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 delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 24.8 and 36.8 min? Give P as a decimal.
μ = 32σ = 6x1 = 24.8x2 = 36.8
P =
2.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 97.5% of them lie (the 97.5th percentile).
μ = 250σ = 5P = 97.5%
x = g
3.Confidence intervalsMedium
A sample of 12 items has mean 47 and standard deviation s = 15. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.
n = 12x̄ = 47s = 1595%t(11) = 2.201
upper limit =
4.Confidence intervalsHard
The diameters of 16 machined shafts have a sample standard deviation of 1.5 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 15, 2.5% of the χ² distribution lies below 6.262 and 2.5% lies above 27.488.
n = 16s = 1.5 µmχ² = 6.262, 27.488
σ = µm
5.Hypothesis tests & confidence intervalsMedium
A sample of 25 students has a mean score of 55.1 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 50.
μ₀ = 50x̄ = 55.1s = 15n = 25
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 81 measurements has mean 199.6. The population standard deviation is 4. Find the lower limit L of the 95% confidence interval for the mean.
x̄ = 199.6σ = 4n = 8195%
L =
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.
The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. A new observation (2.5, 25) is made. Find its residual e = y − ŷ.
n = 6(2.5, 25)
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 “night shift, major”.
row total = 90column total = 28N = 158
E =
10.Chi-square testsHard
A chi-square goodness-of-fit test checks whether counts in 9 categories follow a Poisson distribution whose mean is estimated from the same data. Using the table, find the critical value χ²crit at significance level α = 0.05.
α = 0.05
χ²crit =
3Solutions
1.Normal distribution
Answer: P = 0.6731
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 = (24.8 − 32)/6 = −1.2
z₂ = (36.8 − 32)/6 = 0.8
Area between
P = Φ(0.8) − Φ(−1.2) = 0.7881 − 0.1151 = 0.6731
Answer
P = 0.6731
2.Normal distribution
Answer: x = 259.8 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) = 0.975 → z = 1.96
Back to x
x = 250 + 1.96·5 = 259.8 g
Answer
x = 259.8 g
3.Confidence intervals
Answer: upper limit = 56.53
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 = 11, t = 2.201
Margin of error
E = 2.201·15/√12 = 9.531
Interval
47 ± 9.531 → [37.47, 56.53]
Answer
upper limit = 56.53
4.Confidence intervals
Answer: σ = 1.108 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 = 15·1.5² = 33.75
Limits for σ²
33.75/27.488 = 1.228
33.75/6.262 = 5.39
Limit for σ
σ = √1.228 = 1.108 µm
Answer
σ = 1.108 µm
95% interval for σ: [1.108, 2.322] µm.
5.Hypothesis tests & confidence intervals
Answer: t = 1.7
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 = 15/√25 = 15/5 = 3
Test statistic
t = (55.1 − 50)/3 = 1.7
Answer
t = 1.7
t0.025; df = 24 = 2.064
|t| < 2.064, so H₀ is not rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: L = 198.7
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 = 1.96·4/√81 = 0.8711
Interval limit
L = 199.6 − 0.8711 = 198.7
Answer
L = 198.7
7.Linear regression & correlation
Answer: r = 0.9965
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 = 16163 − 141·653/6 = 817.5
Sxx = 3471 − 141²/6 = 157.5
Syy = 75341 − 653²/6 = 4273
Correlation coefficient
r = 817.5/√(157.5·4273) = 0.9965
Answer
r = 0.9965
8.Linear regression & correlation
Answer: e = 4.005
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 = 109/6 = 18.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −49.5
Regression line
b = −49.5/17.5 = −2.829
a = 18.17 − (−2.829)·3.5 = 28.07
ŷ = 28.07 − 2.829·x
Predicted value
ŷ = 28.07 + (−2.829)·2.5 = 21
Residual
e = 25 − 21 = 4.005
Answer
e = 4.005
The point lies above the line.
9.Chi-square tests
Answer: E = 15.95
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 = 90·28/158 = 15.95
Answer
E = 15.95
10.Chi-square tests
Answer: χ²crit = 14.07
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 = 9 − 1 − 1 = 7
Table
χ²crit(df = 7, α = 0.05) = 14.067
Answer
χ²crit = 14.067
Reject H₀ if the computed χ² is larger than this value.