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 0.8485. Find z to two decimals.
A = 0.8485
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 10% of them lie (the 10th percentile).
μ = 250σ = 5P = 10%
x = g
3.Confidence intervalsMedium
A sample of 11 items has mean 69.5 and standard deviation s = 13.5. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.
n = 11x̄ = 69.5s = 13.590%t(10) = 1.812
lower limit =
4.Confidence intervalsHard
The diameters of 16 machined shafts have a sample standard deviation of 6.9 µm. Assuming the diameters are normally distributed, find the upper 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 = 6.9 µmχ² = 6.262, 27.488
σ = µm
5.Hypothesis tests & confidence intervalsMedium
A sample of 64 students has a mean score of 51.06 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 50.
μ₀ = 50x̄ = 51.06s = 5n = 64
t =
6.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 1.5? The standard deviation is 12. Find the smallest sample size n.
σ = 12E = 1.599%
n =
7.Linear regression & correlationMedium
The table gives 5 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. Predict ŷ for x = 5.5.
n = 6x = 5.5
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 = 1202 × 2
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 = 2403 × 2α = 0.05
χ² =
3Solutions
1.Normal distribution
Answer: z = −1.03
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 − 0.8485 = 0.1515
Read the z-table backwards
Find 0.1515 inside the table and read off z.
z = −1.03
Answer
z = −1.03
2.Normal distribution
Answer: x = 243.6 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) = 0.1 → z = −1.282
Back to x
x = 250 + −1.282·5 = 243.59 g
Answer
x = 243.59 g
3.Confidence intervals
Answer: lower limit = 62.12
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 = 10, t = 1.812
Margin of error
E = 1.812·13.5/√11 = 7.376
Interval
69.5 ± 7.376 → [62.12, 76.88]
Answer
lower limit = 62.12
4.Confidence intervals
Answer: σ = 10.68 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 = 15·6.9² = 714.2
Limits for σ²
714.2/27.488 = 25.98
714.2/6.262 = 114
Limit for σ
σ = √114 = 10.68 µm
Answer
σ = 10.68 µm
95% interval for σ: [5.097, 10.68] µm.
5.Hypothesis tests & confidence intervals
Answer: t = 1.696
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 = 5/√64 = 5/8 = 0.625
Test statistic
t = (51.06 − 50)/0.625 = 1.696
Answer
t = 1.696
t0.025; df = 63 = 1.998
|t| < 1.998, so H₀ is not rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: n = 425
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 = (2.576·12/1.5)² = 424.7
Round up
n = 425
Answer
n = 425
7.Linear regression & correlation
Answer: r = 0.99
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 = 9069 − 95·469/5 = 158
Sxx = 1845 − 95²/5 = 40
Syy = 44629 − 469²/5 = 636.8
Correlation coefficient
r = 158/√(40·636.8) = 0.99
Answer
r = 0.99
8.Linear regression & correlation
Answer: ŷ = 12
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 = 97/6 = 16.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −36.5
Regression line
b = −36.5/17.5 = −2.086
a = 16.17 − (−2.086)·3.5 = 23.47
ŷ = 23.47 − 2.086·x
Prediction
ŷ = 23.47 + (−2.086)·5.5 = 12
Answer
ŷ = 12
9.Chi-square tests
Answer: χ² = 31.11
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.
df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is rejected.
10.Chi-square tests
Answer: χ² = 22.32
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.