Why does the t-distribution have heavier tails than the standard normal distribution?
AThe t-distribution uses a smaller sample size, so each observation has more influence
BReplacing σ with the sample standard deviation s introduces a second source of randomness in the denominator, making extreme values of the ratio more likely
CThe t-distribution is designed to be more conservative by artificially inflating the variance
DThe t-distribution uses degrees of freedom instead of a fixed variance, which spreads the distribution out
The z-score (X̄ − μ)/(σ/√n) has a random numerator and a fixed denominator (σ is known). The t-statistic (X̄ − μ)/(s/√n) has a random numerator AND a random denominator (s fluctuates across samples). When the denominator occasionally takes small values, the ratio can be very large. This extra variability in the denominator, which follows a chi-squared distribution scaled by σ², pushes probability mass into the tails. Option D gestures toward the right idea but 'degrees of freedom replacing fixed variance' is not quite the mechanism.
Question 2 Multiple Choice
A researcher tests a hypothesis about a population mean. The population is normally distributed, but σ is unknown and n = 15. Which test statistic is appropriate?
AZ, because the population is normally distributed
BZ, because n = 15 is large enough to invoke the central limit theorem
CT with 14 degrees of freedom, because σ must be estimated from the sample
DT with 15 degrees of freedom, because the sample size determines the degrees of freedom
The t-statistic is required whenever σ is unknown and must be estimated by s — regardless of whether the population is normal. The degrees of freedom equal n − 1 = 14, not n, because estimating s from the data uses up one degree of freedom (the sample mean is subtracted in the calculation of s). Option A is the common misconception: normality of the population allows exact t-inference, but doesn't let you use z when σ is unknown. Option B is wrong: the CLT approximation might justify z for large samples, but n = 15 is not large.
Question 3 True / False
A 95% confidence interval for a mean constructed with the t-distribution will always be wider than one constructed with the z-distribution using the same data.
TTrue
FFalse
Answer: True
T-critical values are always larger than the corresponding z-critical values (for example, t* ≈ 2.145 vs z* ≈ 1.96 for 95% confidence with 14 df). Since the interval half-width is critical value × standard error, a wider critical value produces a wider interval. This conservatism is appropriate: it reflects the added uncertainty from estimating σ. As degrees of freedom grow, t* → z*, and the intervals converge — reflecting that s becomes an increasingly reliable estimate of σ with larger samples.
Question 4 True / False
When the sample size is large enough, the t-distribution becomes indistinguishable from the standard normal, so there is no practical reason to use t over z for large samples.
TTrue
FFalse
Answer: False
While T(k) → N(0,1) as k → ∞, and the two are indeed very close for large k, there is a principled reason to prefer t: it is the exact correct distribution when σ is estimated, at any sample size. Using z when σ is unknown underclaims uncertainty, producing slightly anti-conservative (narrow) intervals and slightly liberal hypothesis tests. The t-distribution is not an approximation that happens to converge — it is the exact sampling distribution of the standardized mean when σ is replaced by s.
Question 5 Short Answer
Why does replacing the known population standard deviation σ with the sample standard deviation s require switching from the z-distribution to the t-distribution?
Think about your answer, then reveal below.
Model answer: When σ is known, the denominator σ/√n is a fixed constant, so the only randomness in the z-statistic comes from the numerator X̄. When σ is replaced by s, the denominator s/√n is itself a random variable that fluctuates across samples. This second source of randomness means the ratio no longer follows a standard normal distribution. Instead, it follows a t-distribution, which formally arises as the ratio of a standard normal to the square root of an independent chi-squared variable divided by its degrees of freedom.
The chi-squared connection comes from the fact that (n−1)s²/σ² ~ χ²(n−1). The extra randomness in the denominator inflates the tails relative to the normal. As n grows, s converges to σ and its own variability shrinks — the extra randomness in the denominator vanishes, which is why t → z as the degrees of freedom grow. The t-distribution is the honest acknowledgment that estimating σ carries a cost in inferential precision.