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
100.75103.75P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 100.75 and 103.75 points? Give P as a decimal.

μ = 100σ = 15x1 = 100.75x2 = 103.75
P =
2.Normal distributionHard
μ = 250x = ?25%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 25% of them lie (the 25th percentile).

μ = 250σ = 5P = 25%
x = g
3.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E99%

The battery life of two phone models is compared. Model 1: 60 phones, mean 46.5 h, known standard deviation 20 h. Model 2: 40 phones, mean 32 h, known standard deviation 15 h. Find the upper limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 60x̄₁ = 46.5 hσ₁ = 20 hn₂ = 40x̄₂ = 32 hσ₂ = 15 h99%
upper limit = h
4.Confidence intervalsHard
σ = 27, n = ?x̄ − Ex̄ + Ex̄E = 190%

The population standard deviation is 27. 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 1?

σ = 27E = 190%
n =
5.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 504.2 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 504.2s = 10n = 25
t =
6.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 100 measurements has mean 201. The population standard deviation is 12. Find the upper limit U of the 95% confidence interval for the mean.

x̄ = 201σ = 12n = 10095%
U =
7.Linear regression & correlationMedium
xy125222321416515xy24152025

The table gives 5 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.

n = 5
r =
8.Linear regression & correlationHard
xy248454667874109012101xy5106080100ŷ = 51

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 51?

n = 6ŷ = 51
x =
9.Chi-square testsMedium
busbikecarTotalstudents31101960staff29201160Total603030120

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 = 1202 × 3
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 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 = 0.07877

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 = (100.75 − 100)/15 = 0.05
z₂ = (103.75 − 100)/15 = 0.25

Area between

P = Φ(0.25) − Φ(0.05) = 0.5987 − 0.5199 = 0.07877

Answer

P = 0.07877
2.Normal distribution

Answer: x = 246.6 g

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) = 0.25 → z = −0.6745

Back to x

x = 250 + −0.6745·5 = 246.63 g

Answer

x = 246.63 g
3.Confidence intervals

Answer: upper limit = 23.53 h

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.

Point estimate

x̄₁ − x̄₂ = 46.5 − 32 = 14.5

Standard error

SE = √(20²/60 + 15²/40) = 3.506

Margin of error

E = 2.576·3.506 = 9.031

Interval

14.5 ± 9.031 → [5.469, 23.53]

Answer

upper limit = 23.53 h

The interval does not contain 0, so the means differ significantly.

4.Confidence intervals

Answer: n = 1973

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 = (1.645·27/1)² = 1973

Round up

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

n = 1973

Answer

n = 1973
5.Hypothesis tests & confidence intervals

Answer: t = 2.1

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/√25 = 10/5 = 2

Test statistic

t = (504.2 − 500)/2 = 2.1

Answer

t = 2.1
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: U = 203.4

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 = 1.96·12/√100 = 2.352

Interval limit

U = 201 + 2.352 = 203.4

Answer

U = 203.4
7.Linear regression & correlation

Answer: r = −0.9771

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 = 99/5 = 19.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −26
Syy = Σ(y − ȳ)² = 70.8

Correlation coefficient

r = −26/√(10·70.8) = −0.9771

Answer

r = −0.9771
8.Linear regression & correlation

Answer: x = 3.07

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 = 42/6 = 7
ȳ = 434/6 = 72.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 380

Regression line

b = 380/70 = 5.429
a = 72.33 − 5.429·7 = 34.33
ŷ = 34.33 + 5.429·x

Solve for x

x = (51 − 34.33)/5.429 = 3.07

Answer

x = 3.07
9.Chi-square tests

Answer: χ² = 5.533

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

students: 60·60/120 = 30, 60·30/120 = 15, 60·30/120 = 15
staff: 60·60/120 = 30, 60·30/120 = 15, 60·30/120 = 15

Chi-square statistic

χ² = (31 − 30)²/30 + (10 − 15)²/15 + (19 − 15)²/15 + (29 − 30)²/30 + (20 − 15)²/15 + (11 − 15)²/15
χ² = 5.533

Answer

χ² = 5.533

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

10.Chi-square tests

Answer: χ²crit = 15.09

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)·(6 − 1) = 5

Table

χ²crit(df = 5, α = 0.01) = 15.086

Answer

χ²crit = 15.086

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

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