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?

Answer:

1000, 13 , These cases are the birth records in North Carolina in 2004. The number of cases in our sample is 1000

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 habit and weight. What does the plot highlight about the relationship between these two variables?

Answer

The plot b/w habit and weight describes that median weight of the baby is slightly higher in case of non-smoking mothers.

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   6.077   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"))

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

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.

Answer

library(ggplot2)

## Warning: package 'ggplot2' was built under R version 3.5.3

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

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

ggplot(nc, aes(nc$weight))+ geom_bar()

Independence: Since n<10% of the population for the 2 groups are independent. The 1,000 The cases is probably less than 10% of the population to ensure they are simple random samples so independence is reasonable.so we can say that the sample was chosen randomly so observations are independent of each other.

Sample sizes/skew: Sample sizes are greater than 30. Both distributions are more or less symetric. Even though the distribution of differences shown in the boxplots are a bit skewed they seems reasonable for the size of the sample.

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

Answer

H0 Null hypothesis : states that avg weights of babies born to smoking and non-smoking mothers are same

Ha Alternate : Avg weights are different for different habit 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 'openintro' was built under R version 3.5.2

## Warning: package 'BHH2' was built under R version 3.5.3

## 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 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 ($\mu_{nonsmoker} - \mu_{smoker}$) . 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.

Answer

From the below inference function results, We are 95 % Confident that the average length of pregnancies (weeks) is between 38.15 and 38.52. This confidence interval is calculated based on the total sample that consists both smoking and nonsmoking mothers.

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 )

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, conflevel = 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 )

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

Answer

H0 The null hypothesis : The average weight gained by younger mothers are not different from the average weight gained by mature mothers. HA : That avg weight gained by young and mature mothers are different

Samples must be drawn independently from one another. Assumed yes.
Samples must be nearly normal. see below Qplot and bar plot, the data is bit right skewed and the qq-plots is pretty linear.
There must be at least 30 observations. 133 mature and 867 younger.

For our two sided test we get a p-value of 0.1686, which is not less than 0.05, so we conclude that we do not have sufficient evidence to reject the null hypothesis.

levels(nc$mature) # "mature mom" "younger mom"

## [1] "mature mom"  "younger mom"

qplot(data=nc, sample=gained, color=mature)

## Warning: Removed 27 rows containing non-finite values (stat_qq).

ggplot(nc, aes(nc$gained, fill= mature))+ geom_bar() + facet_wrap(~mature)

## Warning: Removed 27 rows containing non-finite values (stat_count).

by(nc$gained, nc$mature, length) # 133 mature, 867 younger

## nc$mature: mature mom
## [1] 133
## -------------------------------------------------------- 
## nc$mature: younger mom
## [1] 867

inference(y=nc$gained, x=nc$mature, est="mean", type="ht", null=0, alternative="twosided",
          method="theoretical", conflevel=0.95, order=c("younger mom", "mature mom"))

## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_younger mom = 844, mean_younger mom = 30.5604, sd_younger mom = 14.3469
## n_mature mom = 129, mean_mature mom = 28.7907, sd_mature mom = 13.4824

## Observed difference between means (younger mom-mature mom) = 1.7697
## 
## H0: mu_younger mom - mu_mature mom = 0 
## HA: mu_younger mom - mu_mature mom != 0 
## Standard error = 1.286 
## Test statistic: Z =  1.376 
## p-value =  0.1686

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, 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

It appears that 35 is the age cutoff for younger and mature mothers since the oldest younger mother is 34 and the youngest mature mother is 35.

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

H0: There is no diffirence in age of monthers for Mature and pre-mature birth. Ha: There is diffirence in age of monthers for Mature and pre-mature birth.

since P vlaue = 0.8266 which is not less than 0.05 , we can’t reject null hypothesi i.e. so it stand true that there is no diffrence in Mother’s age for Mature and and Premature birth of babies.

levels(nc$premie) # "mature mom" "younger mom"

## [1] "full term" "premie"

qplot(data=nc, sample=mage, color=premie)

ggplot(nc, aes(nc$mage, fill= premie))+ geom_bar() + facet_wrap(~premie)

by(nc$mage, nc$premie, length) # 133 mature, 867 younger

## nc$premie: full term
## [1] 846
## -------------------------------------------------------- 
## nc$premie: premie
## [1] 152

inference(y=nc$mage, x=nc$premie, est="mean", type="ht", null=0, alternative="twosided",
          method="theoretical", conflevel=0.95, 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

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.