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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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
76103P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 76 and 103 points? Give P as a decimal.

μ = 100σ = 15x1 = 76x2 = 103
P =
2.Normal distributionHard
235240245250255260265gbottom 95%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 95% is to the left of it. Find the cut-off x.

μ = 250 gσ = 5 g95%
x = g
3.Confidence intervalsMedium
n = 11, sd​ = 0.60LUd̄ = 1.7E = ?90%

The blood pressure of 11 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 1.7 and standard deviation 0.6. Find the margin of error E of the 90% confidence interval for the mean difference.

n = 11d̄ = 1.7sd = 0.690%t(10) = 1.812
E = mmHg
4.Confidence intervalsHard
σ = 16, n = ?x̄ − Ex̄ + Ex̄E = 1.599%

The population standard deviation is 16. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 1.5?

σ = 16E = 1.599%
n =
5.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 53.43 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 53.43s = 10n = 49
t =
6.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 36 measurements has mean 501.5. The population standard deviation is 4. Find the lower limit L of the 90% confidence interval for the mean.

x̄ = 501.5σ = 4n = 3690%
L =
7.Linear regression & correlationMedium
xy37258979791081112613141xy510100

The table gives 6 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 48Σy = 633Σx² = 454Σy² = 69935Σxy = 5531
r =
8.Linear regression & correlationHard
xy17462067237726872994xy2025306080x = 21.5

The table gives 5 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 = 21.5.

n = 5x = 21.5
ŷ =
9.Chi-square testsMedium
busbikecarTotalstudents373035102staff1522845Total525243147

The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “staff, bike”.

row total = 45column total = 52N = 147
E =
10.Chi-square testsHard
passfailTotalplant A4852100plant B142640plant C265480Total88132220dfα = 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 = 0.05 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 = 2203 × 2α = 0.05
χ² =

3Solutions

1.Normal distribution

Answer: P = 0.5245

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 − 100)/15 = −1.6
z₂ = (103 − 100)/15 = 0.2

Area between

P = Φ(0.2) − Φ(−1.6) = 0.5793 − 0.0548 = 0.5245

Answer

P = 0.5245
2.Normal distribution

Answer: x = 258.2 g

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) = 0.95

z from the table

z ≈ 1.64

Back to the original units

x = 250 + 1.64·5 = 258.2 g

Answer

x = 258.2 g
3.Confidence intervals

Answer: E = 0.3278 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 = 10, t = 1.812

Margin of error

E = 1.812·0.6/√11 = 0.3278

Answer

E = 0.3278 mmHg
4.Confidence intervals

Answer: n = 756

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 = (2.576·16/1.5)² = 755

Round up

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

n = 756

Answer

n = 756
5.Hypothesis tests & confidence intervals

Answer: t = 2.401

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/√49 = 10/7 = 1.429

Test statistic

t = (53.43 − 50)/1.429 = 2.401

Answer

t = 2.401
t0.025; df = 48 = 2.011

|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: L = 500.4

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 = 1.645·4/√36 = 1.097

Interval limit

L = 501.5 − 1.097 = 500.4

Answer

L = 500.4
7.Linear regression & correlation

Answer: r = 0.994

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 = 5531 − 48·633/6 = 467
Sxx = 454 − 48²/6 = 70
Syy = 69935 − 633²/6 = 3154

Correlation coefficient

r = 467/√(70·3154) = 0.994

Answer

r = 0.994
8.Linear regression & correlation

Answer: ŷ = 68.4

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 = 115/5 = 23
ȳ = 371/5 = 74.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 348

Regression line

b = 348/90 = 3.867
a = 74.2 − 3.867·23 = −14.73
ŷ = −14.73 + 3.867·x

Prediction

ŷ = −14.73 + 3.867·21.5 = 68.4

Answer

ŷ = 68.4
9.Chi-square tests

Answer: E = 15.92

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 count

E = 45·52/147 = 15.92

Answer

E = 15.92
10.Chi-square tests

Answer: χ² = 4.958

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

plant A: 40, 60
plant B: 16, 24
plant C: 32, 48

Chi-square statistic

χ² = (48 − 40)²/40 + (52 − 60)²/60 + (14 − 16)²/16 + (26 − 24)²/24 + (26 − 32)²/32 + (54 − 48)²/48
χ² = 4.958

Decision

df = 2, χ²crit = 5.991
4.958 < 5.991

Answer

χ² = 4.958

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

ExamProblems · examproblems.com/courses/statistics-2-inference