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 bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 247.75 and 256 g? Give P as a decimal.
μ = 250σ = 5x1 = 247.75x2 = 256
P =
2.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 25% of them lie (the 25th percentile).
μ = 500σ = 40P = 25%
x = h
3.Confidence intervalsMedium
A sample of 10 items has mean 43.5 and standard deviation s = 10. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.
n = 10x̄ = 43.5s = 1090%t(9) = 1.833
lower limit =
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 0.03?
p = 0.5E = 0.0395%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 36 students has a mean score of 116.3 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 116.3s = 10n = 36
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 81 measurements has mean 76.6. The population standard deviation is 12. Find the upper limit U of the 99% confidence interval for the mean.
x̄ = 76.6σ = 12n = 8199%
U =
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. 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 = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 23.
n = 5x = 23
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 = 1802 × 3
χ² =
10.Chi-square testsHard
A chi-square test of independence is carried out on a contingency table with 2 rows and 6 columns (not counting the totals). How many degrees of freedom df does the test have?
2 × 6
df =
3Solutions
1.Normal distribution
Answer: P = 0.5586
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 = (247.75 − 250)/5 = −0.45
z₂ = (256 − 250)/5 = 1.2
Area between
P = Φ(1.2) − Φ(−0.45) = 0.8849 − 0.3264 = 0.5586
Answer
P = 0.5586
2.Normal distribution
Answer: x = 473 h
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) = 0.25 → z = −0.6745
Back to x
x = 500 + −0.6745·40 = 473.02 h
Answer
x = 473.02 h
3.Confidence intervals
Answer: lower limit = 37.7
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 = 9, t = 1.833
Margin of error
E = 1.833·10/√10 = 5.796
Interval
43.5 ± 5.796 → [37.7, 49.3]
Answer
lower limit = 37.7
4.Confidence intervals
Answer: n = 1068
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 = 1.96²·0.5·0.5/0.03² = 1067
Round up
n = 1068
Answer
n = 1068
5.Hypothesis tests & confidence intervals
Answer: t = −2.22
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 = (116.3 − 120)/1.667 = −2.22
Answer
t = −2.22
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 = 80.03
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 = 2.576·12/√81 = 3.435
Interval limit
U = 76.6 + 3.435 = 80.03
Answer
U = 80.03
7.Linear regression & correlation
Answer: r² = 0.982
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 = 205/6 = 34.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 54.5
Syy = Σ(y − ȳ)² = 172.8
Coefficient of determination
r² = 54.5²/(17.5·172.8) = 0.982
Answer
r² = 0.982
About 98.2% of the variation in y is explained by the line.
8.Linear regression & correlation
Answer: ŷ = 92
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 = 110/5 = 22
ȳ = 440/5 = 88
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 160
Regression line
b = 160/40 = 4
a = 88 − 4·22 = 0
ŷ = 0 + 4·x
Prediction
ŷ = 0 + 4·23 = 92
Answer
ŷ = 92
9.Chi-square tests
Answer: χ² = 10.3
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: df = 5
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.