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
245257.5P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 245 and 257.5 g? Give P as a decimal.

μ = 250σ = 5x1 = 245x2 = 257.5
P =
2.Normal distributionHard
235240245250255260265gbottom 75%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 75% is to the left of it. Find the cut-off x.

μ = 250 gσ = 5 g75%
x = g
3.Confidence intervalsMedium
n = 20, sd​ = 3.10LU = ?d̄ = −2.1E99%

The blood pressure of 20 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −2.1 and standard deviation 3.1. Find the upper limit of the 99% confidence interval for the mean difference.

n = 20d̄ = −2.1sd = 3.199%t(19) = 2.861
upper limit = mmHg
4.Confidence intervalsHard
σ = 32, n = ?x̄ − Ex̄ + Ex̄E = 490%

The population standard deviation is 32. 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 4?

σ = 32E = 490%
n =
5.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 123.2 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 123.2s = 10n = 81
t =
6.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 49 measurements has mean 50.2. The population standard deviation is 4. Find the lower limit L of the 99% confidence interval for the mean.

x̄ = 50.2σ = 4n = 4999%
L =
7.Linear regression & correlationMedium
xy34155176396711701376xy510406080

The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 6
r =
8.Linear regression & correlationHard
xy274106108161016xy5101015x = 9

The table gives 5 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 = 9.

n = 5x = 9
ŷ =
9.Chi-square testsMedium
no defectminormajorTotalday shift8392168night shift30163076Total385551144

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 = 76column total = 51N = 144
E =
10.Chi-square testsHard

A chi-square test of independence is carried out on a contingency table with 4 rows and 4 columns (not counting the totals). How many degrees of freedom df does the test have?

4 × 4
df =

3Solutions

1.Normal distribution

Answer: P = 0.7745

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 = (245 − 250)/5 = −1
z₂ = (257.5 − 250)/5 = 1.5

Area between

P = Φ(1.5) − Φ(−1) = 0.9332 − 0.1587 = 0.7745

Answer

P = 0.7745
2.Normal distribution

Answer: x = 253.4 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) = 0.75

z from the table

z ≈ 0.67

Back to the original units

x = 250 + 0.67·5 = 253.4 g

Answer

x = 253.4 g
3.Confidence intervals

Answer: upper limit = −0.1168 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 = 19, t = 2.861

Margin of error

E = 2.861·3.1/√20 = 1.983

Interval

−2.1 ± 1.983 → [−4.083, −0.1168]

Answer

upper limit = −0.1168 mmHg
4.Confidence intervals

Answer: n = 174

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 = (1.645·32/4)² = 173.2

Round up

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

n = 174

Answer

n = 174
5.Hypothesis tests & confidence intervals

Answer: t = 2.88

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 = 10/√81 = 10/9 = 1.111

Test statistic

t = (123.2 − 120)/1.111 = 2.88

Answer

t = 2.88
t0.025; df = 80 = 1.99

|t| > 1.99, so H₀ is rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: L = 48.73

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 = 2.576·4/√49 = 1.472

Interval limit

L = 50.2 − 1.472 = 48.73

Answer

L = 48.73
7.Linear regression & correlation

Answer: r = 0.9702

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 = 48/6 = 8
ȳ = 368/6 = 61.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 236
Syy = Σ(y − ȳ)² = 845.3

Correlation coefficient

r = 236/√(70·845.3) = 0.9702

Answer

r = 0.9702
8.Linear regression & correlation

Answer: ŷ = 15.4

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 = 30/5 = 6
ȳ = 59/5 = 11.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 48

Regression line

b = 48/40 = 1.2
a = 11.8 − 1.2·6 = 4.6
ŷ = 4.6 + 1.2·x

Prediction

ŷ = 4.6 + 1.2·9 = 15.4

Answer

ŷ = 15.4
9.Chi-square tests

Answer: E = 26.92

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 = 76·51/144 = 26.92

Answer

E = 26.92
10.Chi-square tests

Answer: df = 9

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 = (4 − 1)·(4 − 1) = 9

Answer

df = 9

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