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 0.4448. Find z to two decimals.
A = 0.4448
z =
2.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded bottom 60% is to the left of it. Find the cut-off x.
μ = 178 cmσ = 7 cm60%
x = cm
3.Confidence intervalsMedium
A 95% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 8.5. What is the margin of error E₂ if the same study uses 256 observations instead?
95%n₁ = 64E₁ = 8.5n₂ = 256
E₂ =
4.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 10 plots, mean yield 23 kg, standard deviation 3.5 kg. Plot group 2: 11 plots, mean 22 kg, standard deviation 1.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.
A sample of 100 students has a mean score of 119.4 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 119.4s = 4n = 100
t =
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 1? The standard deviation is 20. Find the smallest sample size n.
σ = 20E = 195%
n =
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
8.Linear regression & correlationHard
The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 3?
n = 5ŷ = 3
x =
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 = 1402 × 2
χ² =
10.Chi-square testsHard
A chi-square goodness-of-fit test checks whether counts in 8 categories follow a Poisson distribution whose mean is estimated from the same data. How many degrees of freedom df does the test have?
k = 8
df =
3Solutions
1.Normal distribution
Answer: z = 0.59
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 + 0.4448)/2 = 0.7224
Read the z-table backwards
Find 0.7224 inside the table and read off z.
z = 0.59
Answer
z = 0.59
2.Normal distribution
Answer: x = 179.8 cm
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) = 0.6
z from the table
z ≈ 0.25
Back to the original units
x = 178 + 0.25·7 = 179.8 cm
Answer
x = 179.8 cm
3.Confidence intervals
Answer: E₂ = 4.25
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 = (1.96/1.96)·√(64/256) = 0.5
New margin
E₂ = 8.5·0.5 = 4.25
Answer
E₂ = 4.25
4.Confidence intervals
Answer: E = 1.997 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 = (9·3.5² + 10·1.5²)/19 = 6.987
sp = 2.643
Margin of error
E = 1.729·2.643·√(1/10 + 1/11) = 1.997
Answer
E = 1.997 kg
5.Hypothesis tests & confidence intervals
Answer: t = −1.5
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 = 4/√100 = 4/10 = 0.4
Test statistic
t = (119.4 − 120)/0.4 = −1.5
Answer
t = −1.5
t0.025; df = 99 = 1.984
|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: n = 1537
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 = (1.96·20/1)² = 1537
Round up
n = 1537
Answer
n = 1537
7.Linear regression & correlation
Answer: r² = 0.9793
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 = 110/5 = 22
ȳ = 308/5 = 61.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 258
Syy = Σ(y − ȳ)² = 755.2
Coefficient of determination
r² = 258²/(90·755.2) = 0.9793
Answer
r² = 0.9793
About 97.93% of the variation in y is explained by the line.
8.Linear regression & correlation
Answer: x = 2.9
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 = 30/5 = 6
ȳ = 46/5 = 9.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 80
Regression line
b = 80/40 = 2
a = 9.2 − 2·6 = −2.8
ŷ = −2.8 + 2·x
Solve for x
x = (3 − (−2.8))/2 = 2.9
Answer
x = 2.9
9.Chi-square tests
Answer: χ² = 5
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 = 6
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.