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
178179.4P = ?

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 178 and 179.4 cm? Give P as a decimal.

μ = 178σ = 7x1 = 178x2 = 179.4
P =
2.Normal distributionHard
μ = 32x = ?10%

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 10% of them lie (the 10th percentile).

μ = 32σ = 6P = 10%
x = min
3.Confidence intervalsMedium
n = 21, s = 6.5L = ?Ux̄ = 38.5E90%

A sample of 21 items has mean 38.5 and standard deviation s = 6.5. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.

n = 21x̄ = 38.5s = 6.590%t(20) = 1.725
lower limit =
4.Confidence intervalsHard
σ = 20, n = ?x̄ − Ex̄ + Ex̄E = 490%

The population standard deviation is 20. 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 = 20E = 490%
n =
5.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 122.5 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 122.5s = 15n = 100
t =
6.Hypothesis tests & confidence intervalsHard
σ = 4, n = ?x̄ − Ex̄ + Ex̄E = 0.599%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 0.5? The standard deviation is 4. Find the smallest sample size n.

σ = 4E = 0.599%
n =
7.Linear regression & correlationMedium
xy166119742298251092811531130xy15202530100

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

n = 6Σx = 141Σy = 587Σx² = 3471Σy² = 60807Σxy = 14513
r =
8.Linear regression & correlationHard
xy129234342439543xy24303540x = 3.5

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 3.5.

n = 5x = 3.5
ŷ =
9.Chi-square testsMedium
no defectminormajorTotalday shift10372168night shift3583780Total454558148

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

row total = 68column total = 58N = 148
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 test of independence is carried out on a contingency table with 3 rows and 6 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: P = 0.07926

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 = (178 − 178)/7 = 0
z₂ = (179.4 − 178)/7 = 0.2

Area between

P = Φ(0.2) − Φ(0) = 0.5793 − 0.5 = 0.07926

Answer

P = 0.07926
2.Normal distribution

Answer: x = 24.31 min

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 for the percentile

Φ(z) = 0.1 → z = −1.282

Back to x

x = 32 + −1.282·6 = 24.31 min

Answer

x = 24.31 min
3.Confidence intervals

Answer: lower limit = 36.05

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 = 20, t = 1.725

Margin of error

E = 1.725·6.5/√21 = 2.447

Interval

38.5 ± 2.447 → [36.05, 40.95]

Answer

lower limit = 36.05
4.Confidence intervals

Answer: n = 68

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 = (1.645·20/4)² = 67.65

Round up

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

n = 68

Answer

n = 68
5.Hypothesis tests & confidence intervals

Answer: t = 1.667

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 = 15/√100 = 15/10 = 1.5

Test statistic

t = (122.5 − 120)/1.5 = 1.667

Answer

t = 1.667
t0.025; df = 99 = 1.984

|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: n = 425

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).

Solve for n

n = (2.576·4/0.5)² = 424.7

Round up

n = 425

Answer

n = 425
7.Linear regression & correlation

Answer: r = 0.9849

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 = 14513 − 141·587/6 = 718.5
Sxx = 3471 − 141²/6 = 157.5
Syy = 60807 − 587²/6 = 3379

Correlation coefficient

r = 718.5/√(157.5·3379) = 0.9849

Answer

r = 0.9849
8.Linear regression & correlation

Answer: ŷ = 39.05

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 = 15/5 = 3
ȳ = 187/5 = 37.4

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 = 37.4 − 3.3·3 = 27.5
ŷ = 27.5 + 3.3·x

Prediction

ŷ = 27.5 + 3.3·3.5 = 39.05

Answer

ŷ = 39.05
9.Chi-square tests

Answer: E = 26.65

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 = 68·58/148 = 26.65

Answer

E = 26.65
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 = (3 − 1)·(6 − 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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