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 0.9884. Find z to two decimals.
A = 0.9884
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 top 10% is to the right of it. Find the cut-off x.
μ = 250 gσ = 5 g10%
x = g
3.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 30 phones, mean 124.5 h, known standard deviation 3 h. Model 2: 100 phones, mean 114.5 h, known standard deviation 15 h. Find the upper limit of the 99% confidence interval for μ₁ − μ₂.
The population standard deviation is 16. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 1.5?
σ = 16E = 1.590%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 36 students has a mean score of 101.1 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 100.
μ₀ = 100x̄ = 101.1s = 4n = 36
t =
6.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 200 against H₁: μ > 200 is done at α = 0.10 with a sample of 25; the standard deviation is known to be 7.5. In reality the mean is 203.75. Find the probability of a Type II error β in percent.
μ₀ = 200μ = 203.75σ = 7.5n = 25α = 0.10
β = %
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.
n = 6
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. Predict ŷ for x = 3.5.
n = 6x = 3.5
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 “day shift, minor”.
row total = 55column total = 57N = 136
E =
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 = 2403 × 2α = 0.01
χ² =
3Solutions
1.Normal distribution
Answer: z = −2.27
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 − 0.9884 = 0.0116
Read the z-table backwards
Find 0.0116 inside the table and read off z.
z = −2.27
Answer
z = −2.27
2.Normal distribution
Answer: x = 256.4 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) = 0.9
z from the table
z ≈ 1.28
Back to the original units
x = 250 + 1.28·5 = 256.4 g
Answer
x = 256.4 g
3.Confidence intervals
Answer: upper limit = 14.11 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 = 124.5 − 114.5 = 10
Standard error
SE = √(3²/30 + 15²/100) = 1.597
Margin of error
E = 2.576·1.597 = 4.114
Interval
10 ± 4.114 → [5.886, 14.11]
Answer
upper limit = 14.11 h
The interval does not contain 0, so the means differ significantly.
4.Confidence intervals
Answer: n = 308
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 = (1.645·16/1.5)² = 307.9
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 308
Answer
n = 308
5.Hypothesis tests & confidence intervals
Answer: t = 1.65
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 = 4/√36 = 4/6 = 0.6667
Test statistic
t = (101.1 − 100)/0.6667 = 1.65
Answer
t = 1.65
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: β = 11.16 %
Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).
Critical sample mean
SE = 7.5/√25 = 1.5
x̄c = 200 + 1.282·1.5 = 201.9
z under the true mean
z = (201.9 − 203.75)/1.5 = −1.218
Type II error
power = P(Z > −1.218) = 88.84%
β = 100% − 88.84% = 11.16%
Answer
β = 11.16%
7.Linear regression & correlation
Answer: r = 0.9989
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 = 146/6 = 24.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 226
Syy = Σ(y − ȳ)² = 731.3
Correlation coefficient
r = 226/√(70·731.3) = 0.9989
Answer
r = 0.9989
8.Linear regression & correlation
Answer: ŷ = 43.04
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 = 33/6 = 5.5
ȳ = 321/6 = 53.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 91.5
Regression line
b = 91.5/17.5 = 5.229
a = 53.5 − 5.229·5.5 = 24.74
ŷ = 24.74 + 5.229·x
Prediction
ŷ = 24.74 + 5.229·3.5 = 43.04
Answer
ŷ = 43.04
9.Chi-square tests
Answer: E = 23.05
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 = 55·57/136 = 23.05
Answer
E = 23.05
10.Chi-square tests
Answer: χ² = 1.354
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.