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 heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 174.85 and 188.85 cm? Give P as a decimal.
μ = 178σ = 7x1 = 174.85x2 = 188.85
P =
2.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 5% of them lie (the 5th percentile).
μ = 178σ = 7P = 5%
x = cm
3.Confidence intervalsMedium
A sample of 21 items has mean 15 and standard deviation s = 13.5. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.
n = 21x̄ = 15s = 13.599%t(20) = 2.845
E =
4.Confidence intervalsHard
The diameters of 15 machined shafts have a sample standard deviation of 8 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 14, 2.5% of the χ² distribution lies below 5.629 and 2.5% lies above 26.119.
n = 15s = 8 µmχ² = 5.629, 26.119
σ = µm
5.Hypothesis tests & confidence intervalsMedium
A sample of 36 students has a mean score of 494.8 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 494.8s = 10n = 36
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 36 measurements has mean 121.9. The population standard deviation is 8. Find the upper limit U of the 90% confidence interval for the mean.
x̄ = 121.9σ = 8n = 3690%
U =
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.
n = 6
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 = 65?
n = 6ŷ = 65
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 = 2002 × 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: P = 0.6131
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 = (174.85 − 178)/7 = −0.45
z₂ = (188.85 − 178)/7 = 1.55
Area between
P = Φ(1.55) − Φ(−0.45) = 0.9394 − 0.3264 = 0.6131
Answer
P = 0.6131
2.Normal distribution
Answer: x = 166.5 cm
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) = 0.05 → z = −1.645
Back to x
x = 178 + −1.645·7 = 166.49 cm
Answer
x = 166.49 cm
3.Confidence intervals
Answer: E = 8.381
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 = 20, t = 2.845
Margin of error
E = 2.845·13.5/√21 = 8.381
Answer
E = 8.381
4.Confidence intervals
Answer: σ = 12.62 µm
Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.
Sum of squares
(n − 1)s² = 14·8² = 896
Limits for σ²
896/26.119 = 34.3
896/5.629 = 159.2
Limit for σ
σ = √159.2 = 12.62 µm
Answer
σ = 12.62 µm
95% interval for σ: [5.857, 12.62] µm.
5.Hypothesis tests & confidence intervals
Answer: t = −3.12
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 = 10/√36 = 10/6 = 1.667
Test statistic
t = (494.8 − 500)/1.667 = −3.12
Answer
t = −3.12
t0.025; df = 35 = 2.03
|t| > 2.03, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: U = 124.1
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).
Margin of error
E = 1.645·8/√36 = 2.193
Interval limit
U = 121.9 + 2.193 = 124.1
Answer
U = 124.1
7.Linear regression & correlation
Answer: r = −0.9817
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 = 143/6 = 23.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −31.5
Syy = Σ(y − ȳ)² = 58.83
Correlation coefficient
r = −31.5/√(17.5·58.83) = −0.9817
Answer
r = −0.9817
8.Linear regression & correlation
Answer: x = 4.534
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 = 417/6 = 69.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 81.5
Regression line
b = 81.5/17.5 = 4.657
a = 69.5 − 4.657·5.5 = 43.89
ŷ = 43.89 + 4.657·x
Solve for x
x = (65 − 43.89)/4.657 = 4.534
Answer
x = 4.534
9.Chi-square tests
Answer: χ² = 4.097
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: χ² = 10.11
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.