library(dplyr)
library(plyr)
library(data.table)
library(knitr)
download.file("http://www.openintro.org/stat/data/nc.RData", destfile = "nc.RData")
load("nc.RData")

Exercise 1 What are the cases in this data set? How many cases are there in our sample?

ls()
## [1] "chiTail"   "FTail"     "inference" "nc"        "normTail"
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 white
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  
##                 
##                 
##                 
##

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

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   6.078   7.060   6.829   7.735   9.190
boxplot(nc$weight[nc$habit =="nonsmoker"],nc$weight[nc$habit == "smoker"],xlab="habit",ylab="weight",main="Weight vs. habit", names = c("nonsmoker", "smoker"))

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] 126

Exercise 4 Write the hypotheses for testing if the average weights of babies born to smoking and non-smoking mothers are different. 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 'openintro' was built under R version 3.2.5
## Warning: package 'BHH2' was built under R version 3.2.5
## 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

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.

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

Exercise 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.

#using the custom inference function
inference(y = nc$weeks, est = "mean", type = "ci", 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 )

Exercise 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.

#using the custom inference function
inference(y = nc$weeks, est = "mean", type = "ci", 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 )

Exercise 3 Conduct a hypothesis test evaluating whether the average weight gained by younger mothers is different than the average weight gained by mature mothers.

The null hypothesis is the average weight gained by younger mothers are not different from the average weight gained by mature mothers. Based on the test statics of p-value of 0.8526 there is strong evidence that we fail to reject the null hypothesis that the average weight gained by younger mothers are not different from the average weight gained by mature mothers.

#using the custom inference function
inference(nc$weight, nc$mature, type="ht", est="mean", null=0, method="theoretical", alternative="twosided")
## 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

Exercise 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.

Based on the summary data and the boxplot the age cutoff for younger and mature mothers is 35. Look at the summary data the max age for younger mom is 34, whereas the mininum age for mature mother is 35. In the boxplot for the age of mature mom the minimum value of a data set is 35.

by(nc$mage, nc$mature,summary)
## nc$mature: mature mom
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   35.00   35.00   37.00   37.18   38.00   50.00 
## -------------------------------------------------------- 
## nc$mature: younger mom
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   13.00   21.00   25.00   25.44   30.00   34.00
 boxplot(nc$mage[nc$mature == "mature mom"])

Exercise 5 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.

Interested in researching the relationship between mothers’ ages and the term of their pregnancy. The null hypothesis will be that there is no relationship between mothers’ ages and the term of their pregnancy.

inference(nc$mage, nc$premie, est = "mean", type = "ht", null = 0,alternative = "twosided", method = "theoretical",order = c("full term","premie"))
## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_full term = 846, mean_full term = 27, sd_full term = 6.1444
## n_premie = 152, mean_premie = 26.875, sd_premie = 6.533
## Observed difference between means (full term-premie) = 0.125
## 
## H0: mu_full term - mu_premie = 0 
## HA: mu_full term - mu_premie != 0 
## Standard error = 0.57 
## Test statistic: Z =  0.219 
## p-value =  0.8266

Based on the test statics of p-value of 0.8266 there is strong evidence that we fail to reject the null hypothesis there is no relationship between mothers’ ages and the term of their pregnancy.