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 0.9615. Find z to two decimals.
A = 0.9615
z =
2.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded bottom 10% is to the left of it. Find the cut-off x.
μ = 178 cmσ = 7 cm10%
x = cm
3.Confidence intervalsMedium
A sample of 30 items has mean 39 and standard deviation s = 18. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.
n = 30x̄ = 39s = 1890%t(29) = 1.699
E =
4.Confidence intervalsHard
A pilot study suggests that about 0.7 of products are defective. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.025?
p ≈ 0.7E = 0.02599%
n =
5.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 60 parts has mean length 110.5 mm, and the line's known standard deviation is 16 mm. Line 2: a sample of 100 parts has mean 104.2 mm, known standard deviation 9 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
In a trial, 105 of 150 patients on treatment A recover, and 146 of 200 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 105n₁ = 150x₂ = 146n₂ = 200
z =
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.
n = 5
r =
8.Linear regression & correlationHard
The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 2.5.
n = 5x = 2.5
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 = 1602 × 2
χ² =
10.Chi-square testsHard
The table shows observed counts. Test at α = 0.01 whether the two variables are independent (no continuity correction). Compute the test statistic χ², then compare it with the critical value from the table.
N = 2003 × 2α = 0.01
χ² =
3Solutions
1.Normal distribution
Answer: z = 2.07
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 + 0.9615)/2 = 0.9808
Read the z-table backwards
Find 0.9808 inside the table and read off z.
z = 2.07
Answer
z = 2.07
2.Normal distribution
Answer: x = 169 cm
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
Φ(z) = 0.1
z from the table
z ≈ −1.28
Back to the original units
x = 178 − 1.28·7 = 169 cm
Answer
x = 169 cm
3.Confidence intervals
Answer: E = 5.583
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 = 29, t = 1.699
Margin of error
E = 1.699·18/√30 = 5.583
Answer
E = 5.583
4.Confidence intervals
Answer: n = 2230
Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.
Solve for n
n = 2.576²·0.7·0.3/0.025² = 2230
Round up
n = 2230
Answer
n = 2230
5.Hypothesis tests & confidence intervals
Answer: z = 2.796
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.
Standard error of the difference
SE = √(16²/60 + 9²/100) = √5.077 = 2.253
Test statistic
z = (110.5 − 104.2)/2.253 = 2.796
Answer
z = 2.796
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: z = −0.6167
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 = 105/150 = 0.7
p̂₂ = 146/200 = 0.73
p̂ = 251/350 = 0.7171
Standard error under H₀
SE = √(0.7171·0.2829·(1/150 + 1/200)) = 0.04865
Test statistic
z = (0.7 − 0.73)/0.04865 = −0.6167
Answer
z = −0.6167
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
7.Linear regression & correlation
Answer: r = 0.7165
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 = 15/5 = 3
ȳ = 260/5 = 52
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 27
Syy = Σ(y − ȳ)² = 142
Correlation coefficient
r = 27/√(10·142) = 0.7165
Answer
r = 0.7165
8.Linear regression & correlation
Answer: ŷ = 55.45
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 = 15/5 = 3
ȳ = 290/5 = 58
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 51
Regression line
b = 51/10 = 5.1
a = 58 − 5.1·3 = 42.7
ŷ = 42.7 + 5.1·x
Prediction
ŷ = 42.7 + 5.1·2.5 = 55.45
Answer
ŷ = 55.45
9.Chi-square tests
Answer: χ² = 4.949
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: χ² = 66.37
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.