12  Confidence Intervals in Practice

Author

Dr. Devan Becker

Published

2023-06-27

12.1 Recap

Silly confidence intervals

If \(X\sim N(\mu,\sigma)\), where \(\sigma\) is known, then a \((1-\alpha)\)CI for \(\mu\) based on \(\bar x\) is: \[ \bar x \pm z^*\frac{\sigma}{\sqrt{n}} \] where \(z^*\) is found such that \(P(Z < -z^*) = \alpha/2\),

  • or we could have found \(z^*\) such that \(P(Z > z^*) = \alpha/2\),
  • or \(P(Z < z^*) = 1 - \alpha/2\),
  • or \(P(Z > -z^*) = 1 - \alpha/2\).

A natural question is: why not use \(s\), the sample standard deviation?

To demonstrate why we can’t just use \(s\), I have set up a simulation. I like simulations.

You can safely skip the simulations if you’re the type who wants to just memorize a fact and will be sure to perfectly remember it later on. The upshot is this: since we’re estimating the standard deviation, the normal distribution doesn’t apply. Instead we use the \(t\) distribution whenever we use \(s\).

Simulation Setup

  1. Take random values from the standard normal distribution.
  2. Calculate the mean and sd.
  3. Calculate the 95% confidence interval with \(\sigma\) and with \(s\), both using a \(z\) value.
  4. Record whether the population mean is in the interval.
  5. Count how many intervals contain the population mean.
    • Should be 95% of them!

Before we begin, I want to show some R code for finding confidence intervals. If you’re given that \(\bar x = 7.28\), \(n=15\), \(\sigma = 1.24\), and you want to calculate a 95% CI:1

z_star <- abs(qnorm(0.05/2))
lower_bound <- 7.28 - z_star*1.24/sqrt(15)
upper_bound <- 7.28 + z_star*1.24/sqrt(15)
c(lower_bound, upper_bound)
[1] 6.652485 7.907515

Alternatively, we can use c(-1, 1) to stand in for “\(\pm\)”. The code is a little weird to get your head around, but trust me - it works!

7.28 + c(-1, 1)*z_star*1.24/sqrt(15)
[1] 6.652485 7.907515

Suppose that, unbeknownst to us, the true population mean was 7. To check if this is in our calculated confidence interval, we have to check that it’s larger than the lower bound AND less than the upper bound:

7 > 7.28 - z_star*1.24/sqrt(15) 
[1] TRUE
7 < 7.28 + z_star*1.24/sqrt(15) 
[1] TRUE

This can be combined into code as follows:

(7 > 7.28 - z_star*1.24/sqrt(15)) & (7 < 7.28 + z_star*1.24/sqrt(15))
[1] TRUE

This is enough to set up the simulation. Basically, we’re going to generate a random data set from a known population, then check if the confidence interval contains the true mean. We’ll do this thousands of times, and check which proportion contain the true mean. We’re hoping it’s 95%!

Simulation Code

## Set up empty vectors, to be filled with TRUE or FALSE
## if the population mean is in the interval
sigma_does <- c() # CI based on sigma does contain mu
s_does <- c() # CI based on s does contain mu

pop_sd <- 1
pop_mean <- 0
n <- 15 # sample size

z_star <- abs(qnorm(0.05 / 2))

## You aren't expected to understand "for" loops, but
## you need to be able to find CIs
for (i in 1:100000) { # repeat this code a bunch of times
    new_sample <- rnorm(n = n, mean = pop_mean, sd = pop_sd)
    xbar <- mean(new_sample)
    samp_sd <- sd(new_sample)

    CI_sigma <- xbar + c(-1, 1) * z_star * pop_sd / sqrt(n)
    CI_s <- xbar + c(-1, 1) * z_star * samp_sd / sqrt(n)
    # Do they contain the population mean?
    # in other words, is the lower bound less than pop_mean
    # *and* is the upper bound larger than pop_mean?
    # (Not testable)
    sigma_does[i] <- (CI_sigma[1] < pop_mean) & (CI_sigma[2] > pop_mean)
    s_does[i] <- (CI_s[1] < pop_mean) & (CI_s[2] > pop_mean)
}

