12 Inference II

Confidence intervals are a powerful tool for making sense of data and drawing reliable conclusions about a population. Whether you’re estimating a population mean or proportion, confidence intervals provide a range of values that give you a sense of certainty—backed by statistical rigor—about where the true parameter lies. By mastering this concept, you’ll gain the ability to quantify uncertainty, assess the reliability of your estimates, and make informed decisions based on data. The information here equips you with the foundational formulas, key concepts like confidence levels and significance, and practical R functions to confidently apply these techniques in real-world scenarios.

12.1 Confidence Intervals

A confidence interval provides a range of values that, with a certain level of confidence, contains the population parameter of interest. For proper confidence intervals ensure that the sampling distributions are normal.

A \(95\)% confidence level, indicates that if the interval were constructed many times (from independent samples of the population), it would include the true population parameter \(95\)% of the time.

A significance level (\(\alpha\)) of \(5\)%, means that the confidence interval would would not include the true population parameter \(5\)% of the time.

The interval for the population mean when the population standard deviation is unknown is given by \(\bar x \pm t_{\alpha/2, df} \frac {s}{\sqrt{n}}\), where \(\bar x\) is the point estimate, \(t_{a/2, df} \frac {s}{\sqrt{n}}\) is the margin of error, \(\alpha\) is the allowed probability that the interval does not include \(\mu\), and \(df\) are the degrees of freedom \(n-1\).

The interval for the population proportion mean is given by \(\bar p \pm z_{\alpha/2} \sqrt{\frac {\bar p (1-\bar p)}{n}}\).

12.2 Useful R Functions

The qnorm() and qt() functions calculate quartiles for the normal and \(t\) distributions, respectively.

The if() function creates a conditional statement in R.

12.3 Exercises

The following exercises will help you test your knowledge on Statistical Inference. In particular, the exercises work on:

Simulating confidence intervals.
Estimating confidence intervals in R.
Estimating confidence intervals for proportions.

Try not to peek at the answers until you have formulated your own answer and double-checked your work for any mistakes.

Exercise 1

In this exercise you will be simulating confidence intervals.

Set the seed to \(9\). Create a random sample of 1000 data points and store it in an object called Population. Use the exponential distribution with rate of \(0.02\) to generate the data. Calculate the mean and standard deviation of Population and call them PopMean and PopSD respectively. What are the mean and standard deviation of Population?
Answer

The mean of Population is 48.61. The standard deviation is 47.94.

Start by generating values from the exponential distribution. You can use the rexp() function in R to do this. Setting the seed to 9 yields:
```
set.seed(9)
Population <- rexp(1000, 0.02)
```
The population mean is:
```
(PopMean <- mean(Population))
```
```
[1] 48.61053
```
The standard deviation is:
```
(PopSD <- sd(Population))
```
```
[1] 47.94411
```
Create a for loop (with 10,000 iterations) that takes a sample of 50 points from Population, calculates the mean, and then stores the result in a vector called SampleMeans. What is the mean of the SampleMeans?
Answer

The mean is very close to the population mean 48.83. The standard deviation is 6.83.

In R you can use a for loop to create the vector of sample means.
```
nrep <- 10000
SampleMeans <- c()
for (i in 1:nrep){
  x <- sample(Population, 50, replace = T)
  SampleMeans <- c(SampleMeans, mean(x))
}
```
The mean of SampleMeans is:
```
(xbar <- mean(SampleMeans))
```
```
[1] 48.7005
```
The standard deviation is:
```
(Standard <- sd(SampleMeans))
```
```
[1] 6.827595
```
Create a \(90\)% confidence interval using the first data point in the SampleMeans vector. Does the confidence interval include PopMean?
Answer

The confidence interval is [47.71, 70.17]. Since the population mean is equal to 48.61, the confidence interval does include the population mean.

