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 to the right of z = 0.0143

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0143. Find z to two decimals.

A = 0.0143
z =
2.Normal distributionHard
4 s0X − Y?

Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
3.Confidence intervalsMedium
n = 7, sd​ = 3.20L = ?Ud̄ = 6.6E95%

The blood pressure of 7 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 6.6 and standard deviation 3.2. Find the lower limit of the 95% confidence interval for the mean difference.

n = 7d̄ = 6.6sd = 3.295%t(6) = 2.447
lower limit = mmHg
4.Confidence intervalsHard
σ = 15, n = ?x̄ − Ex̄ + Ex̄E = 290%

The population standard deviation is 15. 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?

σ = 15E = 290%
n =
5.Hypothesis tests & confidence intervalsMedium
−2.1452.145t = ?df = 14

15 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 2.48 and the standard deviation of the differences is 6.4. Find the paired test statistic t for H₀: μd = 0.

n = 15d̄ = 2.48sd = 6.4
t =
6.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 80 against H₁: μ < 80 is done at α = 0.01 with a sample of 25; the standard deviation is known to be 2.5. In reality the mean is 79.15. Find the power 1 − β of the test in percent.

μ₀ = 80μ = 79.15σ = 2.5n = 25α = 0.01
1 − β = %
7.Linear regression & correlationMedium
xy122220322420516618xy24616182022

The table gives 6 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.

n = 6
r =
8.Linear regression & correlationHard
xy2741962383110411247xy5102040x = 5

The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 5.

n = 6x = 5
ŷ =
9.Chi-square testsMedium
no defectminormajorTotalday shift8394289night shift19242568Total276367157

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

row total = 68column total = 67N = 157
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 goodness-of-fit test compares observed counts in 9 categories with fully specified expected proportions (no parameters are estimated from the 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 = 2.19

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 − 0.0143 = 0.9857

Read the z-table backwards

Find 0.9857 inside the table and read off z.

z = 2.19

Answer

z = 2.19
2.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 difference

μ = 52 − 48 = 4 s, σ = 5 s

Standardise

z = (0 − 4)/5 = −0.8

Probability

P = 1 − Φ(−0.8) = 0.7881

Answer

P = 0.7881
3.Confidence intervals

Answer: lower limit = 3.64 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 = 6, t = 2.447

Margin of error

E = 2.447·3.2/√7 = 2.96

Interval

6.6 ± 2.96 → [3.64, 9.56]

Answer

lower limit = 3.64 mmHg
4.Confidence intervals

Answer: n = 153

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 = (1.645·15/2)² = 152.2

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 153

Answer

n = 153
5.Hypothesis tests & confidence intervals

Answer: t = 1.501

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 = 6.4/√15 = 1.652

Test statistic

t = 2.48/1.652 = 1.501

Answer

t = 1.501

With df = 14, the two-sided 5% critical value is 2.145; |t| is smaller, so H₀ is not rejected.

6.Hypothesis tests & confidence intervals

Answer: 1 − β = 26.57 %

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 = 2.5/√25 = 0.5
x̄c = 80 − 2.326·0.5 = 78.84

z under the true mean

z = (78.84 − 79.15)/0.5 = −0.626

Power

power = P(Z < −0.626) = 26.57%
1 − β = 26.57%

Answer

1 − β = 26.57%
7.Linear regression & correlation

Answer: r = −0.7773

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 = 21/6 = 3.5
ȳ = 118/6 = 19.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −17
Syy = Σ(y − ȳ)² = 27.33

Correlation coefficient

r = −17/√(17.5·27.33) = −0.7773

Answer

r = −0.7773
8.Linear regression & correlation

Answer: ŷ = 20.17

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 = 42/6 = 7
ȳ = 168/6 = 28

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 274

Regression line

b = 274/70 = 3.914
a = 28 − 3.914·7 = 0.6
ŷ = 0.6 + 3.914·x

Prediction

ŷ = 0.6 + 3.914·5 = 20.17

Answer

ŷ = 20.17
9.Chi-square tests

Answer: E = 29.02

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 = 68·67/157 = 29.02

Answer

E = 29.02
10.Chi-square tests

Answer: χ²crit = 15.51

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 = 9 − 1 = 8

Table

χ²crit(df = 8, α = 0.05) = 15.507

Answer

χ²crit = 15.507

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

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