Statistics 2: Statistical Inference: cram pack

Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.

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1Formulas and methods

Normal distribution

z-score
z = (x − μ)/σ
Below x
P(X < x) = Φ(z)
Above x
P(X > x) = 1 − Φ(z)
Between
Φ(z2) − Φ(z1)
Percentile
x = μ + z·σ

Method

  1. Convert the value to a z-score: z = (x − μ)/σ.
  2. Look up Φ(z), the area to the left of z.
  3. For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
  4. For a percentile, work backwards: find z for the area, then x = μ + z·σ.

Confidence intervals

σ known
x̄ ± z·σ/√n
σ unknown
x̄ ± t·s/√n, df = n − 1
Critical z
90%: 1.645, 95%: 1.96, 99%: 2.576
Sample size
n = (z·σ/E)², rounded up

Method

  1. Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
  2. Find the critical value for the confidence level.
  3. Margin of error: E = critical value × standard deviation/√n.
  4. Interval: x̄ ± E.

Hypothesis tests & confidence intervals

z-test for a mean
z = (x̄ − μ0)/(σ/√n)
t-test for a mean
t = (x̄ − μ0)/(s/√n), df = n − 1
Two-sided at 5%
reject H0 if |z| > 1.96
Confidence interval
x̄ ± z·σ/√n

Method

  1. State H₀ and H₁ and decide whether the test is one- or two-sided.
  2. Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
  3. Compare with the critical value, or find the p-value.
  4. Reject H₀ if the statistic is beyond the critical value (p below α), and state the conclusion in words.

Linear regression & correlation

Sums of squares
Sxx = Σ(x − x̄)² = Σx² − (Σx)²/n, Sxy = Σxy − ΣxΣy/n
Slope
b = Sxy/Sxx
Intercept
a = ȳ − b·x̄
Correlation
r = Sxy/√(SxxSyy)
Residual
e = y − ŷ

Chi-square tests

Test statistic
χ² = Σ(O − E)²/E
Goodness of fit
E = n·p, df = k − 1 − (estimated parameters)
Independence
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
200 mm199.6 mmX + Y?

X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 199.6 mm. Round z to two decimals.

μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 199.6 mm
P =
2.Normal distributionHard
μ = 500x = ?90%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 90% of them lie (the 90th percentile).

μ = 500σ = 40P = 90%
x = h
3.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E95%

The battery life of two phone models is compared. Model 1: 30 phones, mean 54.5 h, known standard deviation 19 h. Model 2: 100 phones, mean 41 h, known standard deviation 11 h. Find the lower limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 30x̄₁ = 54.5 hσ₁ = 19 hn₂ = 100x̄₂ = 41 hσ₂ = 11 h95%
lower limit = h
4.Confidence intervalsHard
n = 15Lσ = ?s = 8.795%

The diameters of 15 machined shafts have a sample standard deviation of 8.7 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 = 8.7 µmχ² = 5.629, 26.119
σ = µm
5.Hypothesis tests & confidence intervalsMedium
−2.2622.262t = ?df = 9

10 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is −3.19 and the standard deviation of the differences is 2.8. Find the paired test statistic t for H₀: μd = 0.

n = 10d̄ = −3.19sd = 2.8
t =
6.Hypothesis tests & confidence intervalsHard
σ = 10, n = ?x̄ − Ex̄ + Ex̄E = 299%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 10. Find the smallest sample size n.

σ = 10E = 299%
n =
7.Linear regression & correlationMedium
xy28476128161020xy510101520

The table gives 5 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 30Σy = 63Σx² = 220Σy² = 913Σxy = 444
r =
8.Linear regression & correlationHard
xy351446553654764869xy468506070ŷ = 48

The table gives 6 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 = 48?

n = 6ŷ = 48
x =
9.Chi-square testsMedium
busbikecarTotalstudents39131062staff22411881Total615428143

The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “staff, bus”.

row total = 81column total = 61N = 143
E =
10.Chi-square testsHard
dfα = 0.05α = 0.0113.8416.63525.9919.21037.81511.34549.48813.277511.07015.086612.59216.812714.06718.475815.50720.090916.91921.6661018.30723.209

A chi-square test of independence is carried out on a contingency table with 2 rows and 11 columns (not counting the totals). Using the table, find the critical value χ²crit at significance level α = 0.05.

α = 0.05
χ²crit =

3Solutions

1.Normal distribution

Answer: P = 0.7881

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 = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm

Standardise

z = (199.6 − 200)/0.5 = −0.8

Upper tail

P = 1 − Φ(−0.8) = 1 − 0.2119 = 0.7881

Answer

P = 0.7881
2.Normal distribution

Answer: x = 551.3 h

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 = 500 + 1.282·40 = 551.26 h

Answer

x = 551.26 h
3.Confidence intervals

Answer: lower limit = 6.367 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 = 54.5 − 41 = 13.5

Standard error

SE = √(19²/30 + 11²/100) = 3.639

Margin of error

E = 1.96·3.639 = 7.133

Interval

13.5 ± 7.133 → [6.367, 20.63]

Answer

lower limit = 6.367 h

The interval does not contain 0, so the means differ significantly.

4.Confidence intervals

Answer: σ = 13.72 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 = 14·8.7² = 1060

Limits for σ²

1060/26.119 = 40.57
1060/5.629 = 188.3

Limit for σ

σ = √188.3 = 13.72 µm

Answer

σ = 13.72 µm

95% interval for σ: [6.369, 13.72] µm.

5.Hypothesis tests & confidence intervals

Answer: t = −3.603

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 of the mean difference

SE = 2.8/√10 = 0.8854

Test statistic

t = −3.19/0.8854 = −3.603

Answer

t = −3.603

With df = 9, the two-sided 5% critical value is 2.262; |t| is larger, so H₀ is rejected.

6.Hypothesis tests & confidence intervals

Answer: n = 166

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 = (2.576·10/2)² = 165.9

Round up

n = 166

Answer

n = 166
7.Linear regression & correlation

Answer: r = 0.9558

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 = 444 − 30·63/5 = 66
Sxx = 220 − 30²/5 = 40
Syy = 913 − 63²/5 = 119.2

Correlation coefficient

r = 66/√(40·119.2) = 0.9558

Answer

r = 0.9558
8.Linear regression & correlation

Answer: x = 3.529

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 = 337/6 = 56.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 72.5

Regression line

b = 72.5/17.5 = 4.143
a = 56.17 − 4.143·5.5 = 33.38
ŷ = 33.38 + 4.143·x

Solve for x

x = (48 − 33.38)/4.143 = 3.529

Answer

x = 3.529
9.Chi-square tests

Answer: E = 34.55

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 = 81·61/143 = 34.55

Answer

E = 34.55
10.Chi-square tests

Answer: χ²crit = 18.31

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 = (2 − 1)·(11 − 1) = 10

Table

χ²crit(df = 10, α = 0.05) = 18.307

Answer

χ²crit = 18.307

Reject H₀ if the computed χ² is larger than this value.

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