Let’s construct the upper and lower limits of the interval in R.
```
(ll <- SampleMeans[1] + qnorm(0.05) * Standard)
```
```
[1] 47.71385
```
```
(ul <- SampleMeans[1] - qnorm(0.05) * Standard)
```
```
[1] 70.17464
```
Now take the minimum of the SampleMeans vector. Create a new \(90\)% confidence interval. Does the interval include PopMean? Out of the \(10,000\) intervals that you could construct with the vector SampleMeans, how many would you expect to include PopMean?
Answer

The confidence interval is [14.86, 37.32]. This interval does not include the population mean of 48.61. Out of the 10,000 confidence intervals, one would expect about 9,000 to include the population mean.

Let’s find the confidence interval limits using R.
```
(Minll <- min(SampleMeans) + qnorm(0.05) * Standard)
```
```
[1] 14.85631
```
```
(Minul <- min(SampleMeans) - qnorm(0.05) * Standard)
```
```
[1] 37.31709
```
We can confirm in R that about 9,000 of the intervals include PopMean. Once more, let’s use a for loop to construct confidence intervals for each element in SampleMeans and check whether the PopMean is included. The count variable keeps track of how many intervals include the population mean.
```
count = 0

for (i in SampleMeans){
  (ll <- i + qnorm(0.05) * Standard)
  (ul <- i - qnorm(0.05) * Standard)
  if (PopMean <= ul & PopMean >= ll){
    count = count + 1
  }
}

count
```
```
[1] 8978
```

Exercise 2

A random sample of \(24\) observations is used to estimate the population mean. The sample mean is \(104.6\) and the standard deviation is \(28.8\). The population is normally distributed. Construct a \(90\)% and \(95\)% confidence interval for the population mean. How does the confidence level affect the size of the interval?
Answer

The 90% confidence interval is [94.52, 114.67] and the 95% confidence interval is [92.68, 116.76]. The larger the confidence level, the larger the interval.

Let’s construct the intervals using R. Since the population standard deviation is unknown we will use the t-distribution. The interval is constructed as ({x} t_{/2} ).
```
(ul90 <- 104.6 - qt(0.05, 23) * 28.8 / sqrt(24))
```
```
[1] 114.6755
```
```
(ll90 <- 104.6 + qt(0.05, 23) * 28.8 / sqrt(24))
```
```
[1] 94.52453
```
For the 95% confidence interval we adjust the significance level accordingly.
```
(ul95 <- 104.6 - qt(0.025, 23) * 28.8 / sqrt(24))
```
```
[1] 116.7612
```
```
(ll95 <- 104.6 + qt(0.025, 23) * 28.8 / sqrt(24))
```
```
[1] 92.43883
```
A random sample from a normally distributed population yields a mean of \(48.68\) and a standard deviation of \(33.64\). Compute a \(95\)% confidence interval assuming a) that the sample size is \(16\) and b) the sample size is \(25\). What happens to the confidence interval as the sample size increases?
Answer

The confidence interval for a sample size of 16 is [30.75, 66.61]. The confidence interval when the sample size is 25 is [34.79, 62.57]. As the sample size gets larger, the confidence interval gets narrower and more precise.

Let’s use R again to calculate the confidence interval. For a sample size of 16 the interval is:
```
(ul16 <- 48.68 - qt(0.025, 15) * 33.64 / sqrt(16))
```
```
[1] 66.60549
```
```
(ll16 <- 48.68 + qt(0.025, 15) * 33.64 / sqrt(16))
```
```
[1] 30.75451
```
Increasing the sample size to 25 yields:
```
(ul25 <- 48.68 - qt(0.025, 24) * 33.64 / sqrt(25))
```
```
[1] 62.56591
```
```
(ll25 <- 48.68 + qt(0.025, 24) * 33.64 / sqrt(25))
```
```
[1] 34.79409
```

Exercise 3

You will need the sleep data set for this problem. The data is built into R, and displays the effect of two sleep inducing drugs on students. Calculate a \(95\)% confidence interval for group 1 and for group 2. Which drug would you expect to be more effective at increasing sleeping times?

Answer

The 95% confidence interval for group 1 is [-0.53, 2.03]. Let’s first calculate the standard error for group 1.

