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 bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 244.25 and 254.25 g? Give P as a decimal.
μ = 250σ = 5x1 = 244.25x2 = 254.25
P =
2.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 25% of them lie (the 25th percentile).
μ = 250σ = 5P = 25%
x = g
3.Confidence intervalsMedium
The blood pressure of 10 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 7.9 and standard deviation 6.2. Find the margin of error E of the 95% confidence interval for the mean difference.
n = 10d̄ = 7.9sd = 6.295%t(9) = 2.262
E = mmHg
4.Confidence intervalsHard
Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.01?
p = 0.5E = 0.0199%
n =
5.Hypothesis tests & confidence intervalsMedium
11 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 5.89 and the standard deviation of the differences is 9.3. Find the paired test statistic t for H₀: μd = 0.
n = 11d̄ = 5.89sd = 9.3
t =
6.Hypothesis tests & confidence intervalsHard
In a trial, 125 of 250 patients on treatment A recover, and 92 of 150 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 125n₁ = 250x₂ = 92n₂ = 150
z =
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.
n = 6
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. For which value x does the line predict ŷ = 20?
n = 6ŷ = 20
x =
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 “night shift, no defect”.
row total = 81column total = 57N = 141
E =
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 = 2603 × 2α = 0.01
χ² =
3Solutions
1.Normal distribution
Answer: P = 0.6773
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 = (244.25 − 250)/5 = −1.15
z₂ = (254.25 − 250)/5 = 0.85
Area between
P = Φ(0.85) − Φ(−1.15) = 0.8023 − 0.1251 = 0.6773
Answer
P = 0.6773
2.Normal distribution
Answer: x = 246.6 g
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) = 0.25 → z = −0.6745
Back to x
x = 250 + −0.6745·5 = 246.63 g
Answer
x = 246.63 g
3.Confidence intervals
Answer: E = 4.435 mmHg
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.
Critical value
df = 9, t = 2.262
Margin of error
E = 2.262·6.2/√10 = 4.435
Answer
E = 4.435 mmHg
4.Confidence intervals
Answer: n = 16590
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 = 2.576²·0.5·0.5/0.01² = 16589
Round up
n = 16590
Answer
n = 16590
5.Hypothesis tests & confidence intervals
Answer: t = 2.101
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/√11 = 2.804
Test statistic
t = 5.89/2.804 = 2.101
Answer
t = 2.101
With df = 10, the two-sided 5% critical value is 2.228; |t| is smaller, so H₀ is not rejected.
6.Hypothesis tests & confidence intervals
Answer: z = −2.203
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 = 125/250 = 0.5
p̂₂ = 92/150 = 0.6133
p̂ = 217/400 = 0.5425
Standard error under H₀
SE = √(0.5425·0.4575·(1/250 + 1/150)) = 0.05145
Test statistic
z = (0.5 − 0.6133)/0.05145 = −2.203
Answer
z = −2.203
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
7.Linear regression & correlation
Answer: r = 0.9881
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 = 33/6 = 5.5
ȳ = 328/6 = 54.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 113
Syy = Σ(y − ȳ)² = 747.3
Correlation coefficient
r = 113/√(17.5·747.3) = 0.9881
Answer
r = 0.9881
8.Linear regression & correlation
Answer: x = 2.444
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 = 101/6 = 16.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −52.5
Regression line
b = −52.5/17.5 = −3
a = 16.83 − (−3)·3.5 = 27.33
ŷ = 27.33 − 3·x
Solve for x
x = (20 − 27.33)/(−3) = 2.444
Answer
x = 2.444
9.Chi-square tests
Answer: E = 32.74
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 = 81·57/141 = 32.74
Answer
E = 32.74
10.Chi-square tests
Answer: χ² = 9.469
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.