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 0.3035. Find z to two decimals.
A = 0.3035
z =
2.Normal distributionHard
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 97.5% of them lie (the 97.5th percentile).
μ = 32σ = 6P = 97.5%
x = min
3.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 50 phones, mean 127.5 h, known standard deviation 14 h. Model 2: 40 phones, mean 142 h, known standard deviation 4 h. Find the lower limit of the 90% confidence interval for μ₁ − μ₂.
A pilot study suggests that about 0.4 of products are defective. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.04?
p ≈ 0.4E = 0.0495%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 49 students has a mean score of 102.7 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 100.
μ₀ = 100x̄ = 102.7s = 8n = 49
t =
6.Hypothesis tests & confidence intervalsHard
In a trial, 36 of 120 patients on treatment A recover, and 39 of 150 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 36n₁ = 120x₂ = 39n₂ = 150
z =
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 5Σx = 30Σy = 211Σx² = 220Σy² = 9101Σxy = 1354
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. Predict ŷ for x = 5.5.
n = 5x = 5.5
ŷ =
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 = 1202 × 2
χ² =
10.Chi-square testsHard
A chi-square goodness-of-fit test checks whether counts in 6 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 = 0.05.
α = 0.05
χ²crit =
3Solutions
1.Normal distribution
Answer: z = 0.39
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 + 0.3035)/2 = 0.6517
Read the z-table backwards
Find 0.6517 inside the table and read off z.
z = 0.39
Answer
z = 0.39
2.Normal distribution
Answer: x = 43.76 min
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 for the percentile
Φ(z) = 0.975 → z = 1.96
Back to x
x = 32 + 1.96·6 = 43.76 min
Answer
x = 43.76 min
3.Confidence intervals
Answer: lower limit = −17.92 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 = 127.5 − 142 = −14.5
Standard error
SE = √(14²/50 + 4²/40) = 2.078
Margin of error
E = 1.645·2.078 = 3.419
Interval
−14.5 ± 3.419 → [−17.92, −11.08]
Answer
lower limit = −17.92 h
The interval does not contain 0, so the means differ significantly.
4.Confidence intervals
Answer: n = 577
Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.
Solve for n
n = 1.96²·0.4·0.6/0.04² = 576.2
Round up
n = 577
Answer
n = 577
5.Hypothesis tests & confidence intervals
Answer: t = 2.363
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/√49 = 8/7 = 1.143
Test statistic
t = (102.7 − 100)/1.143 = 2.363
Answer
t = 2.363
t0.025; df = 48 = 2.011
|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: z = 0.7292
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 = 36/120 = 0.3
p̂₂ = 39/150 = 0.26
p̂ = 75/270 = 0.2778
Standard error under H₀
SE = √(0.2778·0.7222·(1/120 + 1/150)) = 0.05486
Test statistic
z = (0.3 − 0.26)/0.05486 = 0.7292
Answer
z = 0.7292
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
7.Linear regression & correlation
Answer: r = 0.9918
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 = 1354 − 30·211/5 = 88
Sxx = 220 − 30²/5 = 40
Syy = 9101 − 211²/5 = 196.8
Correlation coefficient
r = 88/√(40·196.8) = 0.9918
Answer
r = 0.9918
8.Linear regression & correlation
Answer: ŷ = 67.25
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 = 328/5 = 65.6
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 = 65.6 − 3.3·5 = 49.1
ŷ = 49.1 + 3.3·x
Prediction
ŷ = 49.1 + 3.3·5.5 = 67.25
Answer
ŷ = 67.25
9.Chi-square tests
Answer: χ² = 3.663
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 not rejected.
10.Chi-square tests
Answer: χ²crit = 9.488
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 = 6 − 1 − 1 = 4
Table
χ²crit(df = 4, α = 0.05) = 9.488
Answer
χ²crit = 9.488
Reject H₀ if the computed χ² is larger than this value.