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
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.
1Formulas and methods
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
- Convert the value to a z-score: z = (x − μ)/σ.
- Look up Φ(z), the area to the left of z.
- For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
- 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
- Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
- Find the critical value for the confidence level.
- Margin of error: E = critical value × standard deviation/√n.
- 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
- State H₀ and H₁ and decide whether the test is one- or two-sided.
- Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
- Compare with the critical value, or find the p-value.
- 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
The same 25 participants complete a memory test before and after caffeine. The scores are interval and roughly normal. Which test is appropriate?
- AChi-square test
- BIndependent samples t-test
- CMann–Whitney U test
- DPaired (related) samples t-test
2.Research methods in psychologyHard
In thematic analysis, the analyst mainly:
- ACounts word frequencies only
- BTests a numerical hypothesis
- CCalculates the mean score of each theme to compare them statistically
- DIdentifies recurring patterns of meaning across qualitative data
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 60, 11, 14, 45, 27, 16, 22.
n = 7
s =
4.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 60. What proportion P of values is below 590?
μ = 500σ = 60x = 590
P =
5.ProbabilityMedium
A fair die is rolled 4 times. What is the probability P of getting at least one six?
n = 4
P =
6.ProbabilityHard
7 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?
n = 7p = 0.2
P =
7.Normal distributionMedium
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 92.5 and 115 points? Give P as a decimal.
μ = 100σ = 15x1 = 92.5x2 = 115
P =
8.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 75% of them lie (the 75th percentile).
μ = 500σ = 40P = 75%
x = h
9.Confidence intervalsMedium
A sample of 7 items has mean 85 and standard deviation s = 17.5. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.
n = 7x̄ = 85s = 17.590%t(6) = 1.943
E =
10.Confidence intervalsHard
A pilot study suggests that about 0.1 of products are defective. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.04?
p ≈ 0.1E = 0.0490%
n =
11.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 120; the population standard deviation is known to be 20. A random sample of 25 observations has a mean of 111.2. Test H₀: μ = 120 against H₁: μ < 120. Find the p-value p.
μ₀ = 120σ = 20n = 25x̄ = 111.2H₁: μ < μ₀
p =
12.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.05 must detect a shift of the mean of 3 units with a power of 90%. The standard deviation is 20. Find the minimum sample size n.
α = 0.05power = 90%δ = 3σ = 20zα = 1.645zβ = 1.282
n =
13.Linear regression & correlationMedium
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
14.Linear regression & correlationHard
The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 58?
n = 6ŷ = 58
x =
15.Chi-square testsMedium
The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “urban, low”.
row total = 65column total = 59N = 155
E =
16.Chi-square testsHard
A chi-square test of independence is carried out on a contingency table with 2 rows and 4 columns (not counting the totals). How many degrees of freedom df does the test have?
2 × 4
df =
3Solutions
1.Research methods in psychology
Answer: D · Paired (related) samples t-test
Why this is the answer
A paired t-test compares means from the same participants under two conditions, using the differences within each person.
Answer
D · Paired (related) samples t-test
2.Research methods in psychology
Answer: D · Identifies recurring patterns of meaning across qualitative data
Why this is the answer
Thematic analysis involves familiarisation, coding, generating, reviewing and naming themes and writing up, focusing on meaning rather than statistics.
Answer
D · Identifies recurring patterns of meaning across qualitative data
3.Descriptive statistics
Answer: s = 18.16
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 = 195/7 = 27.86
Squared deviations
Σ(x − x̄)² = 32.14² + (−16.86)² + (−13.86)² + 17.14² + (−0.8571)² + (−11.86)² + (−5.857)² = 1979
Standard deviation
s = √(1979/6) = 18.16
Answer
s = 18.16
4.Descriptive statistics
Answer: P = 0.9332
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 = 1.5
Standard normal table
Φ(1.5) = 0.9332
P = 0.9332
Answer
P = 0.9332
5.Probability
Answer: P = 0.5177
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.
Complement
P(no six) = (5/6)4 = 0.4823
P = 1 − 0.4823 = 0.5177
Answer
P = 0.5177
6.Probability
Answer: P = 0.5767
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) = 0.87 = 0.2097
P(X = 1) = 7·0.2·0.86 = 0.367
Add them
P = 0.2097 + 0.367 = 0.5767
Answer
P = 0.5767
7.Normal distribution
Answer: P = 0.5328
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 = (92.5 − 100)/15 = −0.5
z₂ = (115 − 100)/15 = 1
Area between
P = Φ(1) − Φ(−0.5) = 0.8413 − 0.3085 = 0.5328
Answer
P = 0.5328
8.Normal distribution
Answer: x = 527 h
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) = 0.75 → z = 0.6745
Back to x
x = 500 + 0.6745·40 = 526.98 h
Answer
x = 526.98 h
9.Confidence intervals
Answer: E = 12.85
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 = 6, t = 1.943
Margin of error
E = 1.943·17.5/√7 = 12.85
Answer
E = 12.85
10.Confidence intervals
Answer: n = 153
Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.
Solve for n
n = 1.645²·0.1·0.9/0.04² = 152.2
Round up
n = 153
Answer
n = 153
11.Hypothesis tests & confidence intervals
Answer: p = 0.0139
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.
Test statistic
SE = σ/√n = 20/√25 = 4
z = (111.2 − 120)/4 = −2.2
Tail area
Φ(2.2) = 0.9861
1 − Φ(2.2) = 0.0139
p-value
p = 0.0139
Answer
p = 0.0139
p ≤ 0.05, so H₀ is rejected at the 5% level.
12.Hypothesis tests & confidence intervals
Answer: n = 381
Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.
Solve for n
n = ((1.645 + 1.282)·20/3)² = 380.8
Round up
n = 381
Answer
n = 381
13.Linear regression & correlation
Answer: r² = 0.9879
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 = 110/5 = 22
ȳ = 444/5 = 88.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 456
Syy = Σ(y − ȳ)² = 2339
Coefficient of determination
r² = 456²/(90·2339) = 0.9879
Answer
r² = 0.9879
About 98.79% of the variation in y is explained by the line.
14.Linear regression & correlation
Answer: x = 3.449
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 = 438/6 = 73
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 128
Regression line
b = 128/17.5 = 7.314
a = 73 − 7.314·5.5 = 32.77
ŷ = 32.77 + 7.314·x
Solve for x
x = (58 − 32.77)/7.314 = 3.449
Answer
x = 3.449
15.Chi-square tests
Answer: E = 24.74
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 = 65·59/155 = 24.74
Answer
E = 24.74
16.Chi-square tests
Answer: df = 3
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 = (2 − 1)·(4 − 1) = 3
Answer
df = 3
ExamProblems · examproblems.com/courses/research-methods-psychology