Lab 2I - R’s Normal Distribution Alphabet

Lab 2I - R's Normal Distribution Alphabet

Directions: Follow along with the slides, completing the questions in blue on your computer, and answering the questions in red in your journal.

Where we're headed

In the last lab, you were able to overlay a normal curve on histograms of data to help you decide if the data's distribution is close to a normal distribution.

– We also saw that calculating the mean of random shuffles also produces differences that are normally distributed.
In this lab, we'll learn how to use some other R functions to:

– Simulate random draws from a normal distribution.

– Calculate probabilities with normal distributions.

Get set up

Start by loading the titanic data and calculate the mean age of people in the data but shuffle their survival status 500 times.

– Assign this data the name shfls.
After creating shfls, use mutate to add a new variable to the dataset. This new variable should have the name diff and should be the mean age of those who survived minus those who died.
Finally, calculate the mean and sd of the diff variable.

– Assign these values the name diff_mean and diff_sd.

Is it normal?

Before we proceed, we need to verify that our diff variable looks approximately normally distributed.

– Is the distribution close to normal? Explain how you determined this. Describe the center and spread of the distribution.

– Compute and write down the mean difference in the age of the actual survivors and the actual non-survivors.

Using the normal model

Since the distribution of our diff variable appears normally distributed, we can use a normal model to estimate the probability of seeing differences that are more extreme than our actual data.
- Draw a sketch of a normal curve. Label the mean age difference, based on your shuffles, and the actual age difference of survivors minus non-survivors from the actual data. Then shade in the areas, under normal the curve, that are smaller than the actual difference.
**Fill in the blanks to calculate the probability of an even smaller difference occurring than our actual difference using a normal model.
```
pnorm(____, mean = diff_mean, sd = ____)
```

Extreme probabilities

The probability you calculated in the previous slide is an estimate for how often we expect to see a difference smaller than the actual one we observed, by chance alone.
If you wanted to instead calculate the probability that the difference would be larger than the one observed, we could run (fill in the blanks):
```
1 - pnorm(____, mean = diff_mean, sd = ____)
```

Simulating normal draws

We can simulate random draws from a normal distribution with the rnorm function.

– Fill in the blanks in the following two lines of code to simulate 100 heights of randomly chosen men. Assume the mean height is 67 inches and the standard deviation is 3 inches.
```
draws <- rnorm(____, mean = ____, sd = ____)
```
– Plot your simulated heights with a histogram.
```
histogram(draws, fit = ____)
```

P's and Q's

We've seen that we can use pnorm to calculate probabilities based on a specified quantity.

– Hence, why we call it "P" norm.
Now we'll see how to do the opposite. That is, calculate the quantity for a specific probability.

– Hence why we'll call this a "Q" norm.
How tall can a man be and still be in the shortest 25% of heights if the mean height is 67 inches with a standard deviation of 3 inches?
```
qnorm(____, mean = ____, sd = ____)
```

On your own

Conduct one of the statistical investigations below:

Using the titanic data:

– Were women on the Titanic typically younger than men?

– Use a histogram, 500 random shuffles and a normal model to answer the question in the bullet above.
Using the cdc data:

– Using 500 random shuffles and a normal model, how much taller would the typical male have to be than the typical female in order for the difference to be in the upper 1% by chance alone?

– How can we use this value to justify the claim that the average Male in our data is taller than the average Female?