## The mean of a bunch of TRUEs and FALSEs is
## the proportion of TRUEs (TRUE == 1, FALSE == 0)
mean(sigma_does)
[1] 0.94887
mean(s_does)
[1] 0.92991

The CI based on \(s\) only contains \(\mu\) 93% of the time! This is a pretty big discrepancy. What happens when you increase the sample size, n?2

The reason for this discrepancy is shown in the next section:

12.2 The Variance has Variance

Recall that the Sampling distribution is all possible values of a statistic when sampling from a population. We’ve covered the sampling distribution for the sample mean: Every time you take a sample, you get a different mean. The distribution of these sample means is \(N(\mu,\sigma/\sqrt{n})\).

The same idea applies to the sample variance! Every time you take a sample, you get a different variance. The sampling distribution is not a normal distribution. In the next section, we’ll demonstrate this fact.

Simulation: sample statistics

I’m going to generate a bunch of samples from a \(N(0, 0.2)\) distribution. I’ll calculate the mean and variance from each distribution, then plot the histogram.

n <- 10
pop_mean <- 0
pop_sd <- 0.2
sample_means <- c()
sample_vars <- c()

for (i in 1:100000) {
    new_sample <- rnorm(n = n, mean = pop_mean, sd = pop_sd)
    sample_means[i] <- mean(new_sample)
    sample_vars[i] <- var(new_sample)
}

par(mfrow = c(1, 2))
hist(sample_means, breaks = 25, freq = FALSE,
    main = "Sampling Dist of Sample Means")
curve(dnorm(x, pop_mean, pop_sd / sqrt(n)), add = TRUE,
    col = 4, lwd = 2)
## (n-1)s^2/sigma^2 follows a chi-square distribution on
## n-1 degrees of freedom. If you understand this, you are
## far too qualified to be taking this course. This fact
## is outside the scope of the course.
hist(sample_vars * (n - 1) / (pop_sd^2), breaks = 25, freq = FALSE,
    main = "Sampling Dist of Sample Vars")
curve(dchisq(x, n - 1), add = TRUE, col = 2, lwd = 2)

As you can tell from the fact that I knew how to draw the correct curve on the plots, the sampling distributions for the mean and variance are well known. Also, the sampling distribution for the variance is skewed, and therefore cannot be normal!

When we use \(\bar x+ z^*s/\sqrt{n}\), \(\bar x\) has variance, but so does \(s\).3 This is why the CI changes. When we know \(\sigma\), the Margin of Error (MoE) is always the same. When the standard deviation changes for each sample, so does the MoE.

Simulation: The Distribution of the Margin of Error

The sampling distribution of the Margin of Error is interesting to look at. This section is entirely optional - you just need to know that each sample has a different margin of error.

n <- 10
pop_mean <- 0
pop_sd <- 0.2
sample_MoEs <- c()
z_star <- abs(qnorm(0.5/2))

for(i in 1:100000){
    new_sample <- rnorm(n=n, mean=pop_mean, sd=pop_sd)
    sample_MoEs[i] <- z_star*sd(new_sample)/sqrt(n)
}

hist(sample_MoEs, breaks = 25,
    main = "Sampling Dist of MoE")
abline(v = z_star*pop_sd/sqrt(n), col = 6, lwd = 2)

The vertical purple line is \(z^*\sigma/\sqrt n\).4 This is just a re-scaling of the sampling distribution of the sample variance, so it’s also skewed! Furthermore, the average MoE using \(s\) is smaller than the MoE using \(\sigma\), even though it’s right-skewed:

c("MoE (sigma)" = z_star*pop_sd/sqrt(n),
    "Average MoE (s)" = mean(sample_MoEs))
    MoE (sigma) Average MoE (s) 
     0.04265848      0.04148352 
c("MoE (sigma)" = z_star*pop_sd/sqrt(n),
    "Median MoE (s)" = median(sample_MoEs))
   MoE (sigma) Median MoE (s) 
    0.04265848     0.04104300

This is why the CI using \(s\) doesn’t capture the true mean as often - it’s giving us smaller intervals!

12.3 Removing the Silliness

The distribution of the sample variance is not important.5 Instead, we care about the confidence intervals.

