Lesson 19: Confidence Intervals II

What We Did: Lessons 17–18

The Central Limit Theorem (CLT): If \(X_1, X_2, \ldots, X_n\) are independent and identically distributed (iid) with mean \(\mu\) and standard deviation \(\sigma\), then for large \(n\):

\[\bar{X} \sim N\left(\mu, \frac{\sigma^2}{n}\right)\]

Standard Error of the Mean: \[\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}\]

When is \(n\) “large enough”?

  • If the population is normal: CLT works for any \(n\)
  • Otherwise: \(n \geq 30\) is the common rule of thumb

To compute probabilities:

  1. Identify \(\mu\) and \(\sigma\)
  2. Compute the standard error: \(SE = \sigma / \sqrt{n}\)
  3. Use pnorm() with the appropriate mean and SD (the SE)
NoteLesson 18: Confidence Intervals I

Confidence Interval for a Mean:

Formula When to Use
Large sample (\(n \geq 30\)) \(\bar{X} \pm z_{\alpha/2} \cdot \dfrac{\sigma}{\sqrt{n}}\) Random sample, independence, \(s \approx \sigma\)
Small sample (\(n < 30\)) \(\bar{X} \pm t_{\alpha/2, n-1} \cdot \dfrac{s}{\sqrt{n}}\) Random sample, independence

Key ideas:

  • Margin of error: \(E = z_{\alpha/2} \cdot \frac{s}{\sqrt{n}}\) (or \(t_{\alpha/2, n-1} \cdot \frac{s}{\sqrt{n}}\))
  • Higher confidence level → wider interval
  • Larger sample size → narrower interval
  • Interpretation: “We are C% confident that the interval captures the true mean \(\mu\).”

What We’re Doing: Lesson 19

Objectives

  • Interpret CIs correctly in context
  • Construct confidence intervals for proportions
  • Assess robustness of CI assumptions

Required Reading

Devore, Section 7.3

Break!

Let’s talk Artificial Intelligence and what it can do for you.

Last Lesson’s Takeaway: Confidence Intervals I

ImportantThe Big Picture from Lesson 18

We learned how to construct confidence intervals — both \(z\)-intervals and \(t\)-intervals. Today we focus on three things:

  1. Interpreting CIs correctly (what they mean and what they DON’T mean)
  2. Confidence intervals for proportions (\(\hat{p}\) instead of \(\bar{X}\))
  3. Robustness of CI assumptions

Quick Review: The \(t\)-Distribution

Last lesson we introduced the \(t\)-distribution. Let’s make sure we’re comfortable with it before moving on.

Why do we need the \(t\)-distribution?

When \(n\) is large (\(n \geq 30\)), \(s \approx \sigma\) and we can use the \(z\)-interval:

\[\bar{X} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}\]

But when \(n\) is small, \(s\) is a less reliable estimate of \(\sigma\) — this introduces extra uncertainty the \(z\)-interval doesn’t account for.

The \(t\)-distribution has heavier tails to account for that extra uncertainty:

\[T = \frac{\bar{X} - \mu}{s / \sqrt{n}} \sim t_{n-1}\]

Key properties:

  • Symmetric and bell-shaped, centered at 0
  • Heavier tails than the normal → wider intervals for small samples
  • Degrees of freedom: \(\nu = n - 1\)
  • As \(df \to \infty\), \(t \to N(0, 1)\) — which is why \(z\) works for large \(n\)
# Compare critical values: t* gets closer to z* as df increases
qt(0.975, df = 4)    # n = 5
[1] 2.776445
qt(0.975, df = 10)   # n = 11
[1] 2.228139
qt(0.975, df = 30)   # n = 31
[1] 2.042272
qnorm(0.975)          # z* (the limit)
[1] 1.959964

Example: Cadet Push-Up Scores

A random sample of \(n = 6\) cadets records the following push-up counts on the ACFT:

\[62, \; 55, \; 71, \; 48, \; 59, \; 65\]

Construct a 95% confidence interval for the true mean push-up score.

