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 0.8385. Find z to two decimals.
A = 0.8385
z =
2.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 20% of them lie (the 20th percentile).
μ = 178σ = 7P = 20%
x = cm
3.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 50 phones, mean 133 h, known standard deviation 16 h. Model 2: 80 phones, mean 128 h, known standard deviation 12 h. Find the margin of error E of the 99% confidence interval for μ₁ − μ₂.
A pilot study suggests that about 0.8 of products are defective. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.04?
p ≈ 0.8E = 0.0495%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 64 students has a mean score of 503.9 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 503.9s = 15n = 64
t =
6.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 3? The standard deviation is 12. Find the smallest sample size n.
σ = 12E = 390%
n =
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
8.Linear regression & correlationHard
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 21.
n = 6x = 21
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 = 1402 × 2
χ² =
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. How many degrees of freedom df does the test have?
k = 5
df =
3Solutions
1.Normal distribution
Answer: z = 1.4
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 + 0.8385)/2 = 0.9192
Read the z-table backwards
Find 0.9192 inside the table and read off z.
z = 1.40
Answer
z = 1.40
2.Normal distribution
Answer: x = 172.1 cm
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 = 178 + −0.8416·7 = 172.11 cm
Answer
x = 172.11 cm
3.Confidence intervals
Answer: E = 6.776 h
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.
Point estimate
x̄₁ − x̄₂ = 133 − 128 = 5
Standard error
SE = √(16²/50 + 12²/80) = 2.631
Margin of error
E = 2.576·2.631 = 6.776
Answer
E = 6.776 h
The interval contains 0, so the data do not show a significant difference.
4.Confidence intervals
Answer: n = 385
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 = 1.96²·0.8·0.2/0.04² = 384.2
Round up
n = 385
Answer
n = 385
5.Hypothesis tests & confidence intervals
Answer: t = 2.08
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 = 15/√64 = 15/8 = 1.875
Test statistic
t = (503.9 − 500)/1.875 = 2.08
Answer
t = 2.08
t0.025; df = 63 = 1.998
|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: n = 44
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).
Solve for n
n = (1.645·12/3)² = 43.3
Round up
n = 44
Answer
n = 44
7.Linear regression & correlation
Answer: r² = 0.9836
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 = 30/5 = 6
ȳ = 90/5 = 18
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 120
Syy = Σ(y − ȳ)² = 366
Coefficient of determination
r² = 120²/(40·366) = 0.9836
Answer
r² = 0.9836
About 98.36% of the variation in y is explained by the line.
8.Linear regression & correlation
Answer: ŷ = 56.1
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 = 138/6 = 23
ȳ = 364/6 = 60.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 160
Regression line
b = 160/70 = 2.286
a = 60.67 − 2.286·23 = 8.095
ŷ = 8.095 + 2.286·x
Prediction
ŷ = 8.095 + 2.286·21 = 56.1
Answer
ŷ = 56.1
9.Chi-square tests
Answer: χ² = 14.7
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: df = 3
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.