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 under the standard normal curve, to the right of z, is 0.3264. Find z to two decimals.
A = 0.3264
z =
2.Normal distributionHard
The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
3.Confidence intervalsMedium
The blood pressure of 8 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 0.5 and standard deviation 0.8. Find the margin of error E of the 95% confidence interval for the mean difference.
n = 8d̄ = 0.5sd = 0.895%t(7) = 2.365
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 0.02?
p = 0.5E = 0.0290%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 64 students has a mean score of 46.1 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 50.
μ₀ = 50x̄ = 46.1s = 12n = 64
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 49 measurements has mean 48.3. The population standard deviation is 5. Find the lower limit L of the 99% confidence interval for the mean.
x̄ = 48.3σ = 5n = 4999%
L =
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 5Σx = 15Σy = 270Σx² = 55Σy² = 14762Σxy = 851
r =
8.Linear regression & correlationHard
The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 70?
n = 6ŷ = 70
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 “rural, low”.
row total = 84column total = 61N = 133
E =
10.Chi-square testsHard
The table shows observed counts. Test at α = 0.05 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 = 0.05
χ² =
3Solutions
1.Normal distribution
Answer: z = 0.45
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 − 0.3264 = 0.6736
Read the z-table backwards
Find 0.6736 inside the table and read off z.
z = 0.45
Answer
z = 0.45
2.Normal distribution
Answer: P = 0.8413
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).
Distribution of the difference
μ = 500 − 490 = 10 g, σ = 10 g
Standardise
z = (0 − 10)/10 = −1
Probability
P = 1 − Φ(−1) = 0.8413
Answer
P = 0.8413
3.Confidence intervals
Answer: E = 0.6689 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 = 7, t = 2.365
Margin of error
E = 2.365·0.8/√8 = 0.6689
Answer
E = 0.6689 mmHg
4.Confidence intervals
Answer: n = 1692
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 = 1.645²·0.5·0.5/0.02² = 1691
Round up
n = 1692
Answer
n = 1692
5.Hypothesis tests & confidence intervals
Answer: t = −2.6
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/√64 = 12/8 = 1.5
Test statistic
t = (46.1 − 50)/1.5 = −2.6
Answer
t = −2.6
t0.025; df = 63 = 1.998
|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: L = 46.46
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).
Margin of error
E = 2.576·5/√49 = 1.84
Interval limit
L = 48.3 − 1.84 = 46.46
Answer
L = 46.46
7.Linear regression & correlation
Answer: r = 0.9611
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 = 851 − 15·270/5 = 41
Sxx = 55 − 15²/5 = 10
Syy = 14762 − 270²/5 = 182
Correlation coefficient
r = 41/√(10·182) = 0.9611
Answer
r = 0.9611
8.Linear regression & correlation
Answer: x = 4.493
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 = 372/6 = 62
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 141
Regression line
b = 141/17.5 = 8.057
a = 62 − 8.057·3.5 = 33.8
ŷ = 33.8 + 8.057·x
Solve for x
x = (70 − 33.8)/8.057 = 4.493
Answer
x = 4.493
9.Chi-square tests
Answer: E = 38.53
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 = 84·61/133 = 38.53
Answer
E = 38.53
10.Chi-square tests
Answer: χ² = 2.517
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.