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` |
fatherâ€™s age in years. |

`mage` |
motherâ€™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?
*There are 1,000 cases.*

`dim(nc)`

`## [1] 1000 13`

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 arenâ€™t sure or want to take a closer look at the data, make a graph.

Consider the possible relationship between a motherâ€™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?*Birth weights for both groups are approximately normal and the average birth weight is normal.*

`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 .

- 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`

.*The conditions appear to be staistified.*

`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.
*H0 = Average weights of babies born to smoking and non-smoking mothers are not different.**HA = 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 '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
```

Letâ€™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 weâ€™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.

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 )
```

`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.`

```
inference(y = nc$weeks,est = "mean", type = "ci", null = 0,
alternative = "twosided", method = "theoretical",
order = c("smoker","nonsmoker"))
```

```
## Single mean
## Summary statistics:
```

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

`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`.`

```
inference(y = nc$weeks,est = "mean", type = "ci", null = 0,
alternative = "twosided", method = "theoretical",
order = c("smoker","nonsmoker"),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 )
```

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

*Average weight gained by younger mothers is not different from the average weight gained by mature mothers.*

```
inference(y = nc$weight, x = nc$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
```

`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.`

`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"])`

`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 testand/or a confidence interval. Answer your question using the `inference` function, report the statistical results, and also provide an explanation in plain language.`

*Question: Does marital status affect weight gained during pregnancy?* *H0:There is no difference in the mean of the weight gained during pregnancy of between married and unmarried mothers.* *HA:There is a difference in the mean of the weight gained during pregnancy between married and unmarried mothers* *Answer: The null hypothesis can not be rejected. There is no evidence based on hypothesis test and confidence interval inference to show a difference between weight gained by married and unmarried mothers during pregnancy.*

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

```
## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_married = 370, mean_married = 29.873, sd_married = 15.2721
## n_not married = 603, mean_not married = 30.6036, sd_not married = 13.5757
```

```
## Observed difference between means (married-not married) = -0.7307
##
## Standard error = 0.9675
## 95 % Confidence interval = ( -2.6269 , 1.1655 )
```

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 Ã‡etinkaya-Rundel from a lab written by the faculty and TAs of UCLA Statistics.