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 0.0582. Find z to two decimals.
A = 0.0582
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 75% is to the right of it. Find the cut-off x.
μ = 250 gσ = 5 g75%
x = g
3.Confidence intervalsMedium
A sample of 30 items has mean 29 and standard deviation s = 20. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.
n = 30x̄ = 29s = 2099%t(29) = 2.756
lower limit =
4.Confidence intervalsHard
The population standard deviation is 38. 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 3?
σ = 38E = 390%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 49 students has a mean score of 79.14 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 79.14s = 10n = 49
t =
6.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.01 must detect a shift of the mean of 3 units with a power of 80%. The standard deviation is 20. Find the minimum sample size n.
The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.
n = 6
r =
8.Linear regression & correlationHard
The table gives 5 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 = 5ŷ = 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 = 2402 × 3
χ² =
10.Chi-square testsHard
A chi-square test of independence is carried out on a contingency table with 2 rows and 3 columns (not counting the totals). How many degrees of freedom df does the test have?
2 × 3
df =
3Solutions
1.Normal distribution
Answer: z = 1.57
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 − 0.0582 = 0.9418
Read the z-table backwards
Find 0.9418 inside the table and read off z.
z = 1.57
Answer
z = 1.57
2.Normal distribution
Answer: x = 246.7 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) = 0.25
z from the table
z ≈ −0.67
Back to the original units
x = 250 − 0.67·5 = 246.7 g
Answer
x = 246.7 g
3.Confidence intervals
Answer: lower limit = 18.94
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 = 2.756
Margin of error
E = 2.756·20/√30 = 10.06
Interval
29 ± 10.06 → [18.94, 39.06]
Answer
lower limit = 18.94
4.Confidence intervals
Answer: n = 435
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 = (1.645·38/3)² = 434.2
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 435
Answer
n = 435
5.Hypothesis tests & confidence intervals
Answer: t = 2.898
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 = 10/√49 = 10/7 = 1.429
Test statistic
t = (79.14 − 75)/1.429 = 2.898
Answer
t = 2.898
t0.025; df = 48 = 2.011
|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: n = 447
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)·20/3)² = 446.1
Round up
n = 447
Answer
n = 447
7.Linear regression & correlation
Answer: r = 0.9787
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 = 21/6 = 3.5
ȳ = 339/6 = 56.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 91.5
Syy = Σ(y − ȳ)² = 499.5
Correlation coefficient
r = 91.5/√(17.5·499.5) = 0.9787
Answer
r = 0.9787
8.Linear regression & correlation
Answer: x = 2.96
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 = 341/5 = 68.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 200
Regression line
b = 200/40 = 5
a = 68.2 − 5·6 = 38.2
ŷ = 38.2 + 5·x
Solve for x
x = (53 − 38.2)/5 = 2.96
Answer
x = 2.96
9.Chi-square tests
Answer: χ² = 15.8
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 rejected.
10.Chi-square tests
Answer: df = 2
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.