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 0.8186. Find z to two decimals.
A = 0.8186
z =
2.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 80% of them lie (the 80th percentile).
μ = 250σ = 5P = 80%
x = g
3.Confidence intervalsMedium
A 95% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 9.5. What is the margin of error E₂ of a 99% interval based on 576 observations?
95%n₁ = 64E₁ = 9.599%n₂ = 576
E₂ =
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 0.01?
p = 0.5E = 0.0195%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 25 students has a mean score of 79.64 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 79.64s = 8n = 25
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 100 measurements has mean 118.9. The population standard deviation is 12. Find the lower limit L of the 90% confidence interval for the mean.
x̄ = 118.9σ = 12n = 10090%
L =
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. 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. Predict ŷ for x = 7.
n = 5x = 7
ŷ =
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 × 3
χ² =
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 = 2203 × 2α = 0.05
χ² =
3Solutions
1.Normal distribution
Answer: z = −0.91
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 − 0.8186 = 0.1814
Read the z-table backwards
Find 0.1814 inside the table and read off z.
z = −0.91
Answer
z = −0.91
2.Normal distribution
Answer: x = 254.2 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) = 0.8 → z = 0.8416
Back to x
x = 250 + 0.8416·5 = 254.21 g
Answer
x = 254.21 g
3.Confidence intervals
Answer: E₂ = 4.162
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.
Ratio of margins
E₂/E₁ = (2.576/1.96)·√(64/576) = 0.4381
New margin
E₂ = 9.5·0.4381 = 4.162
Answer
E₂ = 4.162
4.Confidence intervals
Answer: n = 9604
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 = 1.96²·0.5·0.5/0.01² = 9604
Round up
n = 9604
Answer
n = 9604
5.Hypothesis tests & confidence intervals
Answer: t = 2.9
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 = 8/√25 = 8/5 = 1.6
Test statistic
t = (79.64 − 75)/1.6 = 2.9
Answer
t = 2.9
t0.025; df = 24 = 2.064
|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: L = 116.9
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 = 1.645·12/√100 = 1.974
Interval limit
L = 118.9 − 1.974 = 116.9
Answer
L = 116.9
7.Linear regression & correlation
Answer: r² = 0.9313
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 = 339/6 = 56.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 255
Syy = Σ(y − ȳ)² = 997.5
Coefficient of determination
r² = 255²/(70·997.5) = 0.9313
Answer
r² = 0.9313
About 93.13% of the variation in y is explained by the line.
8.Linear regression & correlation
Answer: ŷ = 18.3
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 = 81/5 = 16.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 84
Regression line
b = 84/40 = 2.1
a = 16.2 − 2.1·6 = 3.6
ŷ = 3.6 + 2.1·x
Prediction
ŷ = 3.6 + 2.1·7 = 18.3
Answer
ŷ = 18.3
9.Chi-square tests
Answer: χ² = 13.65
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 rejected.
10.Chi-square tests
Answer: χ² = 29.89
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.