Welcome to Student's \(t\)-Distribution
In your earlier statistics studies, you worked with the standard normal (\(z\)) distribution to test hypotheses about a population mean. But there was a catch: you assumed you already knew the true population variance (\(\sigma^2\)).
In real life, if you do not know the population mean \(\mu\), you almost certainly do not know the population variance \(\sigma^2\) either! When we estimate \(\sigma^2\) using a sample variance \(S^2\), our calculations introduce extra uncertainty. To account for this, we use Student's \(t\)-distribution.
Don't worry if this seems a bit abstract at first. This chapter will break down exactly when to use the \(t\)-distribution, how to read the CCEA statistical tables, and how to execute hypothesis tests and confidence intervals step by step.
---1. Key Conditions & Underlying Assumptions
Before doing any calculation with the \(t\)-distribution, you must check whether the conditions are satisfied. CCEA examiners frequently award marks specifically for stating these assumptions!
Core Conditions for Using the \(t\)-Distribution:
1. Parent Population is Normally Distributed: The population from which the sample is drawn must follow a normal distribution, \(X \sim \text{N}(\mu, \sigma^2)\).
2. Population Variance is Unknown: The true population variance \(\sigma^2\) is unknown, meaning we must estimate it using the sample data.
3. Sample Size: The sample size \(n\) is typically small, and we replace \(\sigma^2\) with the unbiased sample variance estimate \(S^2\).
Analogy to Keep in Mind: Think of the normal distribution as a rigid suspension bridge built with exact, known measurements (\(\sigma\)). When \(\sigma\) is unknown and estimated from a small sample (\(S\)), there is slight wobbliness. The \(t\)-distribution gives you wider tails (a safer cushion) to account for that extra wobbliness.
Key Takeaway: If \(\sigma^2\) is unknown and the underlying population is normal, always use the \(t\)-distribution, not the \(z\)-distribution.
---2. Degrees of Freedom (\(\nu\)) and the Unbiased Variance Estimator
What are Degrees of Freedom?
The shape of a \(t\)-distribution depends on a parameter called the degrees of freedom, written using the Greek letter \(\nu\) (nu) or simply \(df\).
For a single sample of size \(n\):
\(\nu = n - 1\)
Calculating the Unbiased Estimator of Population Variance (\(S^2\))
Because the sample mean \(\bar{X}\) is calculated from the sample itself, dividing by \(n\) underestimates the true variance. To make the estimator unbiased, we divide by \(n - 1\):
\(S^2 = \frac{1}{n-1}\sum_{i=1}^n (X_i - \bar{X})^2 = \frac{1}{n-1}\left(\sum X_i^2 - \frac{(\sum X_i)^2}{n}\right)\)
The sample standard deviation is simply \(S = \sqrt{S^2}\).
Memory Trick: "Lose one to find \(\nu\)" — whenever you have a single group of size \(n\), subtract \(1\) to find your degrees of freedom \(\nu = n - 1\).
---3. Single-Sample Hypothesis Testing for Population Mean \(\mu\)
When testing a claim about a population mean \(\mu\) from a single sample of size \(n\), follow these structured steps:
Step-by-Step Procedure:
Step 1: State Hypotheses
State the null hypothesis \(H_0\) and alternative hypothesis \(H_1\) clearly:
\(H_0: \mu = \mu_0\)
\(H_1: \mu > \mu_0\) (one-tailed upper), \(\mu < \mu_0\) (one-tailed lower), or \(\mu \neq \mu_0\) (two-tailed).
Step 2: State the Distribution & Degrees of Freedom
Under \(H_0\), the test statistic follows a \(t\)-distribution with \(\nu = n - 1\) degrees of freedom.
Step 3: Calculate the Test Statistic
\(t = \frac{\bar{X} - \mu_0}{S / \sqrt{n}}\)
where \(\bar{X}\) is the sample mean, \(\mu_0\) is the hypothesised mean under \(H_0\), \(S\) is the sample standard deviation, and \(n\) is the sample size.
Step 4: Find the Critical Value from the CCEA Formulae Booklet
Look up the upper-tail critical value \(t_{\nu}\) in the CCEA tables for degrees of freedom \(\nu = n - 1\):
- For a one-tailed test at significance level \(\alpha\), find the column for probability \(p = \alpha\).
- For a two-tailed test at significance level \(\alpha\), find the column for probability \(p = \frac{\alpha}{2}\).
Step 5: Compare and Conclude in Context
Compare your calculated \(t\)-value with the critical value \(t_{\text{crit}}\):
- If \(|t| > t_{\text{crit}}\), reject \(H_0\).
- If \(|t| \leqslant t_{\text{crit}}\), do not reject \(H_0\).
Always write a final conclusion in the context of the original question (e.g., "There is sufficient evidence at the 5% level to suggest that the mean weight of cereal boxes has decreased.").
4. Confidence Intervals for a Population Mean \(\mu\)
A confidence interval gives an estimated range of values likely to contain the unknown population mean \(\mu\).
The Formula
A symmetric \((1 - \alpha)\) confidence interval for \(\mu\) is given by:
\(\bar{x} \pm t_{\text{crit}} \times \frac{s}{\sqrt{n}}\)
where:
- \(\bar{x}\) is the sample mean.
- \(s\) is the unbiased sample standard deviation.
- \(n\) is the sample size.
- \(t_{\text{crit}}\) is the critical value from the \(t\)-distribution with \(\nu = n - 1\) degrees of freedom corresponding to an upper-tail area of \(\frac{\alpha}{2}\) (for example, for a 95% confidence interval, \(\alpha = 0.05\), so the upper-tail area is \(0.025\)).
