Inference for Numerical Data

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")

Exercise 1:

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

Answer:

The cases are newborn babies. There are 1000 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  
##                 
##                 
##                 
##

Exercise 2:

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

boxplot(nc$weight ~ nc$habit, main= "Weight of the Babies by Smoking Habits of the Mothers",
        xlab = "Weight of the Babies", ylab = "Smoking Habits of the Mothers",
horizontal = TRUE)

Answer:

The boxplots show that the medians and interquartile range of the two distributions are very close, but the distribution for nonsmokers have more outliers and is more disperse.

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.

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] 126

Answer:

Both sample sizes are bigger than 30, therefore the conditions for inference should be satisfied.

Exercise 4:

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

Answer:

Null Hypothesis: There is no difference in means of the average weights of babies born between the smoking and non-smoking mother groups.

Alternative Hypothesis: There is a difference in means of the average weights of babies born between the smoking and non-smoking mother groups.

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

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 )

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.

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 )
  1. 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 )
  1. Conduct a hypothesis test evaluating whether the average weight gained by younger mothers is different than the average weight gained by mature mothers.

Answer:

The p-value is 0.1686, therefore there is strong evidence to reject the null hypothesis. We have strong evidence that there is a difference in the means of the average weight gained between younger and mature mothers.

inference(nc$gained, nc$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

  1. 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:

With the following method we can see the range of the ages of the mother divided between the mature and younger groups. The max value for younger moms is 34 years and the min value for mature moms is 35 years.

by (nc$mage, nc$mature, range)
## nc$mature: mature mom
## [1] 35 50
## -------------------------------------------------------- 
## nc$mature: younger mom
## [1] 13 34
  1. 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: Does the mother´s maturity have an effect on babies´average weight at birth?

Null Hypothesis:There is no difference in means of the average weights of babies born between the younger and mature mother groups.

Alternative Hypothesis: There is a difference in means of the average weights of babies born between the younger and mature mother groups.

inference(nc$weight, nc$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 = 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

The p-value is 0.8526, therefore there is weak evidence to reject the null hypothesis. We have weak evidence that there is a difference in the means of the average weight of the baby at birth between younger and mature mothers.