- Basics of Data in R
- Plotting and evaluating one categorical variable
- Plotting and evaluating two categorical variables
- Analyzing shape and center of one quantitative variable
- Analyzing the spread of one quantitative variable
- Relationships between quantitative and categorical data
- Relationships between two quantitative variables
- Final Thoughts on linear regression
- A bit off topic - functions, grep, and colors
- Sampling and Loops
- Confidence Intervals
- Bootstrapping
- More on Bootstrapping
- Hypothesis testing and p-values
- Differences in proportions and statistical thresholds
- Hypothesis testing for means
- Final thoughts on hypothesis testing
- Approximating with a distribution model
- Using the normal model in practice
- Approximating for a single proportion
- Null distribution for a single proportion and limitations
- Approximating for a single mean
- CI and hypothesis tests for a single mean
- Approximating a difference in proportions
- Hypothesis test for a difference in proportions
- Difference in means
- Difference in means - Hypothesis testing and paired differences
- Shortcuts
- Testing categorical variables with Chi-sqare
- Testing proportions in groups
- Comparing the means of many groups
- Linear Regression
- Multiple Regression
- Basic Probability
- Random variables
- Conditional Probability
- Bayesian Analysis

- NFLcombineData.csv
- Data on the Height, Weight, and position type of all 4299 players that participated in the NFL combine during the collection period

- survey.csv
- Data from intro stats students

- mlbCensus2014.csv
- Basic data on every Major League Baseball player on a 25-man roster as of 2014/May/15th

```
# Set working directory
# Note: copy the path from your old script to match your directories
setwd("~/path/to/class/scripts")
# Load data
nflCombine <- read.csv("data/NFLcombineData.csv")
survey <- read.csv("data/survey.csv")
mlb <- read.csv("data/mlbCensus2014.csv")
```

Before we get back to estimating distributions with the normal approximation, let’s recall what it is that we are trying to simulate. For any one sample, we don’t know whether or not our particular sample is representative of the population. If we were able to, we could draw multiple samples from the population and see what they looked like.

So, let’s do that. Our nfl data is the full population of all players that atteneded the combine. We can create a lot of samples from it in order to see what they look like. Because we are talking about proportions, we’ll look at the proportion of participants that are skill players. First, let’s see what the true population parameter is.

```
# Calculate population parameter
popProp <- mean(nflCombine$positionType == "skill")
popProp
```

`## [1] 0.662247`

So, about two-thirds of the players are skill players. If we draw lots of samples, will we alwats see that proportion? We can test that by drawing many samples of 40 players, and measuring the proportion that are skill players.

```
# Initialize variable
sampleProps <- numeric()
# Loop to draw many samples
for(i in 1:12345){
# Save the proportion that are "skill" from a sample
sampleProps[i] <- mean(sample(nflCombine$positionType,
40) == "skill")
}
```

Then, as always, we plot our result to make sure it make sense and to give us an idea of what we found.

```
# Visualize
hist(sampleProps
, xlab = "Sample proportion")
```

As expected, we see a roughly normal distribution that is peaked around our sample proportion (0.662). However, notice that there are several samples that are quite different from that, as low as 0.35 and as high as 0.925. What we are, generally, interested in is where the middle chunk lies. For this, we calculate the quantiles, usually to catch the middle 95%.

```
# Where is the middle 95%?
quantile(sampleProps,
c(0.025, 0.975))
```

```
## 2.5% 97.5%
## 0.525 0.800
```

This tells us that, 95% of the time we sample 40 individuals from this population, our sample proportion will fall between 0.525 and 0.8. We can use this to help us make some inferences soon. If you’d like, you can add lines to your plot to mark those points (note that I also changed the `breaks =`

just for the visual display).

Becuase the distribution is roughly normal, we can also apply the normal model to describe it. All we need to specify the normal is the mean, which we set to the population parameter when we know it, and the standard deviation, which we can calculate from the sample proportions.

```
# Calculate sd
sd(sampleProps)
```

`## [1] 0.0747751`

Of note, this means that we expect 95% of samples to fall within about 0.15 of the true value (two times the sd). That gives us a range between 0.513 and 0.812, which is very similar to what we calculated with `quantile()`

above. Now, let’s add the curve on top of the histogram.

```
curve(dnorm(x,
mean = popProp,
sd = sd(sampleProps)),
add = TRUE,
col = "red3",
lwd = 3)
```

Of note: your line is likely to consitently hit the right edge of the bins on your distribution, rather than the center. That is because when R makes histograms, it uses the function `pretty()`

to pick good looking break points. Unfortunately, those breaks are also at the same place where proportions out of round numbers (like 40), tend to land. Because R counts the values on the right edge of each bin, the position of the bin can look a little bit odd. I manually set (and adjusted) the breaks for the histogram above (not something I expect you to do) for illustrative purposes.

