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
452540P = ?

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 452 and 540 h? Give P as a decimal.

μ = 500σ = 40x1 = 452x2 = 540
P =
2.Normal distributionHard
557085100115130145pointstop 75%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 75% is to the right of it. Find the cut-off x.

μ = 100 pointsσ = 15 points75%
x = points
3.Confidence intervalsMedium
n = 30, s = 2.5LUx̄ = 101.5E = ?99%

A sample of 30 items has mean 101.5 and standard deviation s = 2.5. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 30x̄ = 101.5s = 2.599%t(29) = 2.756
E =
4.Confidence intervalsHard
μ₁ − μ₂, t(12) = 3.0550LUx̄₁ − x̄₂E = ?99%

Two fertilisers are tested. Plot group 1: 7 plots, mean yield 17.5 kg, standard deviation 8 kg. Plot group 2: 7 plots, mean 14 kg, standard deviation 5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 99% confidence interval for μ₁ − μ₂.

n₁ = 7x̄₁ = 17.5s₁ = 8n₂ = 7x̄₂ = 14s₂ = 599%t(12) = 3.055
E = kg
5.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 77.1 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 77.1s = 8n = 64
t =
6.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 50 against H₁: μ > 50 is done at α = 0.01 with a sample of 64; the standard deviation is known to be 4. In reality the mean is 51.15. Find the power 1 − β of the test in percent.

μ₀ = 50μ = 51.15σ = 4n = 64α = 0.01
1 − β = %
7.Linear regression & correlationMedium
xy155617621960216923822577xy152025607080

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.

n = 6
r =
8.Linear regression & correlationHard
xy356454564663768xy46556065x = 5.5

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (5.5, 58) is made. Find its residual e = y − ŷ.

n = 5(5.5, 58)
e =
9.Chi-square testsMedium
coughno coughTotalsmokers152540non-smokers7347120Total8872160

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 = 1602 × 2
χ² =
10.Chi-square testsHard
passfailTotalline 16238100line 254660line 3382260Total15466220dfα = 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 = 0.01 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 = 2203 × 2α = 0.01
χ² =

3Solutions

1.Normal distribution

Answer: P = 0.7263

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-scores

z₁ = (452 − 500)/40 = −1.2
z₂ = (540 − 500)/40 = 1

Area between

P = Φ(1) − Φ(−1.2) = 0.8413 − 0.1151 = 0.7263

Answer

P = 0.7263
2.Normal distribution

Answer: x = 90 points

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

Φ(z) = 0.25

z from the table

z ≈ −0.67

Back to the original units

x = 100 − 0.67·15 = 90 points

Answer

x = 90 points
3.Confidence intervals

Answer: E = 1.258

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.

Critical value

df = n − 1 = 29, t = 2.756

Margin of error

E = 2.756·2.5/√30 = 1.258

Answer

E = 1.258
4.Confidence intervals

Answer: E = 10.89 kg

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.

Pooled standard deviation

sp² = (6·8² + 6·5²)/12 = 44.5
sp = 6.671

Margin of error

E = 3.055·6.671·√(1/7 + 1/7) = 10.89

Answer

E = 10.89 kg
5.Hypothesis tests & confidence intervals

Answer: t = 2.1

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/√64 = 8/8 = 1

Test statistic

t = (77.1 − 75)/1 = 2.1

Answer

t = 2.1
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: 1 − β = 48.96 %

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 = 4/√64 = 0.5
x̄c = 50 + 2.326·0.5 = 51.16

z under the true mean

z = (51.16 − 51.15)/0.5 = 0.026

Power

power = P(Z > 0.026) = 48.96%
1 − β = 48.96%

Answer

1 − β = 48.96%
7.Linear regression & correlation

Answer: r = 0.9108

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 = 406/6 = 67.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 174
Syy = Σ(y − ȳ)² = 521.3

Correlation coefficient

r = 174/√(70·521.3) = 0.9108

Answer

r = 0.9108
8.Linear regression & correlation

Answer: e = −4.65

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 = 25/5 = 5
ȳ = 305/5 = 61

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 33

Regression line

b = 33/10 = 3.3
a = 61 − 3.3·5 = 44.5
ŷ = 44.5 + 3.3·x

Predicted value

ŷ = 44.5 + 3.3·5.5 = 62.65

Residual

e = 58 − 62.65 = −4.65

Answer

e = −4.65

The point lies below the line.

9.Chi-square tests

Answer: χ² = 6.599

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

smokers: 40·88/160 = 22, 40·72/160 = 18
non-smokers: 120·88/160 = 66, 120·72/160 = 54

Chi-square statistic

χ² = (15 − 22)²/22 + (25 − 18)²/18 + (73 − 66)²/66 + (47 − 54)²/54
χ² = 6.599

Answer

χ² = 6.599

df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is rejected.

10.Chi-square tests

Answer: χ² = 15.75

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

line 1: 70, 30
line 2: 42, 18
line 3: 42, 18

Chi-square statistic

χ² = (62 − 70)²/70 + (38 − 30)²/30 + (54 − 42)²/42 + (6 − 18)²/18 + (38 − 42)²/42 + (22 − 18)²/18
χ² = 15.75

Decision

df = 2, χ²crit = 9.210
15.75 > 9.210

Answer

χ² = 15.75

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

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