Research Methods and Statistics for Psychology: cram pack

Formula sheets, the key method steps for every topic and 16 practice problems, with full solutions at the back.

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1Formulas and methods

Descriptive statistics

Mean
x̄ = Σx/n
Sample standard deviation
s = √(Σ(x − x̄)²/(n − 1))
Median
middle value of the sorted data
z-score
z = (x − μ)/σ
Normal distribution
68% within ±1σ, 95% within ±2σ

Probability

Complement
P(not A) = 1 − P(A)
Addition rule
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Independent events
P(A ∩ B) = P(A)·P(B)
Conditional probability
P(A|B) = P(A ∩ B)/P(B)
Binomial
P(X = k) = C(n,k)·pk(1 − p)n−k

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.Research methods in psychologyMedium

Which statistic is appropriate for the relationship between two sets of ranks, such as judges' rankings of paintings?

  1. AIndependent t-test
  2. BSpearman's rho
  3. CPearson's r
  4. DChi-square
2.Research methods in psychologyHard

Which non-parametric test compares two related samples, such as before-and-after scores on an ordinal scale?

  1. AIndependent t-test
  2. BChi-square test of independence
  3. CWilcoxon signed-rank test
  4. DMann–Whitney U test
3.Descriptive statisticsMedium
354045505560

Find the sample standard deviation s of the data: 48, 60, 32, 53, 55, 36.

n = 6
s =
4.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 100 and standard deviation 5. What proportion P of values is above 95?

μ = 100σ = 5x = 95
P =
5.ProbabilityMedium
xp(x)00.110.320.3530.140.15μ = 1.9

For the distribution shown, find the standard deviation σ of X.

x: 0, 1, 2, 3, 4p(x): 0.1, 0.3, 0.35, 0.1, 0.15
σ =
6.ProbabilityHard
n = 12, p = 0.2: P(X ≤ 1) = ?k0123456789101112

12 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?

n = 12p = 0.2
P =
7.Normal distributionMedium
150 kg151.8 kgT = X₁ + … + X₆?

The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 6 of them exceeds 151.8 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 6> 151.8 kg
P =
8.Normal distributionHard
μ = 178x = ?20%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 20% of them lie (the 20th percentile).

μ = 178σ = 7P = 20%
x = cm
9.Confidence intervalsMedium
n = 25, s = 6.5LUx̄ = 42.5E = ?99%

A sample of 25 items has mean 42.5 and standard deviation s = 6.5. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 25x̄ = 42.5s = 6.599%t(24) = 2.797
E =
10.Confidence intervalsHard
σ = 36, n = ?x̄ − Ex̄ + Ex̄E = 290%

The population standard deviation is 36. 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 2?

σ = 36E = 290%
n =
11.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 80 parts has mean length 140 mm, and the line's known standard deviation is 20 mm. Line 2: a sample of 30 parts has mean 132.9 mm, known standard deviation 14 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 80x̄₁ = 140 mmσ₁ = 20 mmn₂ = 30x̄₂ = 132.9 mmσ₂ = 14 mm
z =
12.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 32 of 80 patients on treatment A recover, and 128 of 250 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 32n₁ = 80x₂ = 128n₂ = 250
z =
13.Linear regression & correlationMedium
xy128221322415516611xy246102030

The table gives 6 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.

n = 6
r =
14.Linear regression & correlationHard
xy2941762483110411247xy5102040x = 3

The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 3.

n = 6x = 3
ŷ =
15.Chi-square testsMedium
no defectminormajorTotalday shift294645120night shift152065100Total4466110220

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 =
16.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 compares observed counts in 11 categories with fully specified expected proportions (no parameters are estimated from the data). Using the table, find the critical value χ²crit at significance level α = 0.01.

α = 0.01
χ²crit =

3Solutions

1.Research methods in psychology

Answer: B · Spearman's rho

Why this is the answer

Spearman's rank correlation measures the strength of a monotonic relationship between two ranked or ordinal variables.

Answer

B · Spearman's rho
2.Research methods in psychology

Answer: C · Wilcoxon signed-rank test

Why this is the answer

The Wilcoxon signed-rank test is for paired ordinal data; Mann–Whitney U is its counterpart for independent groups.

