## Warning: package 'ggplot2' was built under R version 3.4.2

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library(dplyr)

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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.

1. What are the cases in this data set? How many cases are there in our sample? Ans: ‘fage’, ‘mage’,‘weeks’, ‘visits’, ‘gianed’, ‘weight’. There are 6 cases in the 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.

Ans: There are varialbes are numeric, ‘fage’, ‘mage’,‘weeks’, ‘visits’, ‘gianed’, ‘weight’; the rest are categorical datas. These numeric variables all have outliers.

boxplot(nc$fage,nc$mage,nc$weeks,nc$visits,nc$gained,nc$weight, names = c('fage', 'mage','weeks', 'visits', 'gianed', 'weight'),las=2, col = c("red","yellow","royalblue2","sienna","pink","green" ))

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.

2. Make a side-by-side boxplot of habit and weight. What does the plot highlight about the relationship between these two variables?

plot_ly(x = ~nc$habit, y= ~nc$weight, type = "box",color = ~nc$habit)

## Warning: Ignoring 1 observations

## Warning in RColorBrewer::brewer.pal(N, "Set2"): minimal value for n is 3, returning requested palette with 3 different levels

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] 6.82873

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

Ans: There is no difference statistcally significant by observation of two means.

Inference

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. Ans:

by(nc$weight, nc$habit, length)

## nc$habit: nonsmoker
## [1] 873
## -------------------------------------------------------- 
## nc$habit: smoker
## [1] 126

by(nc$weight, nc$habit, var)

## nc$habit: nonsmoker
## [1] 2.306391
## -------------------------------------------------------- 
## nc$habit: smoker
## [1] 1.921494

SE= squart(2.306391/873 + 1.921494/126) = squart(0.002642+0.01525)=0.133761

4. Write the hypotheses for testing if the average weights of babies born to smoking and non-smoking mothers are different.

Ans: H0: There is no difference statisticlly significant. Ha: There is difference statisticlly significant.

x1-x2= 7.144273-6.82873 = 0.315543

T=0.315543/0.133761 = 2.359 > 1.96 (95% confident)

p-value = 0.02 which is samller than 0.05, so we reject H0. There is difference statisticlly significant for baby’s weight from nonsmoke and smoke 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")

## Warning: package 'BHH2' was built under R version 3.4.2

## 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.

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.

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 the x variable from the function.

inference(y = nc$weeks, est = "mean", type = "ci", null = 0, 
          alternative = "twosided", method = "theoretical",conflevel = 0.95)

## Single mean 
## Summary statistics:

## mean = 38.3347 ;  sd = 2.9316 ;  n = 998 
## Standard error = 0.0928 
## 95 % Confidence interval = ( 38.1528 , 38.5165 )

• 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", 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 )

• 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$weight, x = nc$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 = 133, mean_mature mom = 7.1256, sd_mature mom = 1.6591
## n_younger mom = 867, mean_younger mom = 7.0972, sd_younger mom = 1.4855

## Observed difference between means (mature mom-younger mom) = 0.0283
## 
## H0: mu_mature mom - mu_younger mom = 0 
## HA: mu_mature mom - mu_younger mom != 0 
## Standard error = 0.152 
## Test statistic: Z =  0.186 
## p-value =  0.8526

Since p-value > 0.05, we accepte H0. There is no difference statisticlly significant for baby’s weight from mature and young mothers.

• 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.

Each person can be considered an indepent data, and two data sets are also indepent.

plot_ly(y = ~nc$mage, x= ~nc$mature, type = "box",color = ~nc$mature)

## Warning in RColorBrewer::brewer.pal(N, "Set2"): minimal value for n is 3, returning requested palette with 3 different levels

• 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.

Ans: The following study is focuse on lowbirthweight, I want to observe the “low birthweight= low” relative to habit of parents.

a<-filter(nc,nc$lowbirthweight=="low")
b<-data.frame(a$weight,a$habit)
b<-na.omit(b)
summary(b)

##     a.weight          a.habit  
##  Min.   :1.000   nonsmoker:92  
##  1st Qu.:3.047   smoker   :18  
##  Median :4.560                 
##  Mean   :4.038                 
##  3rd Qu.:5.175                 
##  Max.   :5.500

inference(y = b$a.weight, x = b$a.habit, est = "mean", type = "ht", null = 0, 
          alternative = "twosided", method = "theoretical", 
          order = c("smoker","nonsmoker"))

## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_smoker = 18, mean_smoker = 4.3589, sd_smoker = 1.1592
## n_nonsmoker = 92, mean_nonsmoker = 3.9758, sd_nonsmoker = 1.4148

## Observed difference between means (smoker-nonsmoker) = 0.3831
## 
## H0: mu_smoker - mu_nonsmoker = 0 
## HA: mu_smoker - mu_nonsmoker != 0 
## Standard error = 0.311 
## Test statistic: T =  1.234 
## Degrees of freedom:  17 
## p-value =  0.234

Since p-value > 0.05, we accepte H0. There is no difference statisticlly significant for low birth weight baby’s from parents who somke and nonsmoke.

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.