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 0.0143. Find z to two decimals.
A = 0.0143
z =
2.Normal distributionHard
Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
3.Confidence intervalsMedium
The blood pressure of 7 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 6.6 and standard deviation 3.2. Find the lower limit of the 95% confidence interval for the mean difference.
n = 7d̄ = 6.6sd = 3.295%t(6) = 2.447
lower limit = mmHg
4.Confidence intervalsHard
The population standard deviation is 15. 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 2?
σ = 15E = 290%
n =
5.Hypothesis tests & confidence intervalsMedium
15 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 2.48 and the standard deviation of the differences is 6.4. Find the paired test statistic t for H₀: μd = 0.
n = 15d̄ = 2.48sd = 6.4
t =
6.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 80 against H₁: μ < 80 is done at α = 0.01 with a sample of 25; the standard deviation is known to be 2.5. In reality the mean is 79.15. Find the power 1 − β of the test in percent.
μ₀ = 80μ = 79.15σ = 2.5n = 25α = 0.01
1 − β = %
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.
n = 6
r =
8.Linear regression & correlationHard
The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 5.
n = 6x = 5
ŷ =
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 “night shift, major”.
row total = 68column total = 67N = 157
E =
10.Chi-square testsHard
A chi-square goodness-of-fit test compares observed counts in 9 categories with fully specified expected proportions (no parameters are estimated from the data). Using the table, find the critical value χ²crit at significance level α = 0.05.
α = 0.05
χ²crit =
3Solutions
1.Normal distribution
Answer: z = 2.19
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 − 0.0143 = 0.9857
Read the z-table backwards
Find 0.9857 inside the table and read off z.
z = 2.19
Answer
z = 2.19
2.Normal distribution
Answer: P = 0.7881
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 = 52 − 48 = 4 s, σ = 5 s
Standardise
z = (0 − 4)/5 = −0.8
Probability
P = 1 − Φ(−0.8) = 0.7881
Answer
P = 0.7881
3.Confidence intervals
Answer: lower limit = 3.64 mmHg
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.
Critical value
df = 6, t = 2.447
Margin of error
E = 2.447·3.2/√7 = 2.96
Interval
6.6 ± 2.96 → [3.64, 9.56]
Answer
lower limit = 3.64 mmHg
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 = (1.645·15/2)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 = 1.501
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 of the mean difference
SE = 6.4/√15 = 1.652
Test statistic
t = 2.48/1.652 = 1.501
Answer
t = 1.501
With df = 14, the two-sided 5% critical value is 2.145; |t| is smaller, so H₀ is not rejected.
6.Hypothesis tests & confidence intervals
Answer: 1 − β = 26.57 %
Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).
Critical sample mean
SE = 2.5/√25 = 0.5
x̄c = 80 − 2.326·0.5 = 78.84
z under the true mean
z = (78.84 − 79.15)/0.5 = −0.626
Power
power = P(Z < −0.626) = 26.57%
1 − β = 26.57%
Answer
1 − β = 26.57%
7.Linear regression & correlation
Answer: r = −0.7773
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 = 118/6 = 19.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −17
Syy = Σ(y − ȳ)² = 27.33
Correlation coefficient
r = −17/√(17.5·27.33) = −0.7773
Answer
r = −0.7773
8.Linear regression & correlation
Answer: ŷ = 20.17
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 = 168/6 = 28
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 274
Regression line
b = 274/70 = 3.914
a = 28 − 3.914·7 = 0.6
ŷ = 0.6 + 3.914·x
Prediction
ŷ = 0.6 + 3.914·5 = 20.17
Answer
ŷ = 20.17
9.Chi-square tests
Answer: E = 29.02
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 = 68·67/157 = 29.02
Answer
E = 29.02
10.Chi-square tests
Answer: χ²crit = 15.51
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.
Degrees of freedom
df = 9 − 1 = 8
Table
χ²crit(df = 8, α = 0.05) = 15.507
Answer
χ²crit = 15.507
Reject H₀ if the computed χ² is larger than this value.