I’m going to write this yet again: since \(\bar X\sim N(\mu,\sigma/\sqrt{n})\)), \[ \frac{\bar X - \mu}{\sigma/\sqrt{n}} \sim N(0, 1) \] That is, you take the sample means, subtract the mean of the means, and divide by the standard error6, and you get a standard normal distribution.7

On the other hand, if we use \(s\) (which has it’s own variance), \[ \frac{\bar X - \mu}{s/\sqrt{n}} \sim t_{n-1} \] where \(n-1\) is the degrees of freedom (or df).8 This is called the \(t\) distribution, and is a lot like the normal distribution but it has higher variance.

Before we move on, notice how the formula with \(\sigma\) results in N(0,1), which does not require any information for our sample. In the \(t\) distribution, we need to know the sample size!

The t distribution

There are two main features of the \(t\) distribution that I want you to know:

  • It’s centered at 0, just like N(0,1).
  • It’s more variable than the normal distribution.

The second point is demonstrated in the following plot:

The red line corresponds to a sample size of 2.9 As the colours move through red to blue, we increase the sample size. At \(df = \infty\), the \(t\) distribution is exactly the same as the N(0,1) distribution. For anything smaller, the \(t\) distribution puts more probability in the tails.

This shows up in the critical values:

abs(qnorm(0.05/2)) # z^*
[1] 1.959964
abs(qt(0.05/2, df = 15 - 1)) # t^* n = 15
[1] 2.144787
abs(qt(0.05/2, df = 30 - 1)) # n = 30
[1] 2.04523
abs(qt(0.05/2, df = 50 - 1)) # n = 50
[1] 2.009575

Note that, just like how qbinom finds the value such of a binomial distribution such that 0.025% of the distribution is to the left and qnorm finds the z-values such that 0.025 is to the left, qt10 finds the t-value.

n_seq <- seq(2, 100, by = 2)
t_seq <- abs(qt(0.05/2, df = n_seq-1))
plot(n_seq, t_seq, type = "b",
    ylab = "abs(qt(0.05/2, df = n - 1))",
    xlab = "n",
    # the code for the title is not important.
    main = bquote("As df -> infinity, t"^"*"*" -> z"^"*"))
abline(h = abs(qnorm(0.05/2)), col = 3, lwd = 2)
## this code just puts a label on the axis - not important
axis(2, abs(qnorm(0.05/2)), "z*", col = 3, font = 2, col.axis = 3)

Since there’s more probability in the tails, you have to go further out to find the point such that 0.025 of the distribution is to the left.11 The \(t\) distribution allows for more variance due to the variance of \(s\), and it does this by having larger critical values.

The \(t\)-distribution

The \(t\) distribution has higher variance than the Normal distribution due to the extra uncertainty in estimating \(s\).

12.4 The \(t\) Confidence Interval

Now that you understand the reasoning behind using wider confidence intervals, I can show you the formula/ \[ \bar x \pm t_{n-1}^*s/\sqrt{n} \]

where \(t^*_{n-1}\) comes from abs(qt(alpha/2, df = n-1)).12

This has the same interpretation as the Z CI: 95% of the intervals constructed this way will contain the true population mean. This does NOT mean that there’s a 95% chance that the interval contains the true mean.

What’s that? Of course, I can demonstrate by simulation! Thanks for asking! The following code is copied and pasted from above, only the critical value has been changed.

## Set up empty vectors, to be filled with TRUE or FALSE
## if the population mean is in the interval
sigma_does <- c() # CI based on sigma does contain mu
s_does <- c() # CI based on s does contain mu

pop_sd <- 1
pop_mean <- 0
n <- 15 # sample size

z_star <- abs(qnorm(0.05/2))
t_star <- abs(qt(0.05/2, n - 1)) # NEW

## You aren't expected to understand "for" loops, but
## you need to be able to find CIs
for(i in 1:100000){ # repeat this code a bunch of times
    new_sample <- rnorm(n = n, mean = pop_mean, sd = pop_sd)
    xbar <- mean(new_sample)
    samp_sd <- sd(new_sample)

