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
A Type I error occurs when a researcher:
- AFails to reject a null hypothesis that is false
- BRejects a null hypothesis that is actually true
- CCollects too much data
- DUses the wrong statistical test
2.Research methods in psychologyHard
Which non-parametric test compares two related samples, such as before-and-after scores on an ordinal scale?
- AMann–Whitney U test
- BIndependent t-test
- CChi-square test of independence
- DWilcoxon signed-rank test
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 12, 25, 55, 51, 54, 18, 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 above 560?
μ = 500σ = 60x = 560
P =
5.ProbabilityMedium
A bag holds 6 red and 4 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
6 / 4
P =
6.ProbabilityHard
8 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?
n = 8p = 0.2
P =
7.Normal distributionMedium
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 20.3 and 42.8 min? Give P as a decimal.
μ = 32σ = 6x1 = 20.3x2 = 42.8
P =
8.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded bottom 10% is to the left of it. Find the cut-off x.
μ = 65 pointsσ = 10 points10%
x = points
9.Confidence intervalsMedium
A sample of 11 items has mean 112 and standard deviation s = 7. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.
n = 11x̄ = 112s = 790%t(10) = 1.812
upper limit =
10.Confidence intervalsHard
The population standard deviation is 31. 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 5?
σ = 31E = 595%
n =
11.Hypothesis tests & confidence intervalsMedium
A sample of 49 students has a mean score of 101.7 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 100.
μ₀ = 100x̄ = 101.7s = 5n = 49
t =
12.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 10. Find the smallest sample size n.
σ = 10E = 199%
n =
13.Linear regression & correlationMedium
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
14.Linear regression & correlationHard
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. A new observation (27.5, 105) is made. Find its residual e = y − ŷ.
n = 5(27.5, 105)
e =
15.Chi-square testsMedium
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 = 1602 × 2
χ² =
16.Chi-square testsHard
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 = 1203 × 2α = 0.05
χ² =
3Solutions
1.Research methods in psychology
Answer: B · Rejects a null hypothesis that is actually true
Why this is the answer
A Type I error claims an effect that does not exist. Its probability equals the significance level (often 0.05). A Type II error misses a real effect.
Answer
B · Rejects a null hypothesis that is actually true
2.Research methods in psychology
Answer: D · 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
D · Wilcoxon signed-rank test
3.Descriptive statistics
Answer: s = 18.69
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 = 237/7 = 33.86
Squared deviations
Σ(x − x̄)² = (−21.86)² + (−8.857)² + 21.14² + 17.14² + 20.14² + (−15.86)² + (−11.86)² = 2095
Standard deviation
s = √(2095/6) = 18.69
Answer
s = 18.69
4.Descriptive statistics
Answer: P = 0.1587
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) = 0.8413
P = 1 − 0.8413 = 0.1587
Answer
P = 0.1587
5.Probability
Answer: P = 0.3333
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.
First and second draw
P = 6/10 · 5/9 = 30/90 = 0.3333
Answer
P = 0.3333
6.Probability
Answer: P = 0.5033
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) = 0.88 = 0.1678
P(X = 1) = 8·0.2·0.87 = 0.3355
Add them
P = 0.1678 + 0.3355 = 0.5033
Answer
P = 0.5033
7.Normal distribution
Answer: P = 0.9385
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 = (20.3 − 32)/6 = −1.95
z₂ = (42.8 − 32)/6 = 1.8
Area between
P = Φ(1.8) − Φ(−1.95) = 0.9641 − 0.02559 = 0.9385
Answer
P = 0.9385
8.Normal distribution
Answer: x = 52.2 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) = 0.1
z from the table
z ≈ −1.28
Back to the original units
x = 65 − 1.28·10 = 52.2 points
Answer
x = 52.2 points
9.Confidence intervals
Answer: upper limit = 115.8
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 = 10, t = 1.812
Margin of error
E = 1.812·7/√11 = 3.824
Interval
112 ± 3.824 → [108.2, 115.8]
Answer
upper limit = 115.8
10.Confidence intervals
Answer: n = 148
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 = (1.96·31/5)² = 147.7
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 148
Answer
n = 148
11.Hypothesis tests & confidence intervals
Answer: t = 2.38
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 = 5/√49 = 5/7 = 0.7143
Test statistic
t = (101.7 − 100)/0.7143 = 2.38
Answer
t = 2.38
t0.025; df = 