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 242.25 and 254 g? Give P as a decimal.
μ = 250σ = 5x1 = 242.25x2 = 254
P =
2.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded top 30% is to the right of it. Find the cut-off x.
μ = 178 cmσ = 7 cm30%
x = cm
3.Confidence intervalsMedium
A sample of 9 items has mean 118.5 and standard deviation s = 15. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.
n = 9x̄ = 118.5s = 1599%t(8) = 3.355
lower limit =
4.Confidence intervalsHard
A pilot study suggests that about 0.15 of products are defective. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.03?
p ≈ 0.15E = 0.0399%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 64 students has a mean score of 67.25 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 67.25s = 20n = 64
t =
6.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 0.5? The standard deviation is 15. Find the smallest sample size n.
σ = 15E = 0.595%
n =
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.
n = 6
r =
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. A new observation (3, 10) is made. Find its residual e = y − ŷ.
n = 5(3, 10)
e =
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 “students, car”.
row total = 91column total = 54N = 173
E =
10.Chi-square testsHard
A chi-square test of independence is carried out on a contingency table with 2 rows and 5 columns (not counting the totals). Using the table, find the critical value χ²crit at significance level α = 0.05.
α = 0.05
χ²crit =
3Solutions
1.Normal distribution
Answer: P = 0.7276
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 = (242.25 − 250)/5 = −1.55
z₂ = (254 − 250)/5 = 0.8
Area between
P = Φ(0.8) − Φ(−1.55) = 0.7881 − 0.06057 = 0.7276
Answer
P = 0.7276
2.Normal distribution
Answer: x = 181.6 cm
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) = 0.7
z from the table
z ≈ 0.52
Back to the original units
x = 178 + 0.52·7 = 181.6 cm
Answer
x = 181.6 cm
3.Confidence intervals
Answer: lower limit = 101.7
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.
Critical value
df = n − 1 = 8, t = 3.355
Margin of error
E = 3.355·15/√9 = 16.78
Interval
118.5 ± 16.78 → [101.7, 135.3]
Answer
lower limit = 101.7
4.Confidence intervals
Answer: n = 941
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 = 2.576²·0.15·0.85/0.03² = 940.1
Round up
n = 941
Answer
n = 941
5.Hypothesis tests & confidence intervals
Answer: t = −3.1
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 = 20/√64 = 20/8 = 2.5
Test statistic
t = (67.25 − 75)/2.5 = −3.1
Answer
t = −3.1
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: n = 3458
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).
Solve for n
n = (1.96·15/0.5)² = 3457
Round up
n = 3458
Answer
n = 3458
7.Linear regression & correlation
Answer: r = 0.985
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 = 126/6 = 21
ȳ = 658/6 = 109.7
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 420
Syy = Σ(y − ȳ)² = 2597
Correlation coefficient
r = 420/√(70·2597) = 0.985
Answer
r = 0.985
8.Linear regression & correlation
Answer: e = −2.35
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 = 100/5 = 20
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 102
Regression line
b = 102/40 = 2.55
a = 20 − 2.55·6 = 4.7
ŷ = 4.7 + 2.55·x
Predicted value
ŷ = 4.7 + 2.55·3 = 12.35
Residual
e = 10 − 12.35 = −2.35
Answer
e = −2.35
The point lies below the line.
9.Chi-square tests
Answer: E = 28.4
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 = 91·54/173 = 28.4
Answer
E = 28.4
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 = (2 − 1)·(5 − 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.