Answer

C · Wilcoxon signed-rank test
3.Descriptive statistics

Answer: s = 11.09

Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.

Mean

x̄ = 284/6 = 47.33

Squared deviations

Σ(x − x̄)² = 0.6667² + 12.67² + (−15.33)² + 5.667² + 7.667² + (−11.33)² = 615.3

Standard deviation

s = √(615.3/5) = 11.09

Answer

s = 11.09
4.Descriptive statistics

Answer: P = 0.8413

Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.

Standardise

z = −1

Standard normal table

Φ(−1) = 0.1587
P = 1 − 0.1587 = 0.8413

Answer

P = 0.8413
5.Probability

Answer: σ = 1.179

Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.

Mean

μ = 0·0.1 + 1·0.3 + 2·0.35 + 3·0.1 + 4·0.15 = 1.9

Mean of the squares

E(X²) = 0·0.1 + 1·0.3 + 4·0.35 + 9·0.1 + 16·0.15 = 5

Variance

Var(X) = 5 − 1.9² = 1.39
σ = √1.39 = 1.179

Answer

σ = 1.179
6.Probability

Answer: P = 0.2749

Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.

Zero and one failure

P(X = 0) = 0.812 = 0.06872
P(X = 1) = 12·0.2·0.811 = 0.2062

Add them

P = 0.06872 + 0.2062 = 0.2749

Answer

P = 0.2749
7.Normal distribution

Answer: P = 0.07078

Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).

Distribution of the total

μT = 150 kg, σT = 0.5·√6 = 1.225 kg

Standardise

z = (151.8 − 150)/1.225 = 1.47

Probability

P = 1 − Φ(1.47) = 0.07078

Answer

P = 0.07078
8.Normal distribution

Answer: x = 172.1 cm

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) = 0.2 → z = −0.8416

Back to x

x = 178 + −0.8416·7 = 172.11 cm

Answer

x = 172.11 cm
9.Confidence intervals

Answer: E = 3.636

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.

Critical value

df = n − 1 = 24, t = 2.797

Margin of error

E = 2.797·6.5/√25 = 3.636

Answer

E = 3.636
10.Confidence intervals

Answer: n = 877

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 = (1.645·36/2)² = 876.8

Round up

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

n = 877

Answer

n = 877
11.Hypothesis tests & confidence intervals

Answer: z = 2.091

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.

Standard error of the difference

SE = √(20²/80 + 14²/30) = √11.53 = 3.396

Test statistic

z = (140 − 132.9)/3.396 = 2.091

Answer

z = 2.091

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: z = −1.745

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 = 128/250 = 0.512
p̂ = 160/330 = 0.4848

Standard error under H₀

SE = √(0.4848·0.5152·(1/80 + 1/250)) = 0.0642

Test statistic

z = (0.4 − 0.512)/0.0642 = −1.745

Answer

z = −1.745

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

13.Linear regression & correlation

Answer: r = −0.9458

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 = 21/6 = 3.5
ȳ = 113/6 = 18.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −53.5
Syy = Σ(y − ȳ)² = 182.8

Correlation coefficient

r = −53.5/√(17.5·182.8) = −0.9458

Answer

r = −0.9458
14.Linear regression & correlation

Answer: ŷ = 12.8

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 = 169/6 = 28.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 269

Regression line

b = 269/70 = 3.843
a = 28.17 − 3.843·7 = 1.267
ŷ = 1.267 + 3.843·x

Prediction

ŷ = 1.267 + 3.843·3 = 12.8

Answer

ŷ = 12.8
15.Chi-square tests

Answer: χ² = 16.65

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: 120·44/220 = 24, 120·66/220 = 36, 120·110/220 = 60
night shift: 100·44/220 = 20, 100·66/220 = 30, 100·110/220 = 50

Chi-square statistic

χ² = (29 − 24)²/24 + (46 − 36)²/36 + (45 − 60)²/60 + (15 − 20)²/20 + (20 − 30)²/30 + (65 − 50)²/50
χ² = 16.65

Answer

χ² = 16.65

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

16.Chi-square tests

Answer: χ²crit = 23.21

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 = 11 − 1 = 10

Table

χ²crit(df = 10, α = 0.01) = 23.209

Answer

χ²crit = 23.209

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

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