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 = 0.2843

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.2843. Find z to two decimals.

A = 0.2843
z =
2.Normal distributionHard
235240245250255260265gtop 40%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 40% is to the right of it. Find the cut-off x.

μ = 250 gσ = 5 g40%
x = g
3.Confidence intervalsMedium
n = 13, s = 11LUx̄ = 24.5E = ?99%

A sample of 13 items has mean 24.5 and standard deviation s = 11. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 13x̄ = 24.5s = 1199%t(12) = 3.055
E =
4.Confidence intervalsHard
μ₁ − μ₂, t(12) = 1.7820L = ?Ux̄₁ − x̄₂E90%

Two fertilisers are tested. Plot group 1: 6 plots, mean yield 70.5 kg, standard deviation 6 kg. Plot group 2: 8 plots, mean 72.5 kg, standard deviation 8.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 6x̄₁ = 70.5s₁ = 6n₂ = 8x̄₂ = 72.5s₂ = 8.590%t(12) = 1.782
lower limit = kg
5.Hypothesis tests & confidence intervalsMedium
−2.0932.093t = ?df = 19

Two teaching methods are compared. Group 1: 13 students, mean score 120, standard deviation 11.5. Group 2: 8 students, mean score 115.3, standard deviation 12. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 13x̄₁ = 120s₁ = 11.5n₂ = 8x̄₂ = 115.3s₂ = 12
t =
6.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 36 measurements has mean 197.6. The population standard deviation is 5. Find the lower limit L of the 90% confidence interval for the mean.

x̄ = 197.6σ = 5n = 3690%
L =
7.Linear regression & correlationMedium
xy3545667779801197xy5106080100

The table gives 5 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 5
r =
8.Linear regression & correlationHard
xy17621972218023832593xy202560708090ŷ = 89

The table gives 5 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 = 89?

n = 5ŷ = 89
x =
9.Chi-square testsMedium
improvednot improvedTotalnew drug293160placebo112940Total4060100

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 = 1002 × 2
χ² =
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 = 0.01.

α = 0.01
χ²crit =

3Solutions

1.Normal distribution

Answer: z = 0.57

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 − 0.2843 = 0.7157

Read the z-table backwards

Find 0.7157 inside the table and read off z.

z = 0.57

Answer

z = 0.57
2.Normal distribution

Answer: x = 251.3 g

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) = 0.6

z from the table

z ≈ 0.25

Back to the original units

x = 250 + 0.25·5 = 251.3 g

Answer

x = 251.3 g
3.Confidence intervals

Answer: E = 9.32

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 = 12, t = 3.055

Margin of error

E = 3.055·11/√13 = 9.32

Answer

E = 9.32
4.Confidence intervals

Answer: lower limit = −9.275 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 = (5·6² + 7·8.5²)/12 = 57.15
sp = 7.559

Margin of error

E = 1.782·7.559·√(1/6 + 1/8) = 7.275

Interval

−2 ± 7.275 → [−9.275, 5.275]

Answer

lower limit = −9.275 kg
5.Hypothesis tests & confidence intervals

Answer: t = 0.895

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.

Pooled variance

sp² = (12·11.5² + 7·12²)/19 = 136.6
sp = 11.69

Standard error

SE = 11.69·√(1/13 + 1/8) = 5.252

Test statistic

t = (120 − 115.3)/5.252 = 0.895

Answer

t = 0.895

With df = 19, the two-sided 5% critical value is 2.093; |t| is smaller, so H₀ is not rejected.

6.Hypothesis tests & confidence intervals

Answer: L = 196.2

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 = 1.645·5/√36 = 1.371

Interval limit

L = 197.6 − 1.371 = 196.2

Answer

L = 196.2
7.Linear regression & correlation

Answer: r = 0.983

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 = 35/5 = 7
ȳ = 374/5 = 74.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 200
Syy = Σ(y − ȳ)² = 1035

Correlation coefficient

r = 200/√(40·1035) = 0.983

Answer

r = 0.983
8.Linear regression & correlation

Answer: x = 24.01

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 = 105/5 = 21
ȳ = 390/5 = 78

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 146

Regression line

b = 146/40 = 3.65
a = 78 − 3.65·21 = 1.35
ŷ = 1.35 + 3.65·x

Solve for x

x = (89 − 1.35)/3.65 = 24.01

Answer

x = 24.01
9.Chi-square tests

Answer: χ² = 4.34

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

new drug: 60·40/100 = 24, 60·60/100 = 36
placebo: 40·40/100 = 16, 40·60/100 = 24

Chi-square statistic

χ² = (29 − 24)²/24 + (31 − 36)²/36 + (11 − 16)²/16 + (29 − 24)²/24
χ² = 4.34

Answer

χ² = 4.34

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

10.Chi-square tests

Answer: χ²crit = 23.21

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 = 0.01) = 23.209

Answer

χ²crit = 23.209

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

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