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 = 0.3264

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.3264. Find z to two decimals.

A = 0.3264
z =
2.Normal distributionHard
10 g0X − Y?

The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
3.Confidence intervalsMedium
n = 7, s = 13L = ?Ux̄ = 101.5E95%

A sample of 7 items has mean 101.5 and standard deviation s = 13. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 7x̄ = 101.5s = 1395%t(6) = 2.447
lower limit =
4.Confidence intervalsHard
σ = 30, n = ?x̄ − Ex̄ + Ex̄E = 395%

The population standard deviation is 30. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 3?

σ = 30E = 395%
n =
5.Hypothesis tests & confidence intervalsMedium
−2.1792.179t = ?df = 12

13 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 6.03 and the standard deviation of the differences is 7.5. Find the paired test statistic t for H₀: μd = 0.

n = 13d̄ = 6.03sd = 7.5
t =
6.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 36 measurements has mean 194.6. The population standard deviation is 12. Find the upper limit U of the 90% confidence interval for the mean.

x̄ = 194.6σ = 12n = 3690%
U =
7.Linear regression & correlationMedium
xy21142162583410441249xy5102040

The table gives 6 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 42Σy = 184Σx² = 364Σy² = 6680Σxy = 1556
r =
8.Linear regression & correlationHard
xy35356177197411831395xy5106080100x = 4

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 4.

n = 6x = 4
ŷ =
9.Chi-square testsMedium
improvednot improvedTotalnew drug4753100placebo8317100Total13070200

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 = 2002 × 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 goodness-of-fit test checks whether counts in 9 categories follow a Poisson distribution whose mean is estimated from the same data. Using the table, find the critical value χ²crit at significance level α = 0.01.

α = 0.01
χ²crit =

3Solutions

1.Normal distribution

Answer: z = 0.45

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 − 0.3264 = 0.6736

Read the z-table backwards

Find 0.6736 inside the table and read off z.

z = 0.45

Answer

z = 0.45
2.Normal distribution

Answer: P = 0.8413

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 = 500 − 490 = 10 g, σ = 10 g

Standardise

z = (0 − 10)/10 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413
3.Confidence intervals

Answer: lower limit = 89.48

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 = 6, t = 2.447

Margin of error

E = 2.447·13/√7 = 12.02

Interval

101.5 ± 12.02 → [89.48, 113.5]

Answer

lower limit = 89.48
4.Confidence intervals

Answer: n = 385

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 = (1.96·30/3)² = 384.2

Round up

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

n = 385

Answer

n = 385
5.Hypothesis tests & confidence intervals

Answer: t = 2.899

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 = 7.5/√13 = 2.08

Test statistic

t = 6.03/2.08 = 2.899

Answer

t = 2.899

With df = 12, the two-sided 5% critical value is 2.179; |t| is larger, so H₀ is rejected.

6.Hypothesis tests & confidence intervals

Answer: U = 197.9

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 = 1.645·12/√36 = 3.29

Interval limit

U = 194.6 + 3.29 = 197.9

Answer

U = 197.9
7.Linear regression & correlation

Answer: r = 0.9945

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 = 1556 − 42·184/6 = 268
Sxx = 364 − 42²/6 = 70
Syy = 6680 − 184²/6 = 1037

Correlation coefficient

r = 268/√(70·1037) = 0.9945

Answer

r = 0.9945
8.Linear regression & correlation

Answer: ŷ = 56.89

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 = 437/6 = 72.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 279

Regression line

b = 279/70 = 3.986
a = 72.83 − 3.986·8 = 40.95
ŷ = 40.95 + 3.986·x

Prediction

ŷ = 40.95 + 3.986·4 = 56.89

Answer

ŷ = 56.89
9.Chi-square tests

Answer: χ² = 28.48

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: 100·130/200 = 65, 100·70/200 = 35
placebo: 100·130/200 = 65, 100·70/200 = 35

Chi-square statistic

χ² = (47 − 65)²/65 + (53 − 35)²/35 + (83 − 65)²/65 + (17 − 35)²/35
χ² = 28.48

Answer

χ² = 28.48

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

10.Chi-square tests

Answer: χ²crit = 18.48

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 − 1 = 7

Table

χ²crit(df = 7, α = 0.01) = 18.475

Answer

χ²crit = 18.475

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

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