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 0.9199. Find z to two decimals.
A = 0.9199
z =
2.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 99% of them lie (the 99th percentile).
μ = 65σ = 10P = 99%
x = points
3.Confidence intervalsMedium
The blood pressure of 15 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 2.1 and standard deviation 7.6. Find the upper limit of the 95% confidence interval for the mean difference.
n = 15d̄ = 2.1sd = 7.695%t(14) = 2.145
upper limit = mmHg
4.Confidence intervalsHard
The population standard deviation is 24. 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 4?
σ = 24E = 495%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 101 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 100.
μ₀ = 100x̄ = 101s = 8n = 100
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 36 measurements has mean 118.3. The population standard deviation is 4. Find the lower limit L of the 99% confidence interval for the mean.
x̄ = 118.3σ = 4n = 3699%
L =
7.Linear regression & correlationMedium
The table gives 5 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 = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (24, 80) is made. Find its residual e = y − ŷ.
n = 6(24, 80)
e =
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 = 2202 × 3
χ² =
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. Using the table, find the critical value χ²crit at significance level α = 0.05.
α = 0.05
χ²crit =
3Solutions
1.Normal distribution
Answer: z = 1.75
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 + 0.9199)/2 = 0.9599
Read the z-table backwards
Find 0.9599 inside the table and read off z.
z = 1.75
Answer
z = 1.75
2.Normal distribution
Answer: x = 88.26 points
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 = 65 + 2.326·10 = 88.26 points
Answer
x = 88.26 points
3.Confidence intervals
Answer: upper limit = 6.309 mmHg
Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.
Critical value
df = 14, t = 2.145
Margin of error
E = 2.145·7.6/√15 = 4.209
Interval
2.1 ± 4.209 → [−2.109, 6.309]
Answer
upper limit = 6.309 mmHg
4.Confidence intervals
Answer: n = 139
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 = (1.96·24/4)² = 138.3
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 139
Answer
n = 139
5.Hypothesis tests & confidence intervals
Answer: t = 1.25
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 = 8/√100 = 8/10 = 0.8
Test statistic
t = (101 − 100)/0.8 = 1.25
Answer
t = 1.25
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: L = 116.6
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).
Margin of error
E = 2.576·4/√36 = 1.717
Interval limit
L = 118.3 − 1.717 = 116.6
Answer
L = 116.6
7.Linear regression & correlation
Answer: r = 0.9669
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 = 11460 − 105·536/5 = 204
Sxx = 2245 − 105²/5 = 40
Syy = 58572 − 536²/5 = 1113
Correlation coefficient
r = 204/√(40·1113) = 0.9669
Answer
r = 0.9669
8.Linear regression & correlation
Answer: e = −2.143
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 = 132/6 = 22
ȳ = 450/6 = 75
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 250
Regression line
b = 250/70 = 3.571
a = 75 − 3.571·22 = −3.571
ŷ = −3.571 + 3.571·x
Predicted value
ŷ = −3.571 + 3.571·24 = 82.14
Residual
e = 80 − 82.14 = −2.143
Answer
e = −2.143
The point lies below the line.
9.Chi-square tests
Answer: χ² = 20.06
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 rejected.
10.Chi-square tests
Answer: χ²crit = 11.07
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 = 7 − 1 − 1 = 5
Table
χ²crit(df = 5, α = 0.05) = 11.070
Answer
χ²crit = 11.070
Reject H₀ if the computed χ² is larger than this value.