Quick Review:
- For a 90% Confidence Interval: Look up \(p = 0.05\) with \(\nu = n - 1\).
- For a 95% Confidence Interval: Look up \(p = 0.025\) with \(\nu = n - 1\).
- For a 99% Confidence Interval: Look up \(p = 0.005\) with \(\nu = n - 1\).
5. Paired Samples / Paired Differences \(t\)-Test
Sometimes observations come in naturally matched pairs (e.g., Before and After measurements on the same individual, or twin studies). Because the two sets of observations are not independent, we analyse the differences between each pair.
Method for Paired Differences:
1. Compute the differences: For each pair \((X_i, Y_i)\), calculate \(d_i = X_i - Y_i\).
2. Check assumption: Assume the population of differences is normally distributed: \(D \sim \text{N}(\mu_d, \sigma_d^2)\).
3. Calculate sample mean and variance of differences:
\(\bar{d} = \frac{\sum d_i}{n}\)
\(S_d^2 = \frac{1}{n-1}\left(\sum d_i^2 - \frac{(\sum d_i)^2}{n}\right)\)
4. Set up hypotheses: Usually \(H_0: \mu_d = 0\) versus \(H_1: \mu_d \neq 0\), \(\mu_d > 0\), or \(\mu_d < 0\).
5. Compute the test statistic:
\(t = \frac{\bar{d} - \mu_{d_0}}{S_d / \sqrt{n}} \sim t_{\nu} \quad \text{with } \nu = n - 1\)
(where \(\mu_{d_0} = 0\) under the standard null hypothesis of no difference).
6. Step-by-Step Worked Example
Problem:
A machine is designed to fill bottles with an average of \(500\text{ ml}\) of olive oil. A quality manager takes a random sample of \(9\) bottles and measures their contents (in ml):
\(498, 502, 495, 497, 501, 496, 494, 499, 496\)
Assuming that the volume of oil in the bottles is normally distributed, test at the \(5\%\) significance level whether the machine is underfilling bottles.
Solution:
Step 1: State hypotheses
Let \(\mu\) be the true mean volume of olive oil per bottle.
\(H_0: \mu = 500\)
\(H_1: \mu < 500\) (one-tailed test)
Step 2: Summary statistics
\(n = 9\)
\(\sum x = 4478 \implies \bar{x} = \frac{4478}{9} \approx 497.556\text{ ml}\)
\(\sum x^2 = 2228112\)
\(s^2 = \frac{1}{9 - 1}\left(2228112 - \frac{(4478)^2}{9}\right) = \frac{1}{8}(2228112 - 2228053.778) = \frac{58.222}{8} \approx 7.2778\)
\(s = \sqrt{7.2778} \approx 2.698\)
Step 3: Calculate test statistic
\(t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{497.556 - 500}{2.698 / \sqrt{9}} = \frac{-2.444}{0.8993} \approx -2.72\)
Step 4: Critical value lookup
Degrees of freedom \(\nu = n - 1 = 9 - 1 = 8\).
For a one-tailed test at \(\alpha = 0.05\), the critical value from the CCEA table is \(t_{8}(0.05) = 1.860\).
Since this is a lower-tail test, the critical region is \(t \leqslant -1.860\).
Step 5: Decision and contextual conclusion
Since calculated \(t = -2.72 < -1.860\), our test statistic falls inside the critical region. We reject \(H_0\).
Conclusion: There is sufficient evidence at the \(5\%\) significance level to suggest that the machine is underfilling bottles.
7. Common Pitfalls & How to Avoid Them
1. Forgetting \(\nu = n - 1\):
Mistake: Looking up \(n\) instead of \(n - 1\) in the statistical table.
Fix: Write \(\nu = n - 1\) explicitly at the top of every \(t\)-test problem.
2. Dividing by \(n\) instead of \((n - 1)\) for Variance:
Mistake: Using the biased population formula \(\frac{1}{n}\sum(x - \bar{x})^2\).
Fix: Always use the sample variance formula with denominator \(n - 1\).
3. Forgetting the Normality Requirement:
Mistake: Stating only the numbers without stating that the population must follow a normal distribution.
Fix: Whenever an exam question asks "State any assumption necessary for this test", write: "The parent population is normally distributed."
4. Confusing One-Tailed and Two-Tailed Probabilities:
Mistake: Using the \(p = 0.05\) column for a 5% two-tailed test.
Fix: For a two-tailed test, halve the significance level: look up \(p = 0.025\).
5. Giving Incomplete Conclusions:
Mistake: Writing just "Reject \(H_0\)".
Fix: Always provide a non-assertive conclusion tied directly to the wording in the question (e.g., "There is evidence to suggest that..."), rounded to 3 significant figures where values are quoted.
8. Chapter Summary Checklist
- Condition: Use the \(t\)-distribution when parent population is normal and \(\sigma^2\) is unknown.
- Degrees of freedom: \(\nu = n - 1\).
- Sample variance: Divide by \(n - 1\) for the unbiased estimator \(S^2\).
- Hypothesis test statistic: \(t = \frac{\bar{X} - \mu_0}{S / \sqrt{n}}\).
- Confidence interval: \(\bar{x} \pm t_{\text{crit}} \times \frac{s}{\sqrt{n}}\).
- Paired differences: Treat \(d_i = X_i - Y_i\) as a single sample with \(\bar{d}\) and \(S_d\).