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 0.3545. Find z to two decimals.
A = 0.3545
z =
2.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 99% of them lie (the 99th percentile).
μ = 500σ = 40P = 99%
x = h
3.Confidence intervalsMedium
A 95% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 6.2. What is the margin of error E₂ of a 99% interval based on 256 observations?
95%n₁ = 64E₁ = 6.299%n₂ = 256
E₂ =
4.Confidence intervalsHard
The population standard deviation is 20. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 2?
σ = 20E = 295%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 198.7 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 200.
μ₀ = 200x̄ = 198.7s = 5n = 100
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 36 measurements has mean 100.8. The population standard deviation is 5. Find the upper limit U of the 99% confidence interval for the mean.
x̄ = 100.8σ = 5n = 3699%
U =
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 5Σx = 15Σy = 290Σx² = 55Σy² = 17338Σxy = 938
r =
8.Linear regression & correlationHard
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 87?
n = 5ŷ = 87
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 “staff, bike”.
row total = 105column total = 87N = 185
E =
10.Chi-square testsHard
A chi-square test of independence is carried out on a contingency table with 2 rows and 6 columns (not counting the totals). How many degrees of freedom df does the test have?
2 × 6
df =
3Solutions
1.Normal distribution
Answer: z = 0.46
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 + 0.3545)/2 = 0.6772
Read the z-table backwards
Find 0.6772 inside the table and read off z.
z = 0.46
Answer
z = 0.46
2.Normal distribution
Answer: x = 593.1 h
Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.
z for the percentile
Φ(z) = 0.99 → z = 2.326
Back to x
x = 500 + 2.326·40 = 593.05 h
Answer
x = 593.05 h
3.Confidence intervals
Answer: E₂ = 4.074
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.
Ratio of margins
E₂/E₁ = (2.576/1.96)·√(64/256) = 0.6571
New margin
E₂ = 6.2·0.6571 = 4.074
Answer
E₂ = 4.074
4.Confidence intervals
Answer: n = 385
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 = (1.96·20/2)² = 384.2
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 385
Answer
n = 385
5.Hypothesis tests & confidence intervals
Answer: t = −2.6
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 = 5/√100 = 5/10 = 0.5
Test statistic
t = (198.7 − 200)/0.5 = −2.6
Answer
t = −2.6
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 = 102.9
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 = 2.576·5/√36 = 2.147
Interval limit
U = 100.8 + 2.147 = 102.9
Answer
U = 102.9
7.Linear regression & correlation
Answer: r = 0.9448
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.
Centred sums
Sxy = 938 − 15·290/5 = 68
Sxx = 55 − 15²/5 = 10
Syy = 17338 − 290²/5 = 518
Correlation coefficient
r = 68/√(10·518) = 0.9448
Answer
r = 0.9448
8.Linear regression & correlation
Answer: x = 17.59
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 = 110/5 = 22
ȳ = 549/5 = 109.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 465
Regression line
b = 465/90 = 5.167
a = 109.8 − 5.167·22 = −3.867
ŷ = −3.867 + 5.167·x
Solve for x
x = (87 − (−3.867))/5.167 = 17.59
Answer
x = 17.59
9.Chi-square tests
Answer: E = 49.38
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 = 105·87/185 = 49.38
Answer
E = 49.38
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.