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
76.75130P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 76.75 and 130 points? Give P as a decimal.

μ = 100σ = 15x1 = 76.75x2 = 130
P =
2.Normal distributionHard
μ = 178x = ?99%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 99% of them lie (the 99th percentile).

μ = 178σ = 7P = 99%
x = cm
3.Confidence intervalsMedium
n = 8, s = 7.5LU = ?x̄ = 106.5E90%

A sample of 8 items has mean 106.5 and standard deviation s = 7.5. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.

n = 8x̄ = 106.5s = 7.590%t(7) = 1.895
upper limit =
4.Confidence intervalsHard
σ = 40, n = ?x̄ − Ex̄ + Ex̄E = 0.590%

The population standard deviation is 40. 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 0.5?

σ = 40E = 0.590%
n =
5.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 112.6 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 112.6s = 12n = 25
t =
6.Hypothesis tests & confidence intervalsHard
σ = 15, n = ?x̄ − Ex̄ + Ex̄E = 0.590%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 0.5? The standard deviation is 15. Find the smallest sample size n.

σ = 15E = 0.590%
n =
7.Linear regression & correlationMedium
xy17662080238726972911332124xy20253080100120

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 = 147Σy = 567Σx² = 3759Σy² = 55879Σxy = 14490
r =
8.Linear regression & correlationHard
xy355460561667766878xy468607080x = 6.5

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 = 6.5.

n = 6x = 6.5
ŷ =
9.Chi-square testsMedium
lowmediumhighTotalurban7161740rural8242860Total154045100

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 × 3
χ² =
10.Chi-square testsHard
passfailTotalage 18–302179100age 31–50204060age 51+3763100Total78182260dfα = 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 = 2603 × 2α = 0.01
χ² =

3Solutions

1.Normal distribution

Answer: P = 0.9167

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 = (76.75 − 100)/15 = −1.55
z₂ = (130 − 100)/15 = 2

Area between

P = Φ(2) − Φ(−1.55) = 0.9772 − 0.06057 = 0.9167

Answer

P = 0.9167
2.Normal distribution

Answer: x = 194.3 cm

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) = 0.99 → z = 2.326

Back to x

x = 178 + 2.326·7 = 194.28 cm

Answer

x = 194.28 cm
3.Confidence intervals

Answer: upper limit = 111.5

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 = 7, t = 1.895

Margin of error

E = 1.895·7.5/√8 = 5.025

Interval

106.5 ± 5.025 → [101.5, 111.5]

Answer

upper limit = 111.5
4.Confidence intervals

Answer: n = 17319

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 = (1.645·40/0.5)² = 17319

Round up

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

n = 17319

Answer

n = 17319
5.Hypothesis tests & confidence intervals

Answer: t = −3.083

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 = 12/√25 = 12/5 = 2.4

Test statistic

t = (112.6 − 120)/2.4 = −3.083

Answer

t = −3.083
t0.025; df = 24 = 2.064

|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: n = 2436

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 = (1.645·15/0.5)² = 2435

Round up

n = 2436

Answer

n = 2436
7.Linear regression & correlation

Answer: r = 0.9949

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 = 14490 − 147·567/6 = 598.5
Sxx = 3759 − 147²/6 = 157.5
Syy = 55879 − 567²/6 = 2298

Correlation coefficient

r = 598.5/√(157.5·2298) = 0.9949

Answer

r = 0.9949
8.Linear regression & correlation

Answer: ŷ = 68.47

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 = 33/6 = 5.5
ȳ = 387/6 = 64.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 69.5

Regression line

b = 69.5/17.5 = 3.971
a = 64.5 − 3.971·5.5 = 42.66
ŷ = 42.66 + 3.971·x

Prediction

ŷ = 42.66 + 3.971·6.5 = 68.47

Answer

ŷ = 68.47
9.Chi-square tests

Answer: χ² = 0.3704

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

urban: 40·15/100 = 6, 40·40/100 = 16, 40·45/100 = 18
rural: 60·15/100 = 9, 60·40/100 = 24, 60·45/100 = 27

Chi-square statistic

χ² = (7 − 6)²/6 + (16 − 16)²/16 + (17 − 18)²/18 + (8 − 9)²/9 + (24 − 24)²/24 + (28 − 27)²/27
χ² = 0.3704

Answer

χ² = 0.3704

df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is not rejected.

10.Chi-square tests

Answer: χ² = 6.508

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

age 18–30: 30, 70
age 31–50: 18, 42
age 51+: 30, 70

Chi-square statistic

χ² = (21 − 30)²/30 + (79 − 70)²/70 + (20 − 18)²/18 + (40 − 42)²/42 + (37 − 30)²/30 + (63 − 70)²/70
χ² = 6.508

Decision

df = 2, χ²crit = 9.210
6.508 < 9.210

Answer

χ² = 6.508

χ² is below the critical value, so H₀ is not rejected: there is no significant evidence of a relationship.

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