    CI_sigma <- xbar + c(-1, 1)*z_star*pop_sd/sqrt(n)
    CI_s <- xbar + c(-1, 1)*t_star*samp_sd/sqrt(n) # NEW
    # Do they contain the population mean?
    # in other words, is the lower bound less than pop_mean
    # *and* is the upper bound larger than pop_mean?
    # (Not testable)
    sigma_does[i] <- (CI_sigma[1] < pop_mean) & (CI_sigma[2] > pop_mean)
    s_does[i] <- (CI_s[1] < pop_mean) & (CI_s[2] > pop_mean)
}

## The mean of a bunch of TRUEs and FALSEs is
## the proportion of TRUEs (TRUE == 1, FALSE == 0)
mean(sigma_does)
[1] 0.95069
mean(s_does)
[1] 0.95091

Now both of them contain the mean 95% of the time!13 The difference between them is that the t CI doesn’t have as much information as the Z CI - the Z CI knows what the population sd is, but the t CI doesn’t. This is kinda magical: using math, we can get the truth with fewer assumptions!

12.5 Examples

  1. \(\bar x = 0.4\), \(n = 100\), \(\sigma = 0.01\), find the 92%CI.
    • This is a bit of a trick: I gave you \(\sigma\)! This always refers to the population standard deviation, so that’s what it is here. The Z CI can be found with the R code:
0.4 + c(-1, 1)*abs(qnorm(0.08/2)) * 0.01/sqrt(100)
[1] 0.3982493 0.4017507
  1. \(\bar x = 0.4\), \(n = 100\), \(s = 0.01\), will a 92%CI be wider than or smaller than the CI from Example 1?
    • We use \(t\) to account for the extra variance we have when we estimate \(s\). More variance means wider tails! The CI will be wider!
0.4 + c(-1, 1)*abs(qt(0.08/2, df = 100-1)) * 0.01/sqrt(100)
[1] 0.3982312 0.4017688

It’s only slightly wider. The sample size is large enough that the variance in the estimate is small.14 Try this again with a smaller \(n\) and see what happens to the difference!

  1. If \(n=16\) and the 95%CI for \(\mu\) is (10, 15), what’s the variance?
    1. A general form of the CI is \(\bar x \pm t^* s/\sqrt{n}\).
      • \(\bar x\) is in the centre, so \(\bar x\) is 12.5
    2. The MoE is 2.5, so \(t^* s/\sqrt{n} = 2.5\).
      • \(t^*\) is qt(0.05/2, 16 - 1) = 2.131
    3. \(2.131s/\sqrt{16} = 2.5\), so \(s = 2.5\sqrt{16}/2.131 = 4.69\)
    4. The variance is \(4.69^2 = 21.9961\)

12.6 Real Example

The following heights were collected from Laurier’s athletics page.

# 
ft <- c(5, 5, 6, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5)
inch <- c(7, 10, 0, 6, 7, 10, 7, 8, 7, 10, 3, 10, 1, 4, 2, 5, 6, 6, 9, 8, 10, 11)

heights <- 30.48 * ft + 2.54 * inch
hist(heights)
abline(v = 162.3, col = "red")

mean(heights)
[1] 170.7573
length(heights)
[1] 22

Assuming that we know the population standard deviation \(\sigma = 7.1\), we can make a 89% CI:

\[ \bar x \pm z^*\frac{\sigma}{\sqrt{n}} \]

mean(heights) + qnorm((1 - 0.89) / 2) * 7.1 / sqrt(length(heights))
[1] 168.338
mean(heights) - qnorm((1 - 0.89) / 2) * 7.1 / sqrt(length(heights))
[1] 173.1765

But is it reasonable to say that the standard deviation of athletic women’s heights is the same as the standard deviation of the heights in the population of all the women?

The t CI

It’s probably silly that we think that the standard deviation for all Canadian women applies here.

