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
−3−2−10123zshaded area between −z and z = 0.2434

Drag z so that the shaded area between −z and z under the standard normal curve is 0.2434. Find z to two decimals.

A = 0.2434
z =
2.Normal distributionHard
10 cm0X − Y?

The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
3.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?99%

The battery life of two phone models is compared. Model 1: 100 phones, mean 67 h, known standard deviation 11 h. Model 2: 50 phones, mean 77.5 h, known standard deviation 10 h. Find the margin of error E of the 99% confidence interval for μ₁ − μ₂.

n₁ = 100x̄₁ = 67 hσ₁ = 11 hn₂ = 50x̄₂ = 77.5 hσ₂ = 10 h99%
E = h
4.Confidence intervalsHard
p ≈ 0.7, n = ?p̂ − Ep̂ + Ep̂E = 0.0495%

A pilot study suggests that about 0.7 of products are defective. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.04?

p ≈ 0.7E = 0.0495%
n =
5.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 30 parts has mean length 122 mm, and the line's known standard deviation is 13 mm. Line 2: a sample of 100 parts has mean 124.8 mm, known standard deviation 9 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 30x̄₁ = 122 mmσ₁ = 13 mmn₂ = 100x̄₂ = 124.8 mmσ₂ = 9 mm
z =
6.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 100 measurements has mean 122.1. The population standard deviation is 15. Find the lower limit L of the 99% confidence interval for the mean.

x̄ = 122.1σ = 15n = 10099%
L =
7.Linear regression & correlationMedium
xy3495557689721178xy51050607080

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.

n = 5Σx = 35Σy = 322Σx² = 285Σy² = 21318Σxy = 2404
r =
8.Linear regression & correlationHard
xy14035055876898011100xy510406080100x = 10

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (10, 93) is made. Find its residual e = y − ŷ.

n = 6(10, 93)
e =
9.Chi-square testsMedium
busbikecarTotalstudents4192878staff184438100Total595366178

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 = 100column total = 59N = 178
E =
10.Chi-square testsHard
passfailTotalplant A2872100plant B166480plant C4060100Total84196280dfα = 0.05α = 0.0113.8416.63525.9919.21037.81511.34549.48813.277511.07015.086612.59216.812714.06718.475815.50720.090916.91921.6661018.30723.209

The table shows observed counts. Test at α = 0.05 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 = 2803 × 2α = 0.05
χ² =

3Solutions

1.Normal distribution

Answer: z = 0.31

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 + 0.2434)/2 = 0.6217

Read the z-table backwards

Find 0.6217 inside the table and read off z.

z = 0.31

Answer

z = 0.31
2.Normal distribution

Answer: P = 0.7389

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 difference

μ = 180 − 170 = 10 cm, σ = 15.62 cm

Standardise

z = (0 − 10)/15.62 = −0.64

Probability

P = 1 − Φ(−0.64) = 0.7389

Answer

P = 0.7389
3.Confidence intervals

Answer: E = 4.615 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 = 67 − 77.5 = −10.5

Standard error

SE = √(11²/100 + 10²/50) = 1.792

Margin of error

E = 2.576·1.792 = 4.615

Answer

E = 4.615 h

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

4.Confidence intervals

Answer: n = 505

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 1.96²·0.7·0.3/0.04² = 504.2

Round up

n = 505

Answer

n = 505
5.Hypothesis tests & confidence intervals

Answer: z = −1.103

Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.

Standard error of the difference

SE = √(13²/30 + 9²/100) = √6.443 = 2.538

Test statistic

z = (122 − 124.8)/2.538 = −1.103

Answer

z = −1.103

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: L = 118.2

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).

Margin of error

E = 2.576·15/√100 = 3.864

Interval limit

L = 122.1 − 3.864 = 118.2

Answer

L = 118.2
7.Linear regression & correlation

Answer: r = 0.9838

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 = 2404 − 35·322/5 = 150
Sxx = 285 − 35²/5 = 40
Syy = 21318 − 322²/5 = 581.2

Correlation coefficient

r = 150/√(40·581.2) = 0.9838

Answer

r = 0.9838
8.Linear regression & correlation

Answer: e = 4.143

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 = 36/6 = 6
ȳ = 396/6 = 66

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 400

Regression line

b = 400/70 = 5.714
a = 66 − 5.714·6 = 31.71
ŷ = 31.71 + 5.714·x

Predicted value

ŷ = 31.71 + 5.714·10 = 88.86

Residual

e = 93 − 88.86 = 4.143

Answer

e = 4.143

The point lies above the line.

9.Chi-square tests

Answer: E = 33.15

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 = 100·59/178 = 33.15

Answer

E = 33.15
10.Chi-square tests

Answer: χ² = 8.762

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 counts

plant A: 30, 70
plant B: 24, 56
plant C: 30, 70

Chi-square statistic

χ² = (28 − 30)²/30 + (72 − 70)²/70 + (16 − 24)²/24 + (64 − 56)²/56 + (40 − 30)²/30 + (60 − 70)²/70
χ² = 8.762

Decision

df = 2, χ²crit = 5.991
8.762 > 5.991

Answer

χ² = 8.762

χ² is above the critical value, so H₀ is rejected: the variables appear to be related.

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