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 0.8836. Find z to two decimals.
A = 0.8836
z =
2.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 30% is to the left of it. Find the cut-off x.
μ = 500 hσ = 40 h30%
x = h
3.Confidence intervalsMedium
A sample of 16 items has mean 47 and standard deviation s = 9.5. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.
n = 16x̄ = 47s = 9.595%t(15) = 2.131
upper limit =
4.Confidence intervalsHard
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 81 students has a mean score of 118.4 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 118.4s = 10n = 81
t =
6.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 1.5? The standard deviation is 20. Find the smallest sample size n.
σ = 20E = 1.595%
n =
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = car age in years and y = price in k€. 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 = 55?
n = 5ŷ = 55
x =
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 “night shift, no defect”.
row total = 82column total = 49N = 130
E =
10.Chi-square testsHard
A chi-square goodness-of-fit test checks whether counts in 7 categories follow a Poisson distribution whose mean is estimated from the same data. How many degrees of freedom df does the test have?
k = 7
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 + 0.8836)/2 = 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 = 479 h
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) = 0.3
z from the table
z ≈ −0.52
Back to the original units
x = 500 − 0.52·40 = 479 h
Answer
x = 479 h
3.Confidence intervals
Answer: upper limit = 52.06
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 = 15, t = 2.131
Margin of error
E = 2.131·9.5/√16 = 5.061
Interval
47 ± 5.061 → [41.94, 52.06]
Answer
upper limit = 52.06
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 = −1.44
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).
Standard error
SE = 10/√81 = 10/9 = 1.111
Test statistic
t = (118.4 − 120)/1.111 = −1.44
Answer
t = −1.44
t0.025; df = 80 = 1.99
|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: n = 683
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 = (1.96·20/1.5)² = 683
Round up
n = 683
Answer
n = 683
7.Linear regression & correlation
Answer: r = −0.9768
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 = 107/6 = 17.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −51.5
Syy = Σ(y − ȳ)² = 158.8
Correlation coefficient
r = −51.5/√(17.5·158.8) = −0.9768
Answer
r = −0.9768
8.Linear regression & correlation
Answer: x = 4.955
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 = 298/5 = 59.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 176
Regression line
b = 176/40 = 4.4
a = 59.6 − 4.4·6 = 33.2
ŷ = 33.2 + 4.4·x
Solve for x
x = (55 − 33.2)/4.4 = 4.955
Answer
x = 4.955
9.Chi-square tests
Answer: E = 30.91
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 = 82·49/130 = 30.91
Answer
E = 30.91
10.Chi-square tests
Answer: df = 5
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.