(se1 <- sd(sleep$extra[sleep$group == 1]) / sqrt(length(sleep$extra[sleep$group == 1])))

[1] 0.5657345

We can now use the standard error to estimate the lower and upper limits of the confidence interval.

(ll1 <- mean(sleep$extra[sleep$group == 1]) + qt(0.025, 9) * se1)

[1] -0.5297804

(ul1 <- mean(sleep$extra[sleep$group == 1]) - qt(0.025, 9) * se1)

[1] 2.02978

The 95% confidence interval for group 2 is [0.90, 3.76].Let’s repeat the procedure for group 2. Start by finding the standard error.

(se2 <- sd(sleep$extra[sleep$group == 2]) / sqrt(length(sleep$extra[sleep$group == 2])))

[1] 0.6331666

Using the standard error we can complete the confidence interval.

(ll2 <- mean(sleep$extra[sleep$group == 2]) + qt(0.025, 9) * se2)

[1] 0.8976775

(ul2 <- mean(sleep$extra[sleep$group == 2]) - qt(0.025, 9) * se2)

[1] 3.762322

Drug 2 is more effective. Drug 2 does not include zero in the interval, and the interval is to the right of zero. It is unlikely that drug 2 has no effect on students’ sleeping time. Additionally, Drug 2’s mean increase in sleeping hours is 2.33 vs. 0.75 for drug 1.

Exercise 4

A random sample of \(100\) observations results in \(40\) successes. Construct a \(90\)% and \(95\)% confidence interval for the population proportion. Can we conclude at either confidence level that the population proportion differs from \(0.5\)?
Answer

The 90% and 95% confidence intervals are [0.319, 0.481], and [0.304, 0.496] respectively. Since they do not include 0.5, we can conclude that the population proportion is significantly different from 0.5.

We can create an object that stores the sample proportion and sample in R:
```
(p <- 0.4)
```
```
[1] 0.4
```
```
(n <- 100)
```
```
[1] 100
```
The 90% confidence interval is given by:
```
(Ex1ll90 <- p + qnorm(0.05) * sqrt(p * (1 - p) / 100))
```
```
[1] 0.319419
```
```
(Ex1ul90 <- p - qnorm(0.05) * sqrt(p * (1 - p) / 100))
```
```
[1] 0.480581
```
The 95% confidence interval is:
```
(Ex1ll95 <- p + qnorm(0.025) * sqrt(p * (1 - p) / 100))
```
```
[1] 0.3039818
```
```
(Ex1ul95 <- p - qnorm(0.025) * sqrt(p * (1 - p) / 100))
```
```
[1] 0.4960182
```
You will need the HairEyeColor data set for this problem. The data is built into R, and displays the distribution of hair and eye color for \(592\) statistics students. Construct a \(95\) confidence interval for the proportion of Hazel eye color students.
Answer

The 95% confidence interval is [0.128, 0.186].

The data can easily be viewed by calling HairEyeColor in R.
```
HairEyeColor
```
```
, , Sex = Male

       Eye
Hair    Brown Blue Hazel Green
  Black    32   11    10     3
  Brown    53   50    25    15
  Red      10   10     7     7
  Blond     3   30     5     8

, , Sex = Female

       Eye
Hair    Brown Blue Hazel Green
  Black    36    9     5     2
  Brown    66   34    29    14
  Red      16    7     7     7
  Blond     4   64     5     8
```
Note that there are three dimensions to this table (Hair, Eye, Sex). We can calculate the proportion of Hazel eye colored students with the following command that makes use of indexing:
```
(p <- (sum(HairEyeColor[, 3, 1]) + sum(HairEyeColor[, 3, 2])) / sum(HairEyeColor))
```
```
[1] 0.1570946
```
Now we can use this proportion to construct the intervals. Recall that for proportions the interval is calculated by ({p} z_{/2} ). The 95% confidence interval is given by:
```
(Ex2ll95 <- p + qnorm(0.025) * sqrt(p * (1 - p) / 592))
```
```
[1] 0.1277818
```
```
(Ex2ul95 <- p - qnorm(0.025) * sqrt(p * (1 - p) / 592))
```
```
[1] 0.1864074
```