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 0.9332. Find z to two decimals.
A = 0.9332
z =
2.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 10% of them lie (the 10th percentile).
μ = 500σ = 40P = 10%
x = h
3.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 30 phones, mean 50.5 h, known standard deviation 15 h. Model 2: 100 phones, mean 52.5 h, known standard deviation 15 h. Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.
The population standard deviation is 13. 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 1.5?
σ = 13E = 1.590%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 121 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 121s = 15n = 100
t =
6.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.01 must detect a shift of the mean of 1.5 units with a power of 90%. The standard deviation is 11. Find the minimum sample size n.
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 6Σx = 21Σy = 173Σx² = 91Σy² = 5063Σxy = 635
r =
8.Linear regression & correlationHard
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 (1.5, 23) is made. Find its residual e = y − ŷ.
n = 6(1.5, 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 = 1602 × 2
χ² =
10.Chi-square testsHard
A chi-square goodness-of-fit test checks whether counts in 8 categories follow a Poisson distribution whose mean is estimated from the same data. How many degrees of freedom df does the test have?
k = 8
df =
3Solutions
1.Normal distribution
Answer: z = −1.5
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 − 0.9332 = 0.0668
Read the z-table backwards
Find 0.0668 inside the table and read off z.
z = −1.50
Answer
z = −1.50
2.Normal distribution
Answer: x = 448.7 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) = 0.1 → z = −1.282
Back to x
x = 500 + −1.282·40 = 448.74 h
Answer
x = 448.74 h
3.Confidence intervals
Answer: E = 5.137 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 = 50.5 − 52.5 = −2
Standard error
SE = √(15²/30 + 15²/100) = 3.122
Margin of error
E = 1.645·3.122 = 5.137
Answer
E = 5.137 h
The interval contains 0, so the data do not show a significant difference.
4.Confidence intervals
Answer: n = 204
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 = (1.645·13/1.5)² = 203.3
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 204
Answer
n = 204
5.Hypothesis tests & confidence intervals
Answer: t = 0.6667
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/√100 = 15/10 = 1.5
Test statistic
t = (121 − 120)/1.5 = 0.6667
Answer
t = 0.6667
t0.025; df = 99 = 1.984
|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: n = 701
Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.
Solve for n
n = ((2.326 + 1.282)·11/1.5)² = 700.1
Round up
n = 701
Answer
n = 701
7.Linear regression & correlation
Answer: r = 0.8152
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 = 635 − 21·173/6 = 29.5
Sxx = 91 − 21²/6 = 17.5
Syy = 5063 − 173²/6 = 74.83
Correlation coefficient
r = 29.5/√(17.5·74.83) = 0.8152
Answer
r = 0.8152
8.Linear regression & correlation
Answer: e = 2.729
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 = 87/6 = 14.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −50.5
Regression line
b = −50.5/17.5 = −2.886
a = 14.5 − (−2.886)·3.5 = 24.6
ŷ = 24.6 − 2.886·x
Predicted value
ŷ = 24.6 + (−2.886)·1.5 = 20.27
Residual
e = 23 − 20.27 = 2.729
Answer
e = 2.729
The point lies above the line.
9.Chi-square tests
Answer: χ² = 1.055
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 not rejected.
10.Chi-square tests
Answer: df = 6
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.