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 0.0694. Find z to two decimals.
A = 0.0694
z =
2.Normal distributionHard
The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
3.Confidence intervalsMedium
A sample of 9 items has mean 65 and standard deviation s = 19. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.
n = 9x̄ = 65s = 1990%t(8) = 1.86
upper limit =
4.Confidence intervalsHard
The population standard deviation is 30. 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 4?
σ = 30E = 490%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 49 students has a mean score of 99.49 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 100.
μ₀ = 100x̄ = 99.49s = 4n = 49
t =
6.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.01 must detect a shift of the mean of 4 units with a power of 80%. The standard deviation is 13. Find the minimum sample size n.
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.
n = 5
r =
8.Linear regression & correlationHard
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (11, 88) is made. Find its residual e = y − ŷ.
n = 6(11, 88)
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 “rural, medium”.
row total = 78column total = 61N = 188
E =
10.Chi-square testsHard
A chi-square test of independence is carried out on a contingency table with 2 rows and 2 columns (not counting the totals). How many degrees of freedom df does the test have?
2 × 2
df =
3Solutions
1.Normal distribution
Answer: z = 1.48
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 − 0.0694 = 0.9306
Read the z-table backwards
Find 0.9306 inside the table and read off z.
z = 1.48
Answer
z = 1.48
2.Normal distribution
Answer: P = 0.7389
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).
Distribution of the difference
μ = 180 − 170 = 10 cm, σ = 15.62 cm
Standardise
z = (0 − 10)/15.62 = −0.64
Probability
P = 1 − Φ(−0.64) = 0.7389
Answer
P = 0.7389
3.Confidence intervals
Answer: upper limit = 76.78
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 = 8, t = 1.86
Margin of error
E = 1.86·19/√9 = 11.78
Interval
65 ± 11.78 → [53.22, 76.78]
Answer
upper limit = 76.78
4.Confidence intervals
Answer: n = 153
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 = (1.645·30/4)² = 152.2
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 153
Answer
n = 153
5.Hypothesis tests & confidence intervals
Answer: t = −0.8925
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 = 4/√49 = 4/7 = 0.5714
Test statistic
t = (99.49 − 100)/0.5714 = −0.8925
Answer
t = −0.8925
t0.025; df = 48 = 2.011
|t| < 2.011, so H₀ is not rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: n = 107
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 + 0.842)·13/4)² = 106
Round up
n = 107
Answer
n = 107
7.Linear regression & correlation
Answer: r = 0.9913
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 = 100/5 = 20
ȳ = 504/5 = 100.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 441
Syy = Σ(y − ȳ)² = 2199
Correlation coefficient
r = 441/√(90·2199) = 0.9913
Answer
r = 0.9913
8.Linear regression & correlation
Answer: e = −5.2
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 = 42/6 = 7
ȳ = 420/6 = 70
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 406
Regression line
b = 406/70 = 5.8
a = 70 − 5.8·7 = 29.4
ŷ = 29.4 + 5.8·x
Predicted value
ŷ = 29.4 + 5.8·11 = 93.2
Residual
e = 88 − 93.2 = −5.2
Answer
e = −5.2
The point lies below the line.
9.Chi-square tests
Answer: E = 25.31
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 = 78·61/188 = 25.31
Answer
E = 25.31
10.Chi-square tests
Answer: df = 1
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.