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 0.9615. Find z to two decimals.
A = 0.9615
z =
2.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 80% of them lie (the 80th percentile).
μ = 100σ = 15P = 80%
x = points
3.Confidence intervalsMedium
A sample of 25 items has mean 106 and standard deviation s = 1.5. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.
n = 25x̄ = 106s = 1.590%t(24) = 1.711
E =
4.Confidence intervalsHard
Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.025?
p = 0.5E = 0.02595%
n =
5.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 50 parts has mean length 143.5 mm, and the line's known standard deviation is 10 mm. Line 2: a sample of 40 parts has mean 149.2 mm, known standard deviation 12 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
A sample of 81 measurements has mean 74.2. The population standard deviation is 5. Find the lower limit L of the 90% confidence interval for the mean.
x̄ = 74.2σ = 5n = 8190%
L =
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.
n = 5
r =
8.Linear regression & correlationHard
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 46?
n = 5ŷ = 46
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 = 2202 × 3
χ² =
10.Chi-square testsHard
A chi-square goodness-of-fit test checks whether counts in 4 categories follow a Poisson distribution whose mean is estimated from the same data. How many degrees of freedom df does the test have?
k = 4
df =
3Solutions
1.Normal distribution
Answer: z = 2.07
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 + 0.9615)/2 = 0.9808
Read the z-table backwards
Find 0.9808 inside the table and read off z.
z = 2.07
Answer
z = 2.07
2.Normal distribution
Answer: x = 112.6 points
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) = 0.8 → z = 0.8416
Back to x
x = 100 + 0.8416·15 = 112.62 points
Answer
x = 112.62 points
3.Confidence intervals
Answer: E = 0.5133
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 = 24, t = 1.711
Margin of error
E = 1.711·1.5/√25 = 0.5133
Answer
E = 0.5133
4.Confidence intervals
Answer: n = 1537
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 = 1.96²·0.5·0.5/0.025² = 1537
Round up
n = 1537
Answer
n = 1537
5.Hypothesis tests & confidence intervals
Answer: z = −2.409
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.
Standard error of the difference
SE = √(10²/50 + 12²/40) = √5.6 = 2.366
Test statistic
z = (143.5 − 149.2)/2.366 = −2.409
Answer
z = −2.409
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: L = 73.29
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 = 1.645·5/√81 = 0.9139
Interval limit
L = 74.2 − 0.9139 = 73.29
Answer
L = 73.29
7.Linear regression & correlation
Answer: r = 0.9737
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 = 120/5 = 24
ȳ = 351/5 = 70.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 276
Syy = Σ(y − ȳ)² = 892.8
Correlation coefficient
r = 276/√(90·892.8) = 0.9737
Answer
r = 0.9737
8.Linear regression & correlation
Answer: x = 4.51
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 = 242/5 = 48.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 49
Regression line
b = 49/10 = 4.9
a = 48.4 − 4.9·5 = 23.9
ŷ = 23.9 + 4.9·x
Solve for x
x = (46 − 23.9)/4.9 = 4.51
Answer
x = 4.51
9.Chi-square tests
Answer: χ² = 5.541
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 = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is not rejected.
10.Chi-square tests
Answer: df = 2
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.