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 under the standard normal curve, to the right of z, is 0.1357. Find z to two decimals.
A = 0.1357
z =
2.Normal distributionHard
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 90% of them lie (the 90th percentile).
μ = 32σ = 6P = 90%
x = min
3.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 80 phones, mean 114 h, known standard deviation 16 h. Model 2: 60 phones, mean 103 h, known standard deviation 17 h. Find the upper limit of the 99% confidence interval for μ₁ − μ₂.
The diameters of 15 machined shafts have a sample standard deviation of 9.4 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 14, 2.5% of the χ² distribution lies below 5.629 and 2.5% lies above 26.119.
n = 15s = 9.4 µmχ² = 5.629, 26.119
σ = µm
5.Hypothesis tests & confidence intervalsMedium
A sample of 64 students has a mean score of 46.7 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 50.
μ₀ = 50x̄ = 46.7s = 12n = 64
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 81 measurements has mean 73.7. The population standard deviation is 4. Find the lower limit L of the 95% confidence interval for the mean.
x̄ = 73.7σ = 4n = 8195%
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. For which value x does the line predict ŷ = 63?
n = 6ŷ = 63
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 = 98column total = 79N = 191
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 = 1.1
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 − 0.1357 = 0.8643
Read the z-table backwards
Find 0.8643 inside the table and read off z.
z = 1.10
Answer
z = 1.10
2.Normal distribution
Answer: x = 39.69 min
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) = 0.9 → z = 1.282
Back to x
x = 32 + 1.282·6 = 39.69 min
Answer
x = 39.69 min
3.Confidence intervals
Answer: upper limit = 18.29 h
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.
Point estimate
x̄₁ − x̄₂ = 114 − 103 = 11
Standard error
SE = √(16²/80 + 17²/60) = 2.831
Margin of error
E = 2.576·2.831 = 7.294
Interval
11 ± 7.294 → [3.706, 18.29]
Answer
upper limit = 18.29 h
The interval does not contain 0, so the means differ significantly.
4.Confidence intervals
Answer: σ = 14.82 µm
Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.
Sum of squares
(n − 1)s² = 14·9.4² = 1237
Limits for σ²
1237/26.119 = 47.36
1237/5.629 = 219.8
Limit for σ
σ = √219.8 = 14.82 µm
Answer
σ = 14.82 µm
95% interval for σ: [6.882, 14.82] µm.
5.Hypothesis tests & confidence intervals
Answer: t = −2.2
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/√64 = 12/8 = 1.5
Test statistic
t = (46.7 − 50)/1.5 = −2.2
Answer
t = −2.2
t0.025; df = 63 = 1.998
|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: L = 72.83
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 = 1.96·4/√81 = 0.8711
Interval limit
L = 73.7 − 0.8711 = 72.83
Answer
L = 72.83
7.Linear regression & correlation
Answer: r = 0.933
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 = 4218 − 90·228/5 = 114
Sxx = 1660 − 90²/5 = 40
Syy = 10770 − 228²/5 = 373.2
Correlation coefficient
r = 114/√(40·373.2) = 0.933
Answer
r = 0.933
8.Linear regression & correlation
Answer: x = 17.91
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 = 120/6 = 20
ȳ = 419/6 = 69.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 229
Regression line
b = 229/70 = 3.271
a = 69.83 − 3.271·20 = 4.405
ŷ = 4.405 + 3.271·x
Solve for x
x = (63 − 4.405)/3.271 = 17.91
Answer
x = 17.91
9.Chi-square tests
Answer: E = 40.53
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 = 98·79/191 = 40.53
Answer
E = 40.53
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.