CHSE Odisha Elective Statistics common mistakes
CHSE Odisha Elective Statistics · Understand the traps and how to avoid them.
Computing the mean, median, or mode from a grouped frequency table by averaging the class mid-points without weighting by frequency, or by taking the mid-point of the modal class as the mode.
Candidates learn the ungrouped formulas first and then transfer them mechanically to grouped data, forgetting that each class represents many observations and that the modal class is only an interval, not a value. The distinction between 'class containing the mode' and 'the mode' collapses under exam pressure.
What to do instead
Always write the formula with the frequency column visible: mean = Σfx/Σf, median = L + [(N/2 − cf)/f]·h, mode = L + [(f1 − f0)/(2f1 − f0 − f2)]·h. Practise ten grouped tables and, for each, explicitly label L, cf, f, h, N before substituting. Check the mean lies inside the range and that median and mode fall in their stated classes.
Choosing between sample and population standard deviation (dividing by n versus n − 1) at random, and reporting a variance in the wrong units or with the wrong symbol.
Textbooks use both conventions, and the divisor n − 1 is often introduced late. Candidates memorise 'σ = √(Σ(x − x̄)²/n)' and apply it to sample data, then lose marks when the examiner expects s with n − 1. Unit confusion arises because variance is in squared units and candidates forget to square-root or to state units.
What to do instead
Adopt a rule and state it in the answer: 'treating this as a sample, I use s² = Σ(x − x̄)²/(n − 1)'. Show the deviation column and its sum, then the divisor explicitly. Always write units: variance in (kg)², standard deviation in kg. When combining two groups, use the pooled formula with n1 + n2 − 2 in the denominator and show the combined mean first.
Interpreting a correlation coefficient of 0.8 as meaning 'x causes 80% of the variation in y', or computing r from grouped or ranked data without applying the correct formula (e.g. using the product-moment formula on ranks instead of Spearman's).
Correlation and causation are conflated in everyday language, and r² (coefficient of determination) is often skipped. For ranked data, candidates see numbers and default to the Pearson formula, ignoring ties and the 1 − 6Σd²/[n(n² − 1)] shortcut, which is invalid when ranks are tied.
What to do instead
State the interpretation explicitly: r measures the strength and direction of linear association only; r² gives the proportion of variance in y explained by x. For ranks, use Spearman and apply the tie correction Σ(t³ − t)/12 before computing. Check the sign of r against the scatter description, and never claim causation from an observational correlation.
Fitting a regression line by eye or by solving normal equations with arithmetic slips, then reporting the regression of x on y when the question asked for y on x (or vice versa).
The two regression lines are easily swapped because both use the same five sums (Σx, Σy, Σx², Σy², Σxy). Candidates also forget that the line of y on x minimises vertical deviations and has slope byx = r·(sy/sx), not the reciprocal. Long hand calculations invite sign and transpose errors.
What to do instead
Label the dependent variable first: 'predicting y from x, so the model is y = a + bx with b = byx = [nΣxy − ΣxΣy]/[nΣx² − (Σx)²]'. Verify that bxy · byx = r² and that both lines pass through (x̄, ȳ). Substitute the means back as a check before finalising the intercept.
Adding probabilities of non-mutually-exclusive events, or treating dependent events as independent, when applying the addition and multiplication rules.
The words 'or' and 'and' are mapped directly to '+' and '×' without checking overlap or dependence. Candidates also forget to subtract P(A ∩ B) in the addition rule and misapply P(A|B) = P(A ∩ B)/P(B) by using P(A) in the numerator.
What to do instead
Before any calculation, write whether events are mutually exclusive, independent, or neither, and draw a two-circle Venn diagram with the overlap labelled. Apply P(A ∪ B) = P(A) + P(B) − P(A ∩ B) and P(A ∩ B) = P(A)·P(B|A). Test independence by checking P(A ∩ B) = P(A)·P(B) numerically.
Confusing the probability mass function with the cumulative distribution function, or substituting into a continuous density without integrating to get a probability.
Discrete and continuous treatments look similar in notation, and candidates plug a value into f(x) for a continuous variable and report it as P(X = x), which is zero. They also forget that a valid pmf/pdf must sum or integrate to 1 and that F(x) is non-decreasing.
What to do instead
State the type first: 'X is continuous, so P(X = x) = 0 and probabilities come from integrals of f'. For any proposed pmf, verify Σp = 1; for a pdf, verify ∫f = 1 over the support. Compute F(x) = P(X ≤ x) by summing or integrating up to x, then differentiate or difference back as a check.
Applying the binomial formula to sampling without replacement from a small finite population, or using the Poisson approximation when n is small and p is not small.
Candidates see 'number of successes in n trials' and default to the binomial, ignoring that trials must be independent with constant p. The Poisson approximation to the binomial is valid only for large n and small p (roughly n ≥ 20, p ≤ 0.05), and the hypergeometric is required for sampling without replacement.
What to do instead
Check the four binomial conditions explicitly: fixed n, two outcomes, constant p, independent trials. If sampling is without replacement from a finite population, use the hypergeometric. State the Poisson approximation conditions and the mean λ = np before using e^(−λ)λ^x/x!. Verify probabilities sum to 1 over the support.
Using the wrong standard error in a hypothesis test or confidence interval - e.g. using σ/√n when σ is unknown and n is small, or the unpooled formula for a difference of means when equal variances are assumed.
The z and t procedures look identical except for the standard error and critical value. Candidates memorise one template and apply it to every scenario, ignoring sample size and whether σ is known. Pooling is forgotten because the pooled variance formula is longer.
What to do instead
Write a decision checklist before computing: is σ known? is n large? are the two samples independent? Then choose z, one-sample t, or pooled two-sample t and write the standard error formula in full. Always state degrees of freedom (n − 1 or n1 + n2 − 2) and look up the critical value from the correct table.
In ANOVA, computing the F ratio with the wrong mean square in the numerator, or failing to state the null hypothesis and degrees of freedom when interpreting a significant F.
The ANOVA table has several sums of squares and mean squares that look alike, and candidates divide MSwithin by MSbetween instead of the reverse. A significant F is also reported as 'all means differ', which the test does not establish.
What to do instead
Build the table in fixed order (SSB, SSW, SST; df; MS = SS/df) and write F = MSB/MSW with df = (k − 1, N − k). State H0: all treatment means equal. If F is significant, follow with a post-hoc comparison (e.g. Tukey or LSD) before naming which pairs differ, and check the assumptions of normality, independence, and equal variances.
Computing a fixed-base index or a chain index with the base year changing silently, or fitting a trend line to time series data without checking the moving-average period against the seasonal pattern.
Index numbers require a single consistent base, and candidates switch base years mid-calculation or forget to convert a series to a common base. In time series, seasonal indices are meaningful only if the moving average length equals the seasonal period (e.g. 4 for quarterly data), a constraint often overlooked.
What to do instead
Write the base year at the top of every column and, when splicing series, convert to a common base by multiplying by the ratio of the two base values. For seasonal decomposition, match the moving-average order to the period, centre the moving average for even periods, and check that seasonal indices sum to the period length (e.g. 4 or 12).