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
49.583.5P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 49.5 and 83.5 points? Give P as a decimal.

μ = 65σ = 10x1 = 49.5x2 = 83.5
P =
2.Normal distributionHard
35455565758595pointsbottom 40%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded bottom 40% is to the left of it. Find the cut-off x.

μ = 65 pointsσ = 10 points40%
x = points
3.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E95%

The battery life of two phone models is compared. Model 1: 50 phones, mean 59.5 h, known standard deviation 3 h. Model 2: 50 phones, mean 57.5 h, known standard deviation 12 h. Find the lower limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 50x̄₁ = 59.5 hσ₁ = 3 hn₂ = 50x̄₂ = 57.5 hσ₂ = 12 h95%
lower limit = h
4.Confidence intervalsHard
σ = 33, n = ?x̄ − Ex̄ + Ex̄E = 0.590%

The population standard deviation is 33. 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 = 33E = 0.590%
n =
5.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 502.8 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 502.8s = 20n = 25
t =
6.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 25 measurements has mean 111. The population standard deviation is 15. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 111σ = 15n = 2595%
L =
7.Linear regression & correlationMedium
xy126227321418517xy242025

The table gives 5 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
8.Linear regression & correlationHard
xy124223318417516xy2420x = 3.5

The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. A new observation (3.5, 23) is made. Find its residual e = y − ŷ.

n = 5(3.5, 23)
e =
9.Chi-square testsMedium
lowmediumhighTotalurban174756120rural23134480Total4060100200

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
passfailTotalline 1142640line 22575100line 3152540Total54126180dfα = 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 = 1803 × 2α = 0.01
χ² =

3Solutions

1.Normal distribution

Answer: P = 0.9073

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 = (49.5 − 65)/10 = −1.55
z₂ = (83.5 − 65)/10 = 1.85

Area between

P = Φ(1.85) − Φ(−1.55) = 0.9678 − 0.06057 = 0.9073

Answer

P = 0.9073
2.Normal distribution

Answer: x = 62.5 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) = 0.4

z from the table

z ≈ −0.25

Back to the original units

x = 65 − 0.25·10 = 62.5 points

Answer

x = 62.5 points
3.Confidence intervals

Answer: lower limit = −1.429 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 = 59.5 − 57.5 = 2

Standard error

SE = √(3²/50 + 12²/50) = 1.749

Margin of error

E = 1.96·1.749 = 3.429

Interval

2 ± 3.429 → [−1.429, 5.429]

Answer

lower limit = −1.429 h

The interval contains 0, so the data do not show a significant difference.

4.Confidence intervals

Answer: n = 11788

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 = (1.645·33/0.5)² = 11787

Round up

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

n = 11788

Answer

n = 11788
5.Hypothesis tests & confidence intervals

Answer: t = 0.7

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 = 20/√25 = 20/5 = 4

Test statistic

t = (502.8 − 500)/4 = 0.7

Answer

t = 0.7
t0.025; df = 24 = 2.064

|t| < 2.064, so H₀ is not rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: L = 105.1

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).

Margin of error

E = 1.96·15/√25 = 5.88

Interval limit

L = 111 − 5.88 = 105.1

Answer

L = 105.1
7.Linear regression & correlation

Answer: r² = 0.8804

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 = 109/5 = 21.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −27
Syy = Σ(y − ȳ)² = 82.8

Coefficient of determination

r² = −27²/(10·82.8) = 0.8804

Answer

r² = 0.8804

About 88.04% of the variation in y is explained by the line.

8.Linear regression & correlation

Answer: e = 4.5

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 = 98/5 = 19.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −22

Regression line

b = −22/10 = −2.2
a = 19.6 − (−2.2)·3 = 26.2
ŷ = 26.2 − 2.2·x

Predicted value

ŷ = 26.2 + (−2.2)·3.5 = 18.5

Residual

e = 23 − 18.5 = 4.5

Answer

e = 4.5

The point lies above the line.

9.Chi-square tests

Answer: χ² = 14.17

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: 120·40/200 = 24, 120·60/200 = 36, 120·100/200 = 60
rural: 80·40/200 = 16, 80·60/200 = 24, 80·100/200 = 40

Chi-square statistic

χ² = (17 − 24)²/24 + (47 − 36)²/36 + (56 − 60)²/60 + (23 − 16)²/16 + (13 − 24)²/24 + (44 − 40)²/40
χ² = 14.17

Answer

χ² = 14.17

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

10.Chi-square tests

Answer: χ² = 2.738

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

line 1: 12, 28
line 2: 30, 70
line 3: 12, 28

Chi-square statistic

χ² = (14 − 12)²/12 + (26 − 28)²/28 + (25 − 30)²/30 + (75 − 70)²/70 + (15 − 12)²/12 + (25 − 28)²/28
χ² = 2.738

Decision

df = 2, χ²crit = 9.210
2.738 < 9.210

Answer

χ² = 2.738

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

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