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 0.9162. Find z to two decimals.
A = 0.9162
z =
2.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 15% is to the left of it. Find the cut-off x.
μ = 250 gσ = 5 g15%
x = g
3.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 80 phones, mean 106 h, known standard deviation 5 h. Model 2: 60 phones, mean 92 h, known standard deviation 9 h. Find the upper limit of the 99% confidence interval for μ₁ − μ₂.
The population standard deviation is 20. 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 4?
σ = 20E = 499%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 197.4 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 200.
μ₀ = 200x̄ = 197.4s = 12n = 100
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 81 measurements has mean 120.7. The population standard deviation is 8. Find the upper limit U of the 95% confidence interval for the mean.
x̄ = 120.7σ = 8n = 8195%
U =
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.
n = 5
r =
8.Linear regression & correlationHard
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 53?
n = 6ŷ = 53
x =
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 compares observed counts in 5 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?
k = 5
df =
3Solutions
1.Normal distribution
Answer: z = −1.38
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 − 0.9162 = 0.0838
Read the z-table backwards
Find 0.0838 inside the table and read off z.
z = −1.38
Answer
z = −1.38
2.Normal distribution
Answer: x = 244.8 g
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) = 0.15
z from the table
z ≈ −1.04
Back to the original units
x = 250 − 1.04·5 = 244.8 g
Answer
x = 244.8 g
3.Confidence intervals
Answer: upper limit = 17.32 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 = 106 − 92 = 14
Standard error
SE = √(5²/80 + 9²/60) = 1.289
Margin of error
E = 2.576·1.289 = 3.321
Interval
14 ± 3.321 → [10.68, 17.32]
Answer
upper limit = 17.32 h
The interval does not contain 0, so the means differ significantly.
4.Confidence intervals
Answer: n = 166
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 = (2.576·20/4)² = 165.9
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 166
Answer
n = 166
5.Hypothesis tests & confidence intervals
Answer: t = −2.167
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 = 12/√100 = 12/10 = 1.2
Test statistic
t = (197.4 − 200)/1.2 = −2.167
Answer
t = −2.167
t0.025; df = 99 = 1.984
|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: U = 122.4
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 = 1.96·8/√81 = 1.742
Interval limit
U = 120.7 + 1.742 = 122.4
Answer
U = 122.4
7.Linear regression & correlation
Answer: r = 0.9979
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 = 88/5 = 17.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 102
Syy = Σ(y − ȳ)² = 261.2
Correlation coefficient
r = 102/√(40·261.2) = 0.9979
Answer
r = 0.9979
8.Linear regression & correlation
Answer: x = 10.83
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 = 277/6 = 46.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 125
Regression line
b = 125/70 = 1.786
a = 46.17 − 1.786·7 = 33.67
ŷ = 33.67 + 1.786·x
Solve for x
x = (53 − 33.67)/1.786 = 10.83
Answer
x = 10.83
9.Chi-square tests
Answer: χ² = 1.886
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 not rejected.
10.Chi-square tests
Answer: df = 4
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.