However, through all of this, we still had to contstruct the null distribution. So, what did we actually gain by applying the normal model?

What we gain from the application of this model is the opportunity to make a short cut. The sampling loops (sampling, bootstrap, and null) that we have been making are relatively simple to implement in R. However, that is only recently true. Through most of the history of statistics, such sampling would have been an immense undertaking. So, statisticians were forced to develop short cuts that would allow them to estimate the shape of sampling distributions.

What we need is to specify the normal distribution we want to use, which only requires a mean and a standard deviation. We have already seen that for a sampling distribution, we can (and should) just use the population proportion as our mean. But, how can we estimate the standard deviation?

It turns out that there is a relatively simple formula to estimate standard error of a proportion estimate which relies on nothing more than knowing the proportion calculated by our sample (p) and the number of samples we collected (n).

\[ SE = \sqrt{\frac{p \times (1-p)}{n}} \]

In R, this looks like:

`SE <- sqrt( ( p*(1-p) ) / n )`

In words, that is the square root of p times 1-p divided by n.

At least some of this should make sense to us. We have already seen that as sample size increases, SE decreases, exactly what the denominator shows. The rest of the derivation is beyond the scope of this course: for now, you will just need to trust that it is valid.

We can compare this shortcut to the NFL sampling data to see how accurate it is.

```
# Calculate the sd by formula
formSD <- sqrt( ( popProp*(1-popProp) ) / 40 )
formSD
```

`## [1] 0.07477899`

Which is incredibly close to the value we calculated from the sampling distribition.

```
# Compare to sd of samples
sd(sampleProps)
```

`## [1] 0.0747751`

And we can see that they are incredibly similar. This means that we can use this calculated standard error to estimate any interval we want. Here, we can compare the 90% interval using the two methods.

```
# Calculate 90% interval
qnorm(c(0.05,0.95),
mean = popProp,
sd = formSD)
```

`## [1] 0.5392465 0.7852475`

```
# Compare
quantile(sampleProps,
c(0.05,0.95))
```

```
## 5% 95%
## 0.525 0.775
```

As above, this tells us that 90% of samples (of size 40) taken from the NFL Combine data will give sample proportions inside these intervals. The slight difference is largely due to the discrete nature of the sample data. In reality, the sample proportions can only take on 41 values (\(\frac{0}{40}\) to \(\frac{40}{40}\)), so there is a limite to where the quantiles can give us cutoffs.

Treating the `THW`

column from the `mlb`

data, create a sampling distribution of the proportion of players that throw left-handed (`mlb$THW == "L"`

). Use a sample size of 50 players. Then, calculate the standard error using the formula above and compare it to the standard deviation of the sampled values.

You should get a standard error of 0.0573.

To calculate a confidence interval (CI) from a sample, we had been using bootstrapping. When we bootstrap, we are using our original sample as representative of the original population. As long as the sample is sort of the same as the original population, the width of our bootstrap distribution will be very similar to the width of the true sampling distribution. This tells us how far we’d have to “reach” to be confident that we are catching the true mean. Since we don’t know which direction we are wrong in (if we are at all), we say that we are confident that our true population mean falls somewhere in that range.

To see this in action, let’s ask what proportion of students that responded to the survey are Freshmen.

```
# Look at the data
table(survey$class)
```

```
##
## Freshman Junior Senior Sophomore
## 21 10 6 19
```

So, we see that 21 of the 56 are Freshmen. Since we are interested in the proportion, that is what we will calculate. Recall that `mean()`

here is just counting up all of the Freshmen, then dividing by the number of students.

```
# Calculate the sample proportion
mySampleProp <- mean(survey$class == "Freshman")
mySampleProp
```

`## [1] 0.375`

Next, we need to construct a bootstrap distribution and look at its spread.

```
# initialize new variable
bootProps <- numeric()
# Loop to draw many samples
for(i in 1:15486){
bootProps[i] <- mean(sample(survey$class,
replace = TRUE) == "Freshman")
}
```

You will notice that this loop looks *a lot* like the loop above. It is just copied with slight changes to work from these data instead. Then, visualize it. I am including the curve directly this time.

```
# Visualize
hist(bootProps,
xlab = "Sample propotion Freshmen",
freq = FALSE)
```

Next, we want to estimate the standard error of the distribution. Let’s jump straight to the formula this time:

```
# Estimate standard error
classSD <- sqrt( ( mySampleProp*(1-mySampleProp)
) / nrow(survey) )
classSD
```

