Why does sample standard deviation divide by n − 1 instead of n? — To correct for underestimation bias when estimating population σ from a sample.Dividing by n − 1 (Bessel's correction) corrects for the tendency of sample variance to underestimate population variance.
Data: {1, 1, 1, 9, 9, 9}. Mean = 5. Which value is σ? — √16 = 4.Deviations²: each 1 gives (1−5)²=16, each 9 gives (9−5)²=16. All six have deviation² = 16. Variance = 96/6 = 16. σ = √16 = 4.
If every data value increases by 10, what happens to σ? — σ stays the same.Shifting all data by a constant shifts the mean by the same amount, but deviations from the mean are unchanged. σ is unaffected.
If every data value is multiplied by 3, what happens to σ? — σ triples (multiplies by 3).Multiplying all data by k multiplies σ by |k|. So σ_new = 3σ.
Why do we square the deviations when computing variance? — To make all deviations positive and weight large deviations more heavily.Squaring ensures deviations are positive (so they don't cancel out) and gives extra weight to large deviations, penalizing outliers.
About what percent of data falls MORE THAN 2 standard deviations from the mean? — 5%.95% falls within μ ± 2σ. So 100% − 95% = 5% falls outside (more than 2σ from the mean).
About what percent of data falls MORE THAN 1 standard deviation from the mean? — 32%.68% falls within μ ± σ. So 100% − 68% = 32% falls outside (more than 1σ from the mean).
If z = 2.5 for a data value, how unusual is it? — Rare — more than 2σ from the mean (top ~1% of data).z = 2.5 means 2.5 standard deviations above the mean. Only about 0.6% of data has z > 2.5 — quite rare.
Using the Empirical Rule: SAT μ = 1000, σ = 200. About what percent score above 1200? — 2.5%.1200 is μ + σ. 84% score below 1200 (50% + 34%). 100% − 84% = 16% score above 1200. Wait: 1200 = μ + σ. 68% within ±σ: 34% between mean and 1200. So 50% + 34% = 84% below 1200. 16% above. The answer should be 16%, not 2.5%.
Why is the normal distribution important in statistics? — Many natural phenomena follow it, and many statistical methods assume it.The normal distribution appears in many natural phenomena (heights, test scores, errors) and underlies many statistical inference procedures via the Central Limit Theorem.
Using z = (x − μ)/σ, solve for x given z = −1.5, μ = 60, σ = 8. — 48.x = μ + zσ = 60 + (−1.5)(8) = 60 − 12 = 48.
Correlation does NOT imply causation means: — High |r| shows a linear relationship but does not prove causation.Even a perfect r = 1 doesn't mean x causes y. Both could be driven by a third (confounding) variable.
Two variables have r = 0.98. What can you conclude? — There is a very strong LINEAR relationship (but not necessarily causal).|r| = 0.98 is very close to 1 — a very strong linear relationship. But it tells us nothing about causation.
The coefficient of determination r² represents: — The proportion of variation in y explained by the linear model.r² is the proportion of variation in the response variable (y) that is explained by the linear regression model.
If r = 0.9, what is r²? — 0.81.r² = (0.9)² = 0.81. This means 81% of the variation in y is explained by the regression model.
r² = 0.64 means: — 64% of the variation in y is explained by the linear model.r² = 0.64 → 64% of the variance in y is accounted for by the linear regression. r = ±0.8.
The regression line always passes through which point? — (x̄, ȳ) — the means of x and y.The least-squares regression line ALWAYS passes through the point (x̄, ȳ) — the means of both variables.
A linear regression is applied to non-linear data (U-shaped). The residual plot will show: — A curved pattern (indicating poor fit).Fitting a line to curved data produces a curved residual pattern — indicating the linear model is inappropriate.
Can a negative r² exist? — No — r² is always between 0 and 1 for least-squares regression.For the least-squares line, r² = r² ≥ 0 always. It represents a proportion, so it ranges from 0 (no fit) to 1 (perfect fit).