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 battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 452 and 540 h? Give P as a decimal.
μ = 500σ = 40x1 = 452x2 = 540
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 top 75% is to the right of it. Find the cut-off x.
μ = 100 pointsσ = 15 points75%
x = points
3.Confidence intervalsMedium
A sample of 30 items has mean 101.5 and standard deviation s = 2.5. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.
n = 30x̄ = 101.5s = 2.599%t(29) = 2.756
E =
4.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 7 plots, mean yield 17.5 kg, standard deviation 8 kg. Plot group 2: 7 plots, mean 14 kg, standard deviation 5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 99% confidence interval for μ₁ − μ₂.
A sample of 64 students has a mean score of 77.1 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 77.1s = 8n = 64
t =
6.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 50 against H₁: μ > 50 is done at α = 0.01 with a sample of 64; the standard deviation is known to be 4. In reality the mean is 51.15. Find the power 1 − β of the test in percent.
μ₀ = 50μ = 51.15σ = 4n = 64α = 0.01
1 − β = %
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.
n = 6
r =
8.Linear regression & correlationHard
The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (5.5, 58) is made. Find its residual e = y − ŷ.
n = 5(5.5, 58)
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
The table shows observed counts. Test at α = 0.01 whether the two variables are independent (no continuity correction). Compute the test statistic χ², then compare it with the critical value from the table.
N = 2203 × 2α = 0.01
χ² =
3Solutions
1.Normal distribution
Answer: P = 0.7263
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 = (452 − 500)/40 = −1.2
z₂ = (540 − 500)/40 = 1
Area between
P = Φ(1) − Φ(−1.2) = 0.8413 − 0.1151 = 0.7263
Answer
P = 0.7263
2.Normal distribution
Answer: x = 90 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) = 0.25
z from the table
z ≈ −0.67
Back to the original units
x = 100 − 0.67·15 = 90 points
Answer
x = 90 points
3.Confidence intervals
Answer: E = 1.258
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 = 29, t = 2.756
Margin of error
E = 2.756·2.5/√30 = 1.258
Answer
E = 1.258
4.Confidence intervals
Answer: E = 10.89 kg
Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.
Pooled standard deviation
sp² = (6·8² + 6·5²)/12 = 44.5
sp = 6.671
Margin of error
E = 3.055·6.671·√(1/7 + 1/7) = 10.89
Answer
E = 10.89 kg
5.Hypothesis tests & confidence intervals
Answer: t = 2.1
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).
Standard error
SE = 8/√64 = 8/8 = 1
Test statistic
t = (77.1 − 75)/1 = 2.1
Answer
t = 2.1
t0.025; df = 63 = 1.998
|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: 1 − β = 48.96 %
Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).
Critical sample mean
SE = 4/√64 = 0.5
x̄c = 50 + 2.326·0.5 = 51.16
z under the true mean
z = (51.16 − 51.15)/0.5 = 0.026
Power
power = P(Z > 0.026) = 48.96%
1 − β = 48.96%
Answer
1 − β = 48.96%
7.Linear regression & correlation
Answer: r = 0.9108
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 = 120/6 = 20
ȳ = 406/6 = 67.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 174
Syy = Σ(y − ȳ)² = 521.3
Correlation coefficient
r = 174/√(70·521.3) = 0.9108
Answer
r = 0.9108
8.Linear regression & correlation
Answer: e = −4.65
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 = 25/5 = 5
ȳ = 305/5 = 61
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 33
Regression line
b = 33/10 = 3.3
a = 61 − 3.3·5 = 44.5
ŷ = 44.5 + 3.3·x
Predicted value
ŷ = 44.5 + 3.3·5.5 = 62.65
Residual
e = 58 − 62.65 = −4.65
Answer
e = −4.65
The point lies below the line.
9.Chi-square tests
Answer: χ² = 6.599
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: χ² = 15.75
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.