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 480 and 574 h? Give P as a decimal.
μ = 500σ = 40x1 = 480x2 = 574
P =
2.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 95% is to the right of it. Find the cut-off x.
μ = 250 gσ = 5 g95%
x = g
3.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 80 phones, mean 141 h, known standard deviation 9 h. Model 2: 80 phones, mean 154 h, known standard deviation 7 h. Find the upper limit of the 90% confidence interval for μ₁ − μ₂.
The population standard deviation is 3. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 0.5?
σ = 3E = 0.590%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 16 students has a mean score of 508.7 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 508.7s = 12n = 16
t =
6.Hypothesis tests & confidence intervalsHard
In a trial, 135 of 300 patients on treatment A recover, and 37 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 135n₁ = 300x₂ = 37n₂ = 120
z =
7.Linear regression & correlationMedium
The table gives 5 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 = 5Σx = 15Σy = 93Σx² = 55Σy² = 1803Σxy = 254
r =
8.Linear regression & correlationHard
The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. A new observation (2.5, 17) is made. Find its residual e = y − ŷ.
n = 6(2.5, 17)
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 = 2002 × 3
χ² =
10.Chi-square testsHard
A chi-square goodness-of-fit test checks whether counts in 7 categories follow a Poisson distribution whose mean is estimated from the same data. Using the table, find the critical value χ²crit at significance level α = 0.01.
α = 0.01
χ²crit =
3Solutions
1.Normal distribution
Answer: P = 0.6593
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 = (480 − 500)/40 = −0.5
z₂ = (574 − 500)/40 = 1.85
Area between
P = Φ(1.85) − Φ(−0.5) = 0.9678 − 0.3085 = 0.6593
Answer
P = 0.6593
2.Normal distribution
Answer: x = 241.8 g
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) = 0.05
z from the table
z ≈ −1.64
Back to the original units
x = 250 − 1.64·5 = 241.8 g
Answer
x = 241.8 g
3.Confidence intervals
Answer: upper limit = −10.9 h
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.
Point estimate
x̄₁ − x̄₂ = 141 − 154 = −13
Standard error
SE = √(9²/80 + 7²/80) = 1.275
Margin of error
E = 1.645·1.275 = 2.097
Interval
−13 ± 2.097 → [−15.1, −10.9]
Answer
upper limit = −10.9 h
The interval does not contain 0, so the means differ significantly.
4.Confidence intervals
Answer: n = 98
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 = (1.645·3/0.5)² = 97.42
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 98
Answer
n = 98
5.Hypothesis tests & confidence intervals
Answer: t = 2.9
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 = 12/√16 = 12/4 = 3
Test statistic
t = (508.7 − 500)/3 = 2.9
Answer
t = 2.9
t0.025; df = 15 = 2.131
|t| > 2.131, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: z = 2.667
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 = 135/300 = 0.45
p̂₂ = 37/120 = 0.3083
p̂ = 172/420 = 0.4095
Standard error under H₀
SE = √(0.4095·0.5905·(1/300 + 1/120)) = 0.05311
Test statistic
z = (0.45 − 0.3083)/0.05311 = 2.667
Answer
z = 2.667
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
7.Linear regression & correlation
Answer: r = −0.924
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 = 254 − 15·93/5 = −25
Sxx = 55 − 15²/5 = 10
Syy = 1803 − 93²/5 = 73.2
Correlation coefficient
r = −25/√(10·73.2) = −0.924
Answer
r = −0.924
8.Linear regression & correlation
Answer: e = −4.695
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 = 21/6 = 3.5
ȳ = 124/6 = 20.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −18
Regression line
b = −18/17.5 = −1.029
a = 20.67 − (−1.029)·3.5 = 24.27
ŷ = 24.27 − 1.029·x
Predicted value
ŷ = 24.27 + (−1.029)·2.5 = 21.7
Residual
e = 17 − 21.7 = −4.695
Answer
e = −4.695
The point lies below the line.
9.Chi-square tests
Answer: χ² = 2.09
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 = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is not rejected.
10.Chi-square tests
Answer: χ²crit = 15.09
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.
Degrees of freedom
df = 7 − 1 − 1 = 5
Table
χ²crit(df = 5, α = 0.01) = 15.086
Answer
χ²crit = 15.086
Reject H₀ if the computed χ² is larger than this value.