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
−3−2−10123zshaded area between −z and z = 0.9199

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9199. Find z to two decimals.

A = 0.9199
z =
2.Normal distributionHard
μ = 65x = ?99%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 99% of them lie (the 99th percentile).

μ = 65σ = 10P = 99%
x = points
3.Confidence intervalsMedium
n = 15, sd​ = 7.60LU = ?d̄ = 2.1E95%

The blood pressure of 15 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 2.1 and standard deviation 7.6. Find the upper limit of the 95% confidence interval for the mean difference.

n = 15d̄ = 2.1sd = 7.695%t(14) = 2.145
upper limit = mmHg
4.Confidence intervalsHard
σ = 24, n = ?x̄ − Ex̄ + Ex̄E = 495%

The population standard deviation is 24. 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 4?

σ = 24E = 495%
n =
5.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 101 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 101s = 8n = 100
t =
6.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 36 measurements has mean 118.3. The population standard deviation is 4. Find the lower limit L of the 99% confidence interval for the mean.

x̄ = 118.3σ = 4n = 3699%
L =
7.Linear regression & correlationMedium
xy17911993211072311325132xy2025100120

The table gives 5 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 = 5Σx = 105Σy = 536Σx² = 2245Σy² = 58572Σxy = 11460
r =
8.Linear regression & correlationHard
xy175919612172237925872792xy20256080x = 24

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. A new observation (24, 80) is made. Find its residual e = y − ŷ.

n = 6(24, 80)
e =
9.Chi-square testsMedium
lowmediumhighTotalurban634017120rural471538100Total1105555220

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 = 2202 × 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 goodness-of-fit test checks whether counts in 7 categories follow a Poisson distribution whose mean is estimated from the same data. Using the table, find the critical value χ²crit at significance level α = 0.05.

α = 0.05
χ²crit =

3Solutions

1.Normal distribution

Answer: z = 1.75

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 of z

Φ(z) = (1 + 0.9199)/2 = 0.9599

Read the z-table backwards

Find 0.9599 inside the table and read off z.

z = 1.75

Answer

z = 1.75
2.Normal distribution

Answer: x = 88.26 points

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 = 65 + 2.326·10 = 88.26 points

Answer

x = 88.26 points
3.Confidence intervals

Answer: upper limit = 6.309 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 = 14, t = 2.145

Margin of error

E = 2.145·7.6/√15 = 4.209

Interval

2.1 ± 4.209 → [−2.109, 6.309]

Answer

upper limit = 6.309 mmHg
4.Confidence intervals

Answer: n = 139

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 = (1.96·24/4)² = 138.3

Round up

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

n = 139

Answer

n = 139
5.Hypothesis tests & confidence intervals

Answer: t = 1.25

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 = 8/√100 = 8/10 = 0.8

Test statistic

t = (101 − 100)/0.8 = 1.25

Answer

t = 1.25
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: L = 116.6

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 = 2.576·4/√36 = 1.717

Interval limit

L = 118.3 − 1.717 = 116.6

Answer

L = 116.6
7.Linear regression & correlation

Answer: r = 0.9669

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 = 11460 − 105·536/5 = 204
Sxx = 2245 − 105²/5 = 40
Syy = 58572 − 536²/5 = 1113

Correlation coefficient

r = 204/√(40·1113) = 0.9669

Answer

r = 0.9669
8.Linear regression & correlation

Answer: e = −2.143

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 = 132/6 = 22
ȳ = 450/6 = 75

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 250

Regression line

b = 250/70 = 3.571
a = 75 − 3.571·22 = −3.571
ŷ = −3.571 + 3.571·x

Predicted value

ŷ = −3.571 + 3.571·24 = 82.14

Residual

e = 80 − 82.14 = −2.143

Answer

e = −2.143

The point lies below the line.

9.Chi-square tests

Answer: χ² = 20.06

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: 120·110/220 = 60, 120·55/220 = 30, 120·55/220 = 30
rural: 100·110/220 = 50, 100·55/220 = 25, 100·55/220 = 25

Chi-square statistic

χ² = (63 − 60)²/60 + (40 − 30)²/30 + (17 − 30)²/30 + (47 − 50)²/50 + (15 − 25)²/25 + (38 − 25)²/25
χ² = 20.06

Answer

χ² = 20.06

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

10.Chi-square tests

Answer: χ²crit = 11.07

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 = 7 − 1 − 1 = 5

Table

χ²crit(df = 5, α = 0.05) = 11.070

Answer

χ²crit = 11.070

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

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