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
Drag z so that the shaded area between −z and z under the standard normal curve is 0.7540. Find z to two decimals.
A = 0.7540
z =
2.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded bottom 80% is to the left of it. Find the cut-off x.
μ = 65 pointsσ = 10 points80%
x = points
3.Confidence intervalsMedium
A 99% confidence interval for a mean, based on 100 observations and a known σ, has a margin of error of 8.3. What is the margin of error E₂ if the same study uses 900 observations instead?
99%n₁ = 100E₁ = 8.3n₂ = 900
E₂ =
4.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 5 plots, mean yield 83 kg, standard deviation 6 kg. Plot group 2: 7 plots, mean 78.5 kg, standard deviation 3.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 90% confidence interval for μ₁ − μ₂.
8 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is −6.25 and the standard deviation of the differences is 9.3. Find the paired test statistic t for H₀: μd = 0.
n = 8d̄ = −6.25sd = 9.3
t =
6.Hypothesis tests & confidence intervalsHard
In a trial, 18 of 120 patients on treatment A recover, and 36 of 200 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 18n₁ = 120x₂ = 36n₂ = 200
z =
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 6Σx = 21Σy = 96Σx² = 91Σy² = 1648Σxy = 293
r =
8.Linear regression & correlationHard
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 8.
n = 5x = 8
ŷ =
9.Chi-square testsMedium
The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “day shift, no defect”.
row total = 70column total = 69N = 126
E =
10.Chi-square testsHard
The table shows observed counts. Test at α = 0.05 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 = 2403 × 2α = 0.05
χ² =
3Solutions
1.Normal distribution
Answer: z = 1.16
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 of z
Φ(z) = (1 + 0.7540)/2 = 0.8770
Read the z-table backwards
Find 0.8770 inside the table and read off z.
z = 1.16
Answer
z = 1.16
2.Normal distribution
Answer: x = 73.4 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 = 65 + 0.84·10 = 73.4 points
Answer
x = 73.4 points
3.Confidence intervals
Answer: E₂ = 2.767
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.
Ratio of margins
E₂/E₁ = (2.576/2.576)·√(100/900) = 0.3333
New margin
E₂ = 8.3·0.3333 = 2.767
Answer
E₂ = 2.767
4.Confidence intervals
Answer: lower limit = −0.4482 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 = (4·6² + 6·3.5²)/10 = 21.75
sp = 4.664
Margin of error
E = 1.812·4.664·√(1/5 + 1/7) = 4.948
Interval
4.5 ± 4.948 → [−0.4482, 9.448]
Answer
lower limit = −0.4482 kg
5.Hypothesis tests & confidence intervals
Answer: t = −1.901
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 of the mean difference
SE = 9.3/√8 = 3.288
Test statistic
t = −6.25/3.288 = −1.901
Answer
t = −1.901
With df = 7, the two-sided 5% critical value is 2.365; |t| is smaller, so H₀ is not rejected.
6.Hypothesis tests & confidence intervals
Answer: z = −0.6937
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.
Sample proportions
p̂₁ = 18/120 = 0.15
p̂₂ = 36/200 = 0.18
p̂ = 54/320 = 0.1688
Standard error under H₀
SE = √(0.1688·0.8313·(1/120 + 1/200)) = 0.04325
Test statistic
z = (0.15 − 0.18)/0.04325 = −0.6937
Answer
z = −0.6937
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
7.Linear regression & correlation
Answer: r = −0.9713
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.
Centred sums
Sxy = 293 − 21·96/6 = −43
Sxx = 91 − 21²/6 = 17.5
Syy = 1648 − 96²/6 = 112
Correlation coefficient
r = −43/√(17.5·112) = −0.9713
Answer
r = −0.9713
8.Linear regression & correlation
Answer: ŷ = 49.8
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 = 35/5 = 7
ȳ = 236/5 = 47.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 104
Regression line
b = 104/40 = 2.6
a = 47.2 − 2.6·7 = 29
ŷ = 29 + 2.6·x
Prediction
ŷ = 29 + 2.6·8 = 49.8
Answer
ŷ = 49.8
9.Chi-square tests
Answer: E = 38.33
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.
Expected count
E = 70·69/126 = 38.33
Answer
E = 38.33
10.Chi-square tests
Answer: χ² = 9.889
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.