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 IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 88.75 and 118 points? Give P as a decimal.
μ = 100σ = 15x1 = 88.75x2 = 118
P =
2.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 25% is to the left of it. Find the cut-off x.
μ = 100 pointsσ = 15 points25%
x = points
3.Confidence intervalsMedium
The blood pressure of 20 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −8.9 and standard deviation 2.1. Find the margin of error E of the 90% confidence interval for the mean difference.
n = 20d̄ = −8.9sd = 2.190%t(19) = 1.729
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 90% confidence interval for the proportion has a margin of error of at most 0.01?
p = 0.5E = 0.0190%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 75.68 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 75.68s = 4n = 100
t =
6.Hypothesis tests & confidence intervalsHard
In a trial, 40 of 200 patients on treatment A recover, and 27 of 80 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 40n₁ = 200x₂ = 27n₂ = 80
z =
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.
n = 6
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 = 15?
n = 5ŷ = 15
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 = 1602 × 2
χ² =
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 = 1803 × 2α = 0.01
χ² =
3Solutions
1.Normal distribution
Answer: P = 0.6583
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 = (88.75 − 100)/15 = −0.75
z₂ = (118 − 100)/15 = 1.2
Area between
P = Φ(1.2) − Φ(−0.75) = 0.8849 − 0.2266 = 0.6583
Answer
P = 0.6583
2.Normal distribution
Answer: x = 90 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) = 0.25
z from the table
z ≈ −0.67
Back to the original units
x = 100 − 0.67·15 = 90 points
Answer
x = 90 points
3.Confidence intervals
Answer: E = 0.8119 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 = 19, t = 1.729
Margin of error
E = 1.729·2.1/√20 = 0.8119
Answer
E = 0.8119 mmHg
4.Confidence intervals
Answer: n = 6766
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 = 1.645²·0.5·0.5/0.01² = 6765
Round up
n = 6766
Answer
n = 6766
5.Hypothesis tests & confidence intervals
Answer: t = 1.7
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 = (75.68 − 75)/0.4 = 1.7
Answer
t = 1.7
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: z = −2.436
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 = 40/200 = 0.2
p̂₂ = 27/80 = 0.3375
p̂ = 67/280 = 0.2393
Standard error under H₀
SE = √(0.2393·0.7607·(1/200 + 1/80)) = 0.05644
Test statistic
z = (0.2 − 0.3375)/0.05644 = −2.436
Answer
z = −2.436
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
7.Linear regression & correlation
Answer: r² = 0.976
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 = 464/6 = 77.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 94
Syy = Σ(y − ȳ)² = 517.3
Coefficient of determination
r² = 94²/(17.5·517.3) = 0.976
Answer
r² = 0.976
About 97.6% of the variation in y is explained by the line.
8.Linear regression & correlation
Answer: x = 7.212
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 = 65/5 = 13
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 66
Regression line
b = 66/40 = 1.65
a = 13 − 1.65·6 = 3.1
ŷ = 3.1 + 1.65·x
Solve for x
x = (15 − 3.1)/1.65 = 7.212
Answer
x = 7.212
9.Chi-square tests
Answer: χ² = 40.11
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: χ² = 6.563
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.