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
247.5254P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 247.5 and 254 g? Give P as a decimal.

μ = 250σ = 5x1 = 247.5x2 = 254
P =
2.Normal distributionHard
35455565758595pointsbottom 90%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded bottom 90% is to the left of it. Find the cut-off x.

μ = 65 pointsσ = 10 points90%
x = points
3.Confidence intervalsMedium
n = 7, sd​ = 6.20LUd̄ = −6.9E = ?95%

The blood pressure of 7 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −6.9 and standard deviation 6.2. Find the margin of error E of the 95% confidence interval for the mean difference.

n = 7d̄ = −6.9sd = 6.295%t(6) = 2.447
E = mmHg
4.Confidence intervalsHard
σ = 33, n = ?x̄ − Ex̄ + Ex̄E = 395%

The population standard deviation is 33. 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 = 33E = 395%
n =
5.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 121.2 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 121.2s = 5n = 49
t =
6.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 32 of 80 patients on treatment A recover, and 48 of 150 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 32n₁ = 80x₂ = 48n₂ = 150
z =
7.Linear regression & correlationMedium
xy352455566666773877xy46850607080

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 6
r =
8.Linear regression & correlationHard
xy1861208022822495269828114xy20256080100120x = 19

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 19.

n = 6x = 19
ŷ =
9.Chi-square testsMedium
no defectminormajorTotalday shift304149120night shift10195180Total4060100200

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 × 3
χ² =
10.Chi-square testsHard

A chi-square goodness-of-fit test compares observed counts in 10 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?

k = 10
df =

3Solutions

1.Normal distribution

Answer: P = 0.4796

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 = (247.5 − 250)/5 = −0.5
z₂ = (254 − 250)/5 = 0.8

Area between

P = Φ(0.8) − Φ(−0.5) = 0.7881 − 0.3085 = 0.4796

Answer

P = 0.4796
2.Normal distribution

Answer: x = 77.8 points

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) = 0.9

z from the table

z ≈ 1.28

Back to the original units

x = 65 + 1.28·10 = 77.8 points

Answer

x = 77.8 points
3.Confidence intervals

Answer: E = 5.734 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 = 6, t = 2.447

Margin of error

E = 2.447·6.2/√7 = 5.734

Answer

E = 5.734 mmHg
4.Confidence intervals

Answer: n = 465

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 = (1.96·33/3)² = 464.8

Round up

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

n = 465

Answer

n = 465
5.Hypothesis tests & confidence intervals

Answer: t = 1.68

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 = 5/√49 = 5/7 = 0.7143

Test statistic

t = (121.2 − 120)/0.7143 = 1.68

Answer

t = 1.68
t0.025; df = 48 = 2.011

|t| < 2.011, so H₀ is not rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: z = 1.213

Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.

Sample proportions

p̂₁ = 32/80 = 0.4
p̂₂ = 48/150 = 0.32
p̂ = 80/230 = 0.3478

Standard error under H₀

SE = √(0.3478·0.6522·(1/80 + 1/150)) = 0.06594

Test statistic

z = (0.4 − 0.32)/0.06594 = 1.213

Answer

z = 1.213

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

7.Linear regression & correlation

Answer: r = 0.9777

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 = 389/6 = 64.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 89.5
Syy = Σ(y − ȳ)² = 478.8

Correlation coefficient

r = 89.5/√(17.5·478.8) = 0.9777

Answer

r = 0.9777
8.Linear regression & correlation

Answer: ŷ = 69.36

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 = 138/6 = 23
ȳ = 530/6 = 88.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 332

Regression line

b = 332/70 = 4.743
a = 88.33 − 4.743·23 = −20.75
ŷ = −20.75 + 4.743·x

Prediction

ŷ = −20.75 + 4.743·19 = 69.36

Answer

ŷ = 69.36
9.Chi-square tests

Answer: χ² = 10.53

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

day shift: 120·40/200 = 24, 120·60/200 = 36, 120·100/200 = 60
night shift: 80·40/200 = 16, 80·60/200 = 24, 80·100/200 = 40

Chi-square statistic

χ² = (30 − 24)²/24 + (41 − 36)²/36 + (49 − 60)²/60 + (10 − 16)²/16 + (19 − 24)²/24 + (51 − 40)²/40
χ² = 10.53

Answer

χ² = 10.53

df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is rejected.

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 = 10 − 1 = 9

Answer

df = 9

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