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 0.3264. Find z to two decimals.
A = 0.3264
z =
2.Normal distributionHard
The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
3.Confidence intervalsMedium
A sample of 7 items has mean 101.5 and standard deviation s = 13. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.
n = 7x̄ = 101.5s = 1395%t(6) = 2.447
lower limit =
4.Confidence intervalsHard
The population standard deviation is 30. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 3?
σ = 30E = 395%
n =
5.Hypothesis tests & confidence intervalsMedium
13 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 6.03 and the standard deviation of the differences is 7.5. Find the paired test statistic t for H₀: μd = 0.
n = 13d̄ = 6.03sd = 7.5
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 36 measurements has mean 194.6. The population standard deviation is 12. Find the upper limit U of the 90% confidence interval for the mean.
x̄ = 194.6σ = 12n = 3690%
U =
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 6Σx = 42Σy = 184Σx² = 364Σy² = 6680Σxy = 1556
r =
8.Linear regression & correlationHard
The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 4.
n = 6x = 4
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 = 2002 × 2
χ² =
10.Chi-square testsHard
A chi-square goodness-of-fit test checks whether counts in 9 categories follow a Poisson distribution whose mean is estimated from the same data. Using the table, find the critical value χ²crit at significance level α = 0.01.
α = 0.01
χ²crit =
3Solutions
1.Normal distribution
Answer: z = 0.45
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 − 0.3264 = 0.6736
Read the z-table backwards
Find 0.6736 inside the table and read off z.
z = 0.45
Answer
z = 0.45
2.Normal distribution
Answer: P = 0.8413
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 = 500 − 490 = 10 g, σ = 10 g
Standardise
z = (0 − 10)/10 = −1
Probability
P = 1 − Φ(−1) = 0.8413
Answer
P = 0.8413
3.Confidence intervals
Answer: lower limit = 89.48
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 = 6, t = 2.447
Margin of error
E = 2.447·13/√7 = 12.02
Interval
101.5 ± 12.02 → [89.48, 113.5]
Answer
lower limit = 89.48
4.Confidence intervals
Answer: n = 385
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 = (1.96·30/3)² = 384.2
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 385
Answer
n = 385
5.Hypothesis tests & confidence intervals
Answer: t = 2.899
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 = 7.5/√13 = 2.08
Test statistic
t = 6.03/2.08 = 2.899
Answer
t = 2.899
With df = 12, the two-sided 5% critical value is 2.179; |t| is larger, so H₀ is rejected.
6.Hypothesis tests & confidence intervals
Answer: U = 197.9
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 = 1.645·12/√36 = 3.29
Interval limit
U = 194.6 + 3.29 = 197.9
Answer
U = 197.9
7.Linear regression & correlation
Answer: r = 0.9945
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 = 1556 − 42·184/6 = 268
Sxx = 364 − 42²/6 = 70
Syy = 6680 − 184²/6 = 1037
Correlation coefficient
r = 268/√(70·1037) = 0.9945
Answer
r = 0.9945
8.Linear regression & correlation
Answer: ŷ = 56.89
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 = 48/6 = 8
ȳ = 437/6 = 72.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 279
Regression line
b = 279/70 = 3.986
a = 72.83 − 3.986·8 = 40.95
ŷ = 40.95 + 3.986·x
Prediction
ŷ = 40.95 + 3.986·4 = 56.89
Answer
ŷ = 56.89
9.Chi-square tests
Answer: χ² = 28.48
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 rejected.
10.Chi-square tests
Answer: χ²crit = 18.48
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 − 1 = 7
Table
χ²crit(df = 7, α = 0.01) = 18.475
Answer
χ²crit = 18.475
Reject H₀ if the computed χ² is larger than this value.