E = row total × column total / grand total, df = (r − 1)(c − 1)
Critical values (α = 0.05)
df 1: 3.841, 2: 5.991, 3: 7.815, 4: 9.488
Decision
reject H0 if χ² > χ²crit
2Practice problems
1.Normal distributionMedium
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 247.75 and 251 g? Give P as a decimal.
μ = 250σ = 5x1 = 247.75x2 = 251
P =
2.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 80% is to the left of it. Find the cut-off x.
μ = 100 pointsσ = 15 points80%
x = points
3.Confidence intervalsMedium
A sample of 12 items has mean 15.5 and standard deviation s = 18. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.
n = 12x̄ = 15.5s = 1895%t(11) = 2.201
lower limit =
4.Confidence intervalsHard
The population standard deviation is 13. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 4?
σ = 13E = 499%
n =
5.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 250; the population standard deviation is known to be 15. A random sample of 100 observations has a mean of 246.4. Test H₀: μ = 250 against H₁: μ ≠ 250. Find the p-value p.
μ₀ = 250σ = 15n = 100x̄ = 246.4H₁: μ ≠ μ₀
p =
6.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 5. Find the smallest sample size n.
σ = 5E = 295%
n =
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.
n = 6
r =
8.Linear regression & correlationHard
The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (2.5, 56) is made. Find its residual e = y − ŷ.
n = 6(2.5, 56)
e =
9.Chi-square testsMedium
The contingency table shows observed counts. Compute the chi-square statistic χ² for the test of independence between the row and column variables. Do not use a continuity correction.
N = 1602 × 2
χ² =
10.Chi-square testsHard
A chi-square test of independence is carried out on a contingency table with 3 rows and 6 columns (not counting the totals). How many degrees of freedom df does the test have?
3 × 6
df =
3Solutions
1.Normal distribution
Answer: P = 0.2529
Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.
z-scores
z₁ = (247.75 − 250)/5 = −0.45
z₂ = (251 − 250)/5 = 0.2
Area between
P = Φ(0.2) − Φ(−0.45) = 0.5793 − 0.3264 = 0.2529
Answer
P = 0.2529
2.Normal distribution
Answer: x = 112.6 points
From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.
Area to the left
Φ(z) = 0.8
z from the table
z ≈ 0.84
Back to the original units
x = 100 + 0.84·15 = 112.6 points
Answer
x = 112.6 points
3.Confidence intervals
Answer: lower limit = 4.063
Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.
Critical value
df = n − 1 = 11, t = 2.201
Margin of error
E = 2.201·18/√12 = 11.44
Interval
15.5 ± 11.44 → [4.063, 26.94]
Answer
lower limit = 4.063
4.Confidence intervals
Answer: n = 71
Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.
Solve for n
n = (2.576·13/4)² = 70.09
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 71
Answer
n = 71
5.Hypothesis tests & confidence intervals
Answer: p = 0.0164
Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.
Test statistic
SE = σ/√n = 15/√100 = 1.5
z = (246.4 − 250)/1.5 = −2.4
Tail area
Φ(2.4) = 0.9918
1 − Φ(2.4) = 0.0082
p-value
p = 2·0.0082 = 0.0164
Answer
p = 0.0164
p ≤ 0.05, so H₀ is rejected at the 5% level.
6.Hypothesis tests & confidence intervals
Answer: n = 25
Testing a mean: the test statistic measures how many standard errors the sample mean lies from the value claimed in H₀: z = (x̄ − μ₀)/(σ/√n) when σ is known, t = (x̄ − μ₀)/(s/√n) when it is estimated from the sample. A confidence interval is x̄ ± (critical value)·(standard error).
Solve for n
n = (1.96·5/2)² = 24.01
Round up
n = 25
Answer
n = 25
7.Linear regression & correlation
Answer: r = 0.9995
Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.
Means
x̄ = 48/6 = 8
ȳ = 299/6 = 49.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 189
Syy = Σ(y − ȳ)² = 510.8
Correlation coefficient
r = 189/√(70·510.8) = 0.9995
Answer
r = 0.9995
8.Linear regression & correlation
Answer: e = −2.724
Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.
Means
x̄ = 27/6 = 4.5
ȳ = 436/6 = 72.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 122
Regression line
b = 122/17.5 = 6.971
a = 72.67 − 6.971·4.5 = 41.3
ŷ = 41.3 + 6.971·x
Predicted value
ŷ = 41.3 + 6.971·2.5 = 58.72
Residual
e = 56 − 58.72 = −2.724
Answer
e = −2.724
The point lies below the line.
9.Chi-square tests
Answer: χ² = 13.33
Chi-square tests: compare observed counts O with the counts E expected under H₀ using χ² = Σ(O − E)²/E. Goodness of fit: E = n·p for each category, df = k − 1 (minus one more for each parameter estimated from the data). Independence in an r × c table: E = row total × column total / grand total, df = (r − 1)(c − 1). Reject H₀ when χ² exceeds the critical value for the chosen α.
df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is rejected.
10.Chi-square tests
Answer: df = 10
Chi-square tests: compare observed counts O with the counts E expected under H₀ using χ² = Σ(O − E)²/E. Goodness of fit: E = n·p for each category, df = k − 1 (minus one more for each parameter estimated from the data). Independence in an r × c table: E = row total × column total / grand total, df = (r − 1)(c − 1). Reject H₀ when χ² exceeds the critical value for the chosen α.