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 exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 49.5 and 83.5 points? Give P as a decimal.
μ = 65σ = 10x1 = 49.5x2 = 83.5
P =
2.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded bottom 40% is to the left of it. Find the cut-off x.
μ = 65 pointsσ = 10 points40%
x = points
3.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 50 phones, mean 59.5 h, known standard deviation 3 h. Model 2: 50 phones, mean 57.5 h, known standard deviation 12 h. Find the lower limit of the 95% confidence interval for μ₁ − μ₂.
The population standard deviation is 33. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 0.5?
σ = 33E = 0.590%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 25 students has a mean score of 502.8 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 502.8s = 20n = 25
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 25 measurements has mean 111. The population standard deviation is 15. Find the lower limit L of the 95% confidence interval for the mean.
x̄ = 111σ = 15n = 2595%
L =
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
8.Linear regression & correlationHard
The table gives 5 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 (3.5, 23) is made. Find its residual e = y − ŷ.
n = 5(3.5, 23)
e =
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 = 0.01 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 = 1803 × 2α = 0.01
χ² =
3Solutions
1.Normal distribution
Answer: P = 0.9073
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.
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
Φ(z) = 0.4
z from the table
z ≈ −0.25
Back to the original units
x = 65 − 0.25·10 = 62.5 points
Answer
x = 62.5 points
3.Confidence intervals
Answer: lower limit = −1.429 h
Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.
Point estimate
x̄₁ − x̄₂ = 59.5 − 57.5 = 2
Standard error
SE = √(3²/50 + 12²/50) = 1.749
Margin of error
E = 1.96·1.749 = 3.429
Interval
2 ± 3.429 → [−1.429, 5.429]
Answer
lower limit = −1.429 h
The interval contains 0, so the data do not show a significant difference.
4.Confidence intervals
Answer: n = 11788
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.
Solve for n
n = (1.645·33/0.5)² = 11787
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 11788
Answer
n = 11788
5.Hypothesis tests & confidence intervals
Answer: t = 0.7
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 = 20/√25 = 20/5 = 4
Test statistic
t = (502.8 − 500)/4 = 0.7
Answer
t = 0.7
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 = 105.1
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 = 1.96·15/√25 = 5.88
Interval limit
L = 111 − 5.88 = 105.1
Answer
L = 105.1
7.Linear regression & correlation
Answer: r² = 0.8804
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 = 15/5 = 3
ȳ = 109/5 = 21.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −27
Syy = Σ(y − ȳ)² = 82.8
Coefficient of determination
r² = −27²/(10·82.8) = 0.8804
Answer
r² = 0.8804
About 88.04% of the variation in y is explained by the line.
8.Linear regression & correlation
Answer: e = 4.5
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 = 15/5 = 3
ȳ = 98/5 = 19.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −22
Regression line
b = −22/10 = −2.2
a = 19.6 − (−2.2)·3 = 26.2
ŷ = 26.2 − 2.2·x
Predicted value
ŷ = 26.2 + (−2.2)·3.5 = 18.5
Residual
e = 23 − 18.5 = 4.5
Answer
e = 4.5
The point lies above the line.
9.Chi-square tests
Answer: χ² = 14.17
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 = 2.738
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.