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
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 76.75 and 130 points? Give P as a decimal.
μ = 100σ = 15x1 = 76.75x2 = 130
P =
2.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 99% of them lie (the 99th percentile).
μ = 178σ = 7P = 99%
x = cm
3.Confidence intervalsMedium
A sample of 8 items has mean 106.5 and standard deviation s = 7.5. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.
n = 8x̄ = 106.5s = 7.590%t(7) = 1.895
upper limit =
4.Confidence intervalsHard
The population standard deviation is 40. 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 0.5?
σ = 40E = 0.590%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 25 students has a mean score of 112.6 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 112.6s = 12n = 25
t =
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 0.5? The standard deviation is 15. Find the smallest sample size n.
σ = 15E = 0.590%
n =
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.
The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 6.5.
n = 6x = 6.5
ŷ =
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 = 1002 × 3
χ² =
10.Chi-square testsHard
The table shows observed counts. Test at α = 0.01 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 = 2603 × 2α = 0.01
χ² =
3Solutions
1.Normal distribution
Answer: P = 0.9167
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-scores
z₁ = (76.75 − 100)/15 = −1.55
z₂ = (130 − 100)/15 = 2
Area between
P = Φ(2) − Φ(−1.55) = 0.9772 − 0.06057 = 0.9167
Answer
P = 0.9167
2.Normal distribution
Answer: x = 194.3 cm
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 = 178 + 2.326·7 = 194.28 cm
Answer
x = 194.28 cm
3.Confidence intervals
Answer: upper limit = 111.5
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 = 7, t = 1.895
Margin of error
E = 1.895·7.5/√8 = 5.025
Interval
106.5 ± 5.025 → [101.5, 111.5]
Answer
upper limit = 111.5
4.Confidence intervals
Answer: n = 17319
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 = (1.645·40/0.5)² = 17319
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 17319
Answer
n = 17319
5.Hypothesis tests & confidence intervals
Answer: t = −3.083
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 = 12/√25 = 12/5 = 2.4
Test statistic
t = (112.6 − 120)/2.4 = −3.083
Answer
t = −3.083
t0.025; df = 24 = 2.064
|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: n = 2436
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 = (1.645·15/0.5)² = 2435
Round up
n = 2436
Answer
n = 2436
7.Linear regression & correlation
Answer: r = 0.9949
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 = 14490 − 147·567/6 = 598.5
Sxx = 3759 − 147²/6 = 157.5
Syy = 55879 − 567²/6 = 2298
Correlation coefficient
r = 598.5/√(157.5·2298) = 0.9949
Answer
r = 0.9949
8.Linear regression & correlation
Answer: ŷ = 68.47
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 = 33/6 = 5.5
ȳ = 387/6 = 64.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 69.5
Regression line
b = 69.5/17.5 = 3.971
a = 64.5 − 3.971·5.5 = 42.66
ŷ = 42.66 + 3.971·x
Prediction
ŷ = 42.66 + 3.971·6.5 = 68.47
Answer
ŷ = 68.47
9.Chi-square tests
Answer: χ² = 0.3704
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 = 6.508
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.