With a different sample, we would have gotten a different sample standard deviation! We need to account for this extra source of variance while creating the CI - we do this with the t distribution.

mean(heights) + qt((1 - 0.89) / 2, df = length(heights) - 1) * sd(heights) / sqrt(length(heights))
[1] 168.0833
mean(heights) - qt((1 - 0.89) / 2, df = length(heights) - 1) * sd(heights) / sqrt(length(heights))
[1] 173.4313

We’ll rarely have to go through this calculation again in this class. Instead, R does the calculations for us and we do the hard work (interpretations).

t.test(heights, conf.level = 0.89)

    One Sample t-test

data:  heights
t = 106.57, df = 21, p-value < 2.2e-16
alternative hypothesis: true mean is not equal to 0
89 percent confidence interval:
 168.0833 173.4313
sample estimates:
mean of x 
 170.7573

Bonus - Webscraping

library(rvest)
library(dplyr)
library(ggplot2)

# Read in web pages
basketball <- read_html("https://laurierathletics.com/sports/womens-basketball/roster")
soccer <- read_html("https://laurierathletics.com/sports/wsoc/roster")
hockey <- read_html("https://laurierathletics.com/sports/whock/roster")

# Extract the stats tables
bstats <- html_table(basketball)[[3]]
sstats <- html_table(soccer)[[3]]
hstats <- html_table(hockey)[[3]]

# Combine into one data frame
bstats$Sport <- "Basketball"
hstats$Sport <- "Hockey"
sstats$Sport <- "Soccer"
# There are, like 5 things here that experienced R users might not know.
stats <- bind_rows(bstats, hstats, sstats) |>
    select(Number = No., Pos = Pos., Height = Ht.,
        Year = `Academic Year`, Major, Sport,
        EligYear = `Elig. Yr.`) |>
    tidyr::separate(Height, sep = "-",
        into = c("ft", "inch")) |>
    mutate(Height = as.numeric(ft) * 30.48 + as.numeric(inch) * 2.24) |>
    select(-ft, -inch)
head(stats)
# A tibble: 6 × 7
  Number Pos   Year  Major         Sport      EligYear Height
   <int> <chr> <chr> <chr>         <chr>         <int>  <dbl>
1     14 G     4th   Business      Basketball       NA   168.
2      9 F     2nd   Kin/Phys. Ed. Basketball       NA   175.
3     11 F     3rd   Poli Sci      Basketball       NA   183.
4      4 G     2nd   Philosophy    Basketball       NA   166.
5      5 G     3rd   Business      Basketball       NA   168.
6     13 G     2nd   Kin/Phys. Ed. Basketball       NA   175.

The code above gets all of the heights from the hockey, soccer, and basketball teams (thank you Laurier Athletecs for having well-structured wed pages!).

Let’s take a minute to explore these data:

ggplot(stats) +
    aes(x = Sport, y = Height) +
    geom_boxplot() +
    labs(title = "Heights by sport")

For now, let’s focus on the basketball team. The following code finds a 95%CI. It also tests the hypothesis that the true mean height is 0, so we’ll ignore that for now.

# Just the heights of basketball players
ballers <- stats$Height[stats$Sport == "Basketball"]
mean(ballers)
[1] 172.1771
sd(ballers)
[1] 4.641504
length(ballers)
[1] 14
ggplot() +
    aes(x = ballers) +
    geom_histogram(bins = 15, fill = "dodgerblue", colour = 1) +
    geom_vline(xintercept = 162.3, colour = "red", linewidth = 2)

In the histogram above, our data look somewhat normal. If this sample is representative of the population, then we might guess that the population15 looks somewhat normal. However, a sample size of 14 is probably too small to make reliable conclusions.

I put a red line at the height of 163.2, the height of women in the general population. It seems like 163.2 would be not be a reasonable value in this sample, but could it be a reasonable guess at the overall mean? Let’s use a CI and a t-test to find out:

t.test(ballers, conf.level = 0.95)

    One Sample t-test

data:  ballers
t = 138.8, df = 13, p-value < 2.2e-16
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
 169.4972 174.8571
sample estimates:
mean of x 
 172.1771

A 95% CI is (169.5, 174.9), which clearly does not contain 163.2. Since we’re 95% confident that the true population mean is in our CI, so 162.3 is unlikely to be the true mean.

To test the hypothesis that this is greater than the general population, we still do a t-test. We just have to tell R what kind of test we want to do. As always, we have to set the significance level. Since we’re not skeptical about the alternate hypothesis, let’s set it to 10%. This says that we don’t need particularly strong evidence before we reject the null hypothesis.

