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 between −z and z under the standard normal curve is 0.4843. Find z to two decimals.
A = 0.4843
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 20% of them lie (the 20th percentile).
μ = 500σ = 40P = 20%
x = h
3.Confidence intervalsMedium
A sample of 21 items has mean 68.5 and standard deviation s = 19.5. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.
n = 21x̄ = 68.5s = 19.595%t(20) = 2.086
lower limit =
4.Confidence intervalsHard
The population standard deviation is 19. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 5?
σ = 19E = 599%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 36 students has a mean score of 74.25 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 74.25s = 5n = 36
t =
6.Hypothesis tests & confidence intervalsHard
In a trial, 50 of 100 patients on treatment A recover, and 135 of 300 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 50n₁ = 100x₂ = 135n₂ = 300
z =
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 = 59Σx² = 364Σy² = 827Σxy = 544
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 = 7.
n = 6x = 7
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 = 77column total = 40N = 175
E =
10.Chi-square testsHard
A chi-square goodness-of-fit test checks whether counts in 5 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 = 0.65
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 + 0.4843)/2 = 0.7422
Read the z-table backwards
Find 0.7422 inside the table and read off z.
z = 0.65
Answer
z = 0.65
2.Normal distribution
Answer: x = 466.3 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) = 0.2 → z = −0.8416
Back to x
x = 500 + −0.8416·40 = 466.34 h
Answer
x = 466.34 h
3.Confidence intervals
Answer: lower limit = 59.62
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 = 20, t = 2.086
Margin of error
E = 2.086·19.5/√21 = 8.876
Interval
68.5 ± 8.876 → [59.62, 77.38]
Answer
lower limit = 59.62
4.Confidence intervals
Answer: n = 96
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 = (2.576·19/5)² = 95.82
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 96
Answer
n = 96
5.Hypothesis tests & confidence intervals
Answer: t = −0.9
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).
Standard error
SE = 5/√36 = 5/6 = 0.8333
Test statistic
t = (74.25 − 75)/0.8333 = −0.9
Answer
t = −0.9
t0.025; df = 35 = 2.03
|t| < 2.03, so H₀ is not rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: z = 0.8685
Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.
Sample proportions
p̂₁ = 50/100 = 0.5
p̂₂ = 135/300 = 0.45
p̂ = 185/400 = 0.4625
Standard error under H₀
SE = √(0.4625·0.5375·(1/100 + 1/300)) = 0.05757
Test statistic
z = (0.5 − 0.45)/0.05757 = 0.8685
Answer
z = 0.8685
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
7.Linear regression & correlation
Answer: r = 0.9966
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 = 544 − 42·59/6 = 131
Sxx = 364 − 42²/6 = 70
Syy = 827 − 59²/6 = 246.8
Correlation coefficient
r = 131/√(70·246.8) = 0.9966
Answer
r = 0.9966
8.Linear regression & correlation
Answer: ŷ = 18.83
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 = 113/6 = 18.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 187
Regression line
b = 187/70 = 2.671
a = 18.83 − 2.671·7 = 0.1333
ŷ = 0.1333 + 2.671·x
Prediction
ŷ = 0.1333 + 2.671·7 = 18.83
Answer
ŷ = 18.83
9.Chi-square tests
Answer: E = 17.6
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.
Expected count
E = 77·40/175 = 17.6
Answer
E = 17.6
10.Chi-square tests
Answer: χ²crit = 11.35
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 = 5 − 1 − 1 = 3
Table
χ²crit(df = 3, α = 0.01) = 11.345
Answer
χ²crit = 11.345
Reject H₀ if the computed χ² is larger than this value.