\[\bar{x} = \frac{62 + 55 + 71 + 48 + 59 + 65}{6} = \frac{360}{6} = 60\]

\[s = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n-1}} = \sqrt{\frac{(62-60)^2 + (55-60)^2 + (71-60)^2 + (48-60)^2 + (59-60)^2 + (65-60)^2}{5}} = 8\]

\[\bar{x} \pm t_{\alpha/2, \, n-1} \cdot \frac{s}{\sqrt{n}}\]

\[60 \pm t_{0.025, \, 5} \cdot \frac{8}{\sqrt{6}}\]

qt(0.025, df = 5)   # lower critical value
[1] -2.570582
qt(0.975, df = 5)   # upper critical value
[1] 2.570582
t_star <- qt(0.975, df = n - 1)
se <- s / sqrt(n)
margin_t <- t_star * se
c(lower = xbar - margin_t, upper = xbar + margin_t)
   lower    upper 
51.60451 68.39549

\[60 \pm 2.571 \cdot \frac{8}{\sqrt{6}} = 60 \pm 8.4 = \left(51.6, \; 68.4\right)\]

TipInterpretation

We are 95% confident that the true mean push-up score for all cadets is between 51.6 and 68.4.

What if we had used the \(z\)-distribution instead?

z_star <- qnorm(0.975)
margin_z <- z_star * se
c(lower = xbar - margin_z, upper = xbar + margin_z)
   lower    upper 
53.59878 66.40122

\[60 \pm 1.96 \cdot \frac{8}{\sqrt{6}} = 60 \pm 6.4 = \left(53.6, \; 66.4\right)\]

WarningThe z-interval is dangerously narrow here
Critical Value Lower Bound Upper Bound Interval Width
\(t\)-interval (\(df = 5\)) \(2.571\) \(51.6\) \(68.4\) \(16.79\)
\(z\)-interval \(1.96\) \(53.6\) \(66.4\) \(12.8\)

The \(z\)-interval is narrower — but that’s not a good thing! With only 6 observations, \(s\) is an unreliable estimate of \(\sigma\). The \(z\)-interval ignores this extra uncertainty, making it overconfident. It claims 95% confidence but its actual coverage rate is less than 95%. The \(t\)-interval honestly accounts for our uncertainty.

Confidence Intervals for Proportions

We’ve now seen CIs for means in two settings: large samples (\(z\)-interval) and small samples (\(t\)-interval). Now we want to build a CI for a population proportion \(p\).

The good news? It’s the exact same idea. Every CI has the same structure:

\[\text{estimate} \pm \text{critical value} \times \text{standard error}\]

For means, the estimate was \(\bar{x}\) and the SE was \(\frac{s}{\sqrt{n}}\) (or \(\frac{\sigma}{\sqrt{n}}\)). For proportions, we just swap in the right estimate and SE:

  • Estimate: \(\hat{p} = \frac{X}{n}\) (sample proportion, where \(X\) = number of “successes”)
  • Standard error: \(\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\)

Why does this work? By the CLT, for large \(n\):

\[\hat{p} \sim N\left(p, \; \frac{p(1-p)}{n}\right)\]

Since \(n\) is large, we use the \(z\)-distribution (just like the large-sample mean CI).

ImportantConditions for the Proportion CI

The normal approximation is valid when both:

  • \(n\hat{p} \geq 10\) (at least 10 successes)
  • \(n(1-\hat{p}) \geq 10\) (at least 10 failures)

These are the success-failure conditions — they play the same role as “\(n \geq 30\)” for means.

The One-Proportion \(z\)-Interval

ImportantConfidence Interval for a Proportion

\[\hat{p} \pm z_{\alpha/2} \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\]

where:

  • \(\hat{p} = X/n\) = sample proportion
  • \(n\) = sample size
  • \(z_{\alpha/2}\) = critical value from the standard normal

Example: ACFT Pass Rate

A random sample of \(n = 200\) cadets shows that 172 passed the ACFT on the first attempt. Construct a 95% confidence interval for the true pass rate.

\[\hat{p} = \frac{172}{200} = 0.86\]

Check conditions: \(n\hat{p} = 200(0.86) = 172 \geq 10\) ✓ and \(n(1-\hat{p}) = 200(0.14) = 28 \geq 10\) ✓

\[\hat{p} \pm z_{\alpha/2} \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\]

\[0.86 \pm z_{0.025} \cdot \sqrt{\frac{0.86 \times 0.14}{200}}\]

qnorm(0.025)   # lower critical value
[1] -1.959964
qnorm(0.975)   # upper critical value
[1] 1.959964

\[0.86 \pm 1.96 \cdot \sqrt{\frac{0.86 \times 0.14}{200}} = 0.86 \pm 1.96 \times 0.0245 = 0.86 \pm 0.048\]

\[\left(0.812, \; 0.908\right)\]

TipInterpretation

We are 95% confident that the true proportion of cadets who pass the ACFT on the first attempt is between 81.2% and 90.8%.