We are testing whether the basketball players are taller than the general population. In symbols:

\[ H_0: \mu = 163.2 \text{ versus } H_A: \mu > 163.2 \]

The R code to test this is below:

t.test(ballers, mu = 162.3, alternative = "greater")

    One Sample t-test

data:  ballers
t = 7.9623, df = 13, p-value = 1.177e-06
alternative hypothesis: true mean is greater than 162.3
95 percent confidence interval:
 169.9803      Inf
sample estimates:
mean of x 
 172.1771

Our p-value is 0.00000177. This is a very small p-value, which means we have strong evidence against the null. We reject the null hypothesis, and conclude that the average of female varisty basketball players is taller than the general population (an unsurprising result, but nice to have confirmation).

12.7 Summary

This lesson could have been two sentences: The sample standard deviation has variance, so each confidence interval based on \(s\) is slightly different. To account for this, we use the \(t\) distribution. Then again, when someone tells me their name at a party I immediately forget it. Hopefully this long-winded exploration helps you understand why these facts are true and how they’re relevant to the course.

Note that all of the best practices for inference still apply! We can still get smaller intervals by taking better samples with larger sample sizes, and we still have to be careful to never speak of a calculated confidence interval in terms of chance.

The \(t\) confidence interval is actually used in practice. We saw some code that calculates the interval for us in this lecture, and we’ll never have to use qt() again! (Except possibly to demonstrate knowledge on tests.)

12.8 Self-Study Questions

  1. Explain the confidence interval that we found in the basketball player example.
    • For feedback, try the following in ChatGPT, Google Gemini, or Bing Chat: “In a sample of basketball players, a 95%CI was calculated as (169.5, 174.9). My explanation is: !!!!!put your explanation here!!!!!. Please rate my explanation as if it’s an answer on an exam in an introductory statistics course.”
  2. Explain very clearly why we checked whether 162.3 was in our interval.
  3. Repeat the example of creating a CI for basketball players, except for hockey players. The code below creates the data for hockey players. If you’d prefer to use a calculator, I also provide the mean and sd (you’ll still need to calculate critical value from the \(t\) distribution).
pucks <- stats$Height[stats$Sport == "Hockey"]
mean(pucks)
[1] 165.2128
sd(pucks)
[1] 4.55038

A plot of the values is below to help with your interpretation:

  1. In 2014, Dr. S. Hooker published a paper claiming to find a link between vaccines and autism, but only for African American boys. In the paper, Dr. Hooker tested 35 p-values and found 2 were less than the significance level of 0.05.
    1. Using the binomial distribution, find the probability of at least 2 successes in 35 trials when the probability is 0.05.
    2. Interpret the probability in part (a) in terms of the vaccines study.
    3. The 35 trials came from not just testing for a link, but splitting up the data by race, gender, and vaccine timing so that there were more hypothesis tests. This is known as “p-hacking”. Explain why this will often lead to statistically significant results, even when the null hypothesis is true.
Q4 Solution 
1 - pbinom(1, size = 35, prob = 0.05)
[1] 0.5279735

This calculation assumes that all of the null hypotheses are true (as we should do when calculating p-values). Under this assumption, there’s about a 50% chance of getting 2 or more significant results, even if there’s nothing going on. This is vastly different than the “5% risk of rejecting a true null” that we get when setting our significance level.

Use this app to play around with p-hacking and see why it’s so dangerous!

12.9 Crowdsourced Questions

The following questions are added from the Winter 2024 section of ST231 at Wilfrid Laurier University. The students submitted questions for bonus marks, and I have included them here (with permission, and possibly minor modifications).

  1. Which of the following statements is false about degrees of freedom (df) for a one-sample t-test?
    1. df = n - 1, where n is the number of items in your sample.
    2. Degrees of freedom refer to how many values in your data can change without breaking any rules.
    3. The higher the degrees of freedom, the more the t-distribution resembles a normal distribution.
    4. Degrees of freedom always equal the number of items in your dataset.
Solution

D is false because degrees of freedom (df) are calculated based on the sample size minus the number of parameters estimated. For example, in a simple t-test, df = n - 1, where n is the sample size. Degrees of freedom are not simply equal to the number of items but reflect the number of values that are free to vary in the calculation of a statistic, taking into account any constraints such as parameters estimated.

