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 0.2358. Find z to two decimals.
A = 0.2358
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
A sample of 30 items has mean 90 and standard deviation s = 2.5. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.
n = 30x̄ = 90s = 2.595%t(29) = 2.045
upper limit =
4.Confidence intervalsHard
The diameters of 11 machined shafts have a sample standard deviation of 5.6 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 10, 2.5% of the χ² distribution lies below 3.247 and 2.5% lies above 20.483.
n = 11s = 5.6 µmχ² = 3.247, 20.483
σ = µm
5.Hypothesis tests & confidence intervalsMedium
A sample of 25 students has a mean score of 102.1 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 100.
μ₀ = 100x̄ = 102.1s = 15n = 25
t =
6.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.01 must detect a shift of the mean of 4 units with a power of 80%. The standard deviation is 21. Find the minimum sample size n.
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. 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. A new observation (6, 55) is made. Find its residual e = y − ŷ.
n = 6(6, 55)
e =
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 × 3
χ² =
10.Chi-square testsHard
The table shows observed counts. Test at α = 0.05 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 = 0.05
χ² =
3Solutions
1.Normal distribution
Answer: z = 0.3
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 + 0.2358)/2 = 0.6179
Read the z-table backwards
Find 0.6179 inside the table and read off z.
z = 0.30
Answer
z = 0.30
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 = 90.93
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 = 2.045
Margin of error
E = 2.045·2.5/√30 = 0.9334
Interval
90 ± 0.9334 → [89.07, 90.93]
Answer
upper limit = 90.93
4.Confidence intervals
Answer: σ = 9.828 µm
Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.
Sum of squares
(n − 1)s² = 10·5.6² = 313.6
Limits for σ²
313.6/20.483 = 15.31
313.6/3.247 = 96.58
Limit for σ
σ = √96.58 = 9.828 µm
Answer
σ = 9.828 µm
95% interval for σ: [3.913, 9.828] µm.
5.Hypothesis tests & confidence intervals
Answer: t = 0.7
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/√25 = 15/5 = 3
Test statistic
t = (102.1 − 100)/3 = 0.7
Answer
t = 0.7
t0.025; df = 24 = 2.064
|t| < 2.064, so H₀ is not rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: n = 277
Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.
Solve for n
n = ((2.326 + 0.842)·21/4)² = 276.6
Round up
n = 277
Answer
n = 277
7.Linear regression & correlation
Answer: r = 0.9925
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 = 141/6 = 23.5
ȳ = 631/6 = 105.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 826.5
Syy = Σ(y − ȳ)² = 4403
Correlation coefficient
r = 826.5/√(157.5·4403) = 0.9925
Answer
r = 0.9925
8.Linear regression & correlation
Answer: e = −1.5
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 = 36/6 = 6
ȳ = 339/6 = 56.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 227
Regression line
b = 227/70 = 3.243
a = 56.5 − 3.243·6 = 37.04
ŷ = 37.04 + 3.243·x
Predicted value
ŷ = 37.04 + 3.243·6 = 56.5
Residual
e = 55 − 56.5 = −1.5
Answer
e = −1.5
The point lies below the line.
9.Chi-square tests
Answer: χ² = 0.8444
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 = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is not rejected.
10.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.