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.

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?

The cases in this data set are individual pregnancies which includes data on the mother, baby(ies), and father.

`nrow(nc)`

`## [1] 1000`

There 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)`

Each data set has some outliers, appearing as the dots above or beneath the whiskers. Notable are pregnancy duration in weeks (3) has many low outliers. Weight gained by mother (5) has many high outliers.

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?

```
suppressMessages(suppressWarnings(library('ggplot2')))
df <- data.frame(nc$habit, nc$weight)
ggplot(aes(y = nc.weight , x = nc.habit, fill = nc.habit), data = df) + geom_boxplot()
```

Smokers tend to have a lower birth weight.

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 .

- 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
```

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

\(H_o\) is that there is no difference in the mean of the weights for the two populations. \(H_A\) 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 'openintro' was built under R version 3.4.1`

`## Warning: package 'BHH2' was built under R version 3.4.1`

```
## 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.

```
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 \(H_o\) and accept \(H_A\).

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 )
```

- 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")
```

```
## 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`

.

```
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. Note the difference between the upper and lower boundary is smaller than the 95% CI.

- 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$gained, x = nc$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 )
```

Since the 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.

`summary(nc$mage)`

```
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 13 22 27 27 32 50
```

`by(nc$mage,nc$mature, length)`

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

```
#We know that 75% of women are below 32 and noone is above 50. We need to find an age that splits the data at 133/1000.
ge_35 <- nc$mage >= 35.0 # this will give true if >= 35, false if > 35
table(ge_35) #Table will report the total true and the total false.
```

```
## ge_35
## FALSE TRUE
## 867 133
```

I used summary() to constrain the mother’s age. The by() command allowed me to see how many mature moms there were, 133. I then tested the mother’s age data against a condition that had to be above Q3 which had a cutoff of 32 years. I selected 35 years as a start point based on my own experiences as a father (recalling pregnancy risk statistics when consulting the doctor), and that is the cutoff age.

- 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.

Is there a difference in mean mother’s age for premature pregnancies compared to full termed pregnancies? Mother’s age ‘mage’ is the numerical data, premature status ‘premie’ is the categorical data. \(H_o\) is that there is no difference in mean age of the populations. \(H_A\) is that there is a difference in the mean age of the two populations.

```
inference(y = nc$mage, x = nc$premie, est = "mean", type = "ci", null = 0,
alternative = "twosided", method = "theoretical")
```

```
## 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
##
## Standard error = 0.5705
## 95 % Confidence interval = ( -0.9931 , 1.2431 )
```

Since the confidence interval spans 0, there is no difference in the age of the two populations. We accept \(H_o\).

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.