  1. A survey was carried out to estimate the average daily screen time of university students. A 90% confidence interval for the mean daily screen time was found to be (3 hours, 5 hours). How should this confidence interval be interpreted?
    1. The true average daily screen time of all university students is between 3 hours and 5 hours.
    2. There is a 90% confidence that the average daily screen time for university students is within the 3 to 5 hours range.
    3. 90% of university students have a daily screen time between 3 and 5 hours.
    4. There is a 10% chance that the true average daily screen time for university students is not between 3 and 5 hours.
Solution

b is the correct answer - The 90% confidence interval means that if we were to replicate this survey multiple times, each time calculating a new 90% confidence interval for the mean daily screen time from different samples, about 90% of these intervals would be expected to contain the true mean daily screen time for the population. It demonstrates our level of confidence in the interval’s ability to capture the true population parameter, not the distribution of individual screen times.

  1. Suppose a random sample of 100 students is taken from a large university. The average number of hours these students spend studying per week is found to be 15 hours, with a standard deviation of 3 hours. Construct a 95% confidence interval for the mean number of hours all students at the university spend studying per week.
Solution

To construct a confidence interval for the population mean, we’ll use the formula:

- Confidence interval = Sample mean ± Margin of error

where the margin of error is calculated as:

- Margin of error = Z × standard deviation / sqrt(Sample size) ​

For a 95% confidence interval, the critical value Z is 1.96 (based on the standard normal distribution).

Plugging in the values:

Margin of error = 1.96 × 3 / sqrt(100) = 1.96 × 3/10 = 0.588

So the confidence interval is:

15 + 0.588 = 15.588

15 - 0.588 = 14.412

In R:

15 - qnorm((1 - 0.95) / 2) * 3 / sqrt(100)
[1] 15.58799
15 + qnorm((1 - 0.95) / 2) * 3 / sqrt(100)
[1] 14.41201
  1. A random sample of 50 students at Laurier have a mean GPA of 3.2, and a standard deviation of 0.5. determine a 95% confidence interval for the population mean GPA of all of the students attending Laurier.
Solution

CI = Sample Mean +/- Margin of Error

Margin of Error = Z * sd/sqrt(n)

Using r or a z table we find that the Z value for a 95% confidence interval is 1.96

Margin of Error = 1.96 * 0.5/sqrt(50)

Margin of Error = 0.138

Now we can Calculate the 95% confidence interval

(3.2-0.138, 3.2+0.138)

CI: (3.062, 3.338)

Therefore, the population mean will fall into the range from 3.062 to 3.338 with 95% confidence.

In R:

3.2 + qnorm(((1 - 0.95) / 2)) * 0.5 / sqrt(50)
[1] 3.06141
3.2 - qnorm(((1 - 0.95) / 2)) * 0.5 / sqrt(50)
[1] 3.33859

  1. You’ll need to do this sort of thing on a test/assignment.↩︎

  2. Re-run the code and try it!↩︎

  3. Both are random variables.↩︎

  4. Recall that this never changes since \(\sigma\) is fixed.↩︎

  5. And very complicated.↩︎

  6. the standard deviation of the sampling distribution↩︎

  7. The word “standard” shows up way too much. Statisticians are bad at naming things.↩︎

  8. This is another example of statisticians being bad at naming things.↩︎

  9. The degrees of freedom is \(n-1\).↩︎

  10. The person reading this is a cutie.↩︎

  11. Try this for other \(\alpha\) values and larger \(n\).↩︎

  12. Note: I’m not even going to bother writing out the \(P()\) notation for \(t^*_{n-1}\) because you’ll never use it. You’ll only ever need to find \(t^*_{n-1}\) in this course.↩︎

  13. This means it’s working!↩︎

  14. Recall: For both the sample mean and the sample proportion, the variance of the sampling distribution decreases as \(n\) increases.↩︎

  15. In this example, the population refers to the population of female varsity basketball players.↩︎