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 0.9700. Find z to two decimals.
A = 0.9700
z =
2.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 25% is to the right of it. Find the cut-off x.
μ = 100 pointsσ = 15 points25%
x = points
3.Confidence intervalsMedium
A sample of 25 items has mean 58 and standard deviation s = 6. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.
n = 25x̄ = 58s = 690%t(24) = 1.711
E =
4.Confidence intervalsHard
The population standard deviation is 12. 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 0.5?
σ = 12E = 0.599%
n =
5.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 50; the population standard deviation is known to be 12. A random sample of 64 observations has a mean of 51.89. Test H₀: μ = 50 against H₁: μ ≠ 50. Find the p-value p.
μ₀ = 50σ = 12n = 64x̄ = 51.89H₁: μ ≠ μ₀
p =
6.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 2.5? The standard deviation is 4. Find the smallest sample size n.
σ = 4E = 2.590%
n =
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 = 20Σy = 363Σx² = 90Σy² = 26587Σxy = 1500
r =
8.Linear regression & correlationHard
The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 68?
n = 5ŷ = 68
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 = 1802 × 2
χ² =
10.Chi-square testsHard
A chi-square goodness-of-fit test compares observed counts in 6 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?
k = 6
df =
3Solutions
1.Normal distribution
Answer: z = 2.17
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 + 0.9700)/2 = 0.9850
Read the z-table backwards
Find 0.9850 inside the table and read off z.
z = 2.17
Answer
z = 2.17
2.Normal distribution
Answer: x = 110.1 points
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) = 0.75
z from the table
z ≈ 0.67
Back to the original units
x = 100 + 0.67·15 = 110.1 points
Answer
x = 110.1 points
3.Confidence intervals
Answer: E = 2.053
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 = 24, t = 1.711
Margin of error
E = 1.711·6/√25 = 2.053
Answer
E = 2.053
4.Confidence intervals
Answer: n = 3823
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 = (2.576·12/0.5)² = 3822
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 3823
Answer
n = 3823
5.Hypothesis tests & confidence intervals
Answer: p = 0.2077
Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.
Test statistic
SE = σ/√n = 12/√64 = 1.5
z = (51.89 − 50)/1.5 = 1.26
Tail area
Φ(1.26) = 0.8962
1 − Φ(1.26) = 0.1038
p-value
p = 2·0.1038 = 0.2077
Answer
p = 0.2077
p > 0.05, so H₀ is not rejected at the 5% level.
6.Hypothesis tests & confidence intervals
Answer: n = 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).
Solve for n
n = (1.645·4/2.5)² = 6.927
Round up
n = 7
Answer
n = 7
7.Linear regression & correlation
Answer: r = 0.994
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 = 1500 − 20·363/5 = 48
Sxx = 90 − 20²/5 = 10
Syy = 26587 − 363²/5 = 233.2
Correlation coefficient
r = 48/√(10·233.2) = 0.994
Answer
r = 0.994
8.Linear regression & correlation
Answer: x = 4.08
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 = 25/5 = 5
ȳ = 366/5 = 73.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 226
Regression line
b = 226/40 = 5.65
a = 73.2 − 5.65·5 = 44.95
ŷ = 44.95 + 5.65·x
Solve for x
x = (68 − 44.95)/5.65 = 4.08
Answer
x = 4.08
9.Chi-square tests
Answer: χ² = 6.857
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 = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is rejected.
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.