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
X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 117.8 s. Round z to two decimals.
μX = 60, σX = 5μY = 45, σY = 4X + Y > 117.8 s
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 97.5% of them lie (the 97.5th percentile).
μ = 178σ = 7P = 97.5%
x = cm
3.Confidence intervalsMedium
A sample of 25 items has mean 110.5 and standard deviation s = 12.5. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.
n = 25x̄ = 110.5s = 12.599%t(24) = 2.797
upper limit =
4.Confidence intervalsHard
The population standard deviation is 34. 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 2.5?
σ = 34E = 2.590%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 75.7 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 75.7s = 10n = 100
t =
6.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 80 against H₁: μ < 80 is done at α = 0.01 with a sample of 100; the standard deviation is known to be 15. In reality the mean is 74.3. Find the power 1 − β of the test in percent.
μ₀ = 80μ = 74.3σ = 15n = 100α = 0.01
1 − β = %
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.
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 = 62?
n = 5ŷ = 62
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 × 3
χ² =
10.Chi-square testsHard
A chi-square test of independence is carried out on a contingency table with 3 rows and 4 columns (not counting the totals). Using the table, find the critical value χ²crit at significance level α = 0.01.
α = 0.01
χ²crit =
3Solutions
1.Normal distribution
Answer: P = 0.02275
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).
Distribution of the sum
μ = 105 s, σ = √(5² + 4²) = 6.403 s
Standardise
z = (117.8 − 105)/6.403 = 2
Upper tail
P = 1 − Φ(2) = 1 − 0.9772 = 0.02275
Answer
P = 0.02275
2.Normal distribution
Answer: x = 191.7 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) = 0.975 → z = 1.96
Back to x
x = 178 + 1.96·7 = 191.72 cm
Answer
x = 191.72 cm
3.Confidence intervals
Answer: upper limit = 117.5
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 = 24, t = 2.797
Margin of error
E = 2.797·12.5/√25 = 6.992
Interval
110.5 ± 6.992 → [103.5, 117.5]
Answer
upper limit = 117.5
4.Confidence intervals
Answer: n = 501
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 = (1.645·34/2.5)² = 500.5
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 501
Answer
n = 501
5.Hypothesis tests & confidence intervals
Answer: t = 0.7
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).
Standard error
SE = 10/√100 = 10/10 = 1
Test statistic
t = (75.7 − 75)/1 = 0.7
Answer
t = 0.7
t0.025; df = 99 = 1.984
|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: 1 − β = 92.98 %
Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).
Critical sample mean
SE = 15/√100 = 1.5
x̄c = 80 − 2.326·1.5 = 76.51
z under the true mean
z = (76.51 − 74.3)/1.5 = 1.474
Power
power = P(Z < 1.474) = 92.98%
1 − β = 92.98%
Answer
1 − β = 92.98%
7.Linear regression & correlation
Answer: r = 0.9865
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 = 1724 − 25·306/5 = 194
Sxx = 165 − 25²/5 = 40
Syy = 19694 − 306²/5 = 966.8
Correlation coefficient
r = 194/√(40·966.8) = 0.9865
Answer
r = 0.9865
8.Linear regression & correlation
Answer: x = 3.536
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 = 20/5 = 4
ȳ = 326/5 = 65.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 69
Regression line
b = 69/10 = 6.9
a = 65.2 − 6.9·4 = 37.6
ŷ = 37.6 + 6.9·x
Solve for x
x = (62 − 37.6)/6.9 = 3.536
Answer
x = 3.536
9.Chi-square tests
Answer: χ² = 2.979
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: χ²crit = 16.81
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.
Degrees of freedom
df = (3 − 1)·(4 − 1) = 6
Table
χ²crit(df = 6, α = 0.01) = 16.812
Answer
χ²crit = 16.812
Reject H₀ if the computed χ² is larger than this value.