Interpreting Confidence Intervals

What Does “95% Confident” Mean?

WarningCommon Misconceptions

WRONG: “There is a 95% probability that the true parameter is in this interval.”

WRONG: “95% of the data falls in this interval.”

WRONG: “If we sample again, there’s a 95% chance the new sample statistic falls in this interval.”

ImportantCorrect Interpretation

RIGHT: “We are 95% confident that the interval from LB to UB captures the true parameter.”

RIGHT: “If we repeated this sampling process many times, about 95% of the resulting intervals would contain the true parameter.”

The parameter (\(\mu\) or \(p\)) is fixed — it’s not random. The interval is random (because \(\bar{X}\) or \(\hat{p}\) is random). The confidence level describes the method’s long-run success rate, not the probability that any single interval contains the parameter.

Each horizontal line is a 95% CI from a new random sample of n = 45. Black intervals contain the true mean; red intervals miss it. About 95% of intervals capture μ.

Parameters vs. Statistics

Parameter (population) Statistic (sample)
Mean \(\mu\) \(\bar{x}\)
Standard deviation \(\sigma\) \(s\)
Proportion \(p\) \(\hat{p}\)

Parameters are fixed but unknown — statistics are what we compute from data to estimate them. Confidence intervals use sample statistics to build a plausible range for the parameter.

The Takeaway for Today

ImportantConfidence Interval Formulas
Formula Conditions Critical Value
Large sample mean (\(n \geq 30\)) \(\bar{X} \pm z_{\alpha/2} \cdot \dfrac{\sigma}{\sqrt{n}}\) Random sample, independence, \(s \approx \sigma\) \(z_{\alpha/2}\) from qnorm()
Small sample mean (\(n < 30\)) \(\bar{X} \pm t_{\alpha/2, n-1} \cdot \dfrac{s}{\sqrt{n}}\) Random sample, independence, population ~ Normal \(t_{\alpha/2, n-1}\) from qt()
Proportion \(\hat{p} \pm z_{\alpha/2} \cdot \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\) Random sample, independence, \(n\hat{p} \geq 10\), \(n(1-\hat{p}) \geq 10\) \(z_{\alpha/2}\) from qnorm()
ImportantInterpreting CIs

“We are C% confident that [interval] captures the true [parameter in context].”

The confidence level describes the method’s long-run success rate, not the probability that any single interval contains the parameter.

Board Problems

Problem 1: AFT Sprint-Drag-Carry

A random sample of \(n = 12\) cadets has the following Sprint-Drag-Carry times (in seconds):

\[105, \; 112, \; 98, \; 121, \; 115, \; 103, \; 110, \; 108, \; 99, \; 117, \; 106, \; 114\]

  1. What is the point estimate for the true mean Sprint-Drag-Carry time?

  2. Construct a 95% confidence interval for the true mean time.

  3. Now construct a 90% confidence interval. How does it compare to the 95% interval?

times <- c(105, 112, 98, 121, 115, 103, 110, 108, 99, 117, 106, 114)
n <- length(times)
xbar <- mean(times)
s <- sd(times)
c(xbar = xbar, s = round(s, 2), n = n)
  xbar      s      n 
109.00   7.15  12.00

  1. The point estimate is \(\bar{x} = 109\) seconds.

  2. Since \(n = 12 < 30\), use the \(t\)-interval with \(df = 11\):

\[109 \pm t_{0.025, \, 11} \cdot \frac{7.15}{\sqrt{12}}\]

t_star_95 <- qt(0.975, df = n - 1)
se <- s / sqrt(n)
margin_95 <- t_star_95 * se
c(lower = xbar - margin_95, upper = xbar + margin_95)
   lower    upper 
104.4585 113.5415

\[\left(104.46, \; 113.54\right)\]

We are 95% confident the true mean Sprint-Drag-Carry time is between 104.5 and 113.5 seconds.