`## [1] 0.06469365`

How does this compare to the standard error from our sample values?

```
# Calculate sd from samples
sd(bootProps)
```

`## [1] 0.06405583`

Quite close. So, we can use this estimated standard error to add the curve to the plot.

```
# Add the normal curve
curve(dnorm(x,
mean = mySampleProp,
sd = sd(bootProps)),
col = "green3", lwd = 3,
add = TRUE)
```

Finally, what we started off trying to do: let’s calculate the 90% confidence interval using each method. We use `quantile()`

to assess the distribution and `qnorm()`

to work from the normal model.

```
# show the CI cutoffs using the distribution
quantile(bootProps,
c(0.05, 0.95))
```

```
## 5% 95%
## 0.2678571 0.4821429
```

```
# Compare to CI from normal model
qnorm(c(0.05, 0.95),
mean = mySampleProp,
sd = classSD)
```

`## [1] 0.2685884 0.4814116`

This tells us that, because we are assuming the population is roughly similar to the sample, we expect that the same shape would have been created by the population, so 90% of the time our sample should not be more wrong than the width of the CI. In simpler terms: it tells us that our true population proportion is likely to fall within the CI 90% of the time.

Specifically for this class, we are 90% confident that the true population proportion of Freshman in intro stats courses at Viterbo is between 0.269 and 0.481.

Based on the `sugaryDrink`

column of the `survey`

data, calculate the 99% confidence interval for the proportion of intro statistics students at Viterbo that call a “sugary carbonated beverage” a “Pop”.

You do *not* need to use a bootstrap sample, though you can do one if you want to confirm your answer. You should get a confidence interval of 0.241 to 0.58

In all of the above examples, we have used the the `mean =`

and `sd =`

arguments to `qnorm()`

to specify which normal model we were talking about. However, historically, that was not possible. If you wanted to know a quantile cut off, you looked it up in a massive table (See this Wikipedia example). The table only had enteries at a selected points, and so was less specific.

Further, because there are an infinite number of possible normal distributions (any mean and any standard deviation), only one table was printed: the standard normal. This meant that to use it, it was neccessary to convert values to values back and forth from the normal model.

This has some advantages, and is also necessary for working with some of the distributions we will encounter soon. The biggest advantage is that it shows that the cut-offs for a given CI are always very closely related.

The value we find in this way is called the z-critical value, as it is the cutoff point on the z-scale (the abbreviation for the standard normal). It is generally written as z*.

So, if we want to caluculate a 90% CI, we can find the z* values using the standard normal:

```
# Find z* values
zCrit <- qnorm( c(0.05,0.95))
zCrit
```

`## [1] -1.644854 1.644854`

This tells us how many standard deviations above and below the mean we need to include to capture 90% of the distribution. To use this on our bootstrap example from above, we simply add and subtract 1.645 standard deviations from the mean.

```
# Calculate CI from z*
mySampleProp + zCrit*formSD
```

`## [1] 0.2519995 0.4980005`

Which we can see is exactly the same as if we calculate it directly.

```
qnorm(c(0.05,0.95),
mean = mySampleProp,
sd = formSD)
```

`## [1] 0.2519995 0.4980005`

Mathematically we can see why this works because the z score of a value can be calculated as:

\[ z_i = \frac{x_i - \bar{x}}{\sigma_x} \]

Where x is a numeric variable, `i`

represents the index (so \(x_i\) is the i^{th} value of x), \(\bar{x}\) is the mean of x, and \(\sigma_x\) is the standard deviation of x.

This means that we can solve for x_{i} to get:

\[ x_i = \bar{x} + (z_i \times \sigma_x) \]

Finally, we can check on our 95% rule this way. What are the z* values for a 95% CI?

```
# Find z* values for 95% CI
qnorm( c(0.025,0.975))
```

`## [1] -1.959964 1.959964`

These values are close, but not *exactly* the value of 2 we were using. So, 95% CI’s calculated directly will be more accurate than our rule of thumb of 2 times the standard error. However, that rule is still a useful first approximation.

It may seem that this is an odd thing to have to do when we can already calculate CI’s directly. However, the z* values play a crucial role in some calculations that we will see soon, and a closely related form is required when we start working with means. In addition, the historical use of these approaches ensures that you will encounter some mention of critical values in other statistics (either resources or publications you read).

Calculate the z-critical (z*) values for a 99% CI. Use them to calculate a 99% CI of the intro statistics students that use the word “Pop”. Compare it to your previous try it our to confirm that it worked.