  1. For 90% CI, \(t^*_{0.05, \, 11} = 1.796\):
t_star_90 <- qt(0.95, df = n - 1)
margin_90 <- t_star_90 * se
c(lower = xbar - margin_90, upper = xbar + margin_90)
   lower    upper 
105.2944 112.7056

\[\left(105.29, \; 112.71\right)\]

The 90% interval is narrower — less confidence means a smaller critical value and a tighter interval.

Problem 2: Weapon Qualification

In a random sample of \(n = 180\) Soldiers, 153 qualified as “marksman” or better on the M4 range.

  1. What is the point estimate for the true proportion of Soldiers who qualify?

  2. Construct a 95% confidence interval for the true qualification rate.

  3. Now construct a 99% confidence interval. How does it compare?

phat <- 153 / 180
n <- 180
  1. The point estimate is \(\hat{p} = \frac{153}{180} = 0.85\).

Check conditions: \(n\hat{p} = 153 \geq 10\) ✓ and \(n(1-\hat{p}) = 27 \geq 10\) ✓

  1. 95% CI:

\[0.85 \pm 1.96 \cdot \sqrt{\frac{0.85 \times 0.15}{180}}\]

se <- sqrt(phat * (1 - phat) / n)
z_95 <- qnorm(0.975)
margin_95 <- z_95 * se
c(lower = phat - margin_95, upper = phat + margin_95)
    lower     upper 
0.7978365 0.9021635

\[\left(0.798, \; 0.902\right)\]

We are 95% confident that the true qualification rate is between 79.8% and 90.2%.

  1. 99% CI (\(z^* = 2.576\)):
z_99 <- qnorm(0.995)
margin_99 <- z_99 * se
c(lower = phat - margin_99, upper = phat + margin_99)
    lower     upper 
0.7814455 0.9185545

\[\left(0.781, \; 0.919\right)\]

The 99% interval is wider — more confidence requires a larger critical value, casting a wider net.

Problem 3: Vehicle Maintenance

The motor pool tracks the time (in hours) to complete routine maintenance on tactical vehicles. Historical data suggests \(\sigma = 2.5\) hours. A random sample of \(n = 40\) maintenance jobs gives \(\bar{x} = 8.3\) hours.

  1. What is the point estimate for the true mean maintenance time?

  2. Construct a 95% confidence interval for the true mean maintenance time.

  3. Now construct a 90% confidence interval. How does it compare?

  1. The point estimate is \(\bar{x} = 8.3\) hours.

  2. Since \(n = 40 \geq 30\) and \(\sigma\) is known, use the \(z\)-interval:

\[8.3 \pm 1.96 \cdot \frac{2.5}{\sqrt{40}}\]

xbar <- 8.3; sigma <- 2.5; n <- 40
z_95 <- qnorm(0.975)
se <- sigma / sqrt(n)
margin_95 <- z_95 * se
c(lower = xbar - margin_95, upper = xbar + margin_95)
   lower    upper 
7.525256 9.074744

\[\left(7.53, \; 9.07\right)\]

We are 95% confident the true mean maintenance time is between 7.53 and 9.07 hours.

  1. For 90% CI (\(z^* = 1.645\)):
z_90 <- qnorm(0.95)
margin_90 <- z_90 * se
c(lower = xbar - margin_90, upper = xbar + margin_90)
   lower    upper 
7.649815 8.950185

\[\left(7.65, \; 8.95\right)\]

The 90% interval is narrower — lower confidence means a smaller \(z^*\) and a tighter interval.

Before You Leave

Today

  • Interpretation: “We are C% confident that [interval] captures [parameter in context]”
  • Proportion CI: \(\hat{p} \pm z_{\alpha/2} \sqrt{\hat{p}(1-\hat{p})/n}\) with success-failure conditions
  • Robustness: Large \(n\) is forgiving; small \(n\) needs normality; outliers are dangerous

Any questions?

Next Lesson

Lesson 20: Intro to Hypothesis Testing

  • State \(H_0\) and \(H_a\) for practical questions
  • Compute test statistics and \(p\)-values
  • Link significance, Type I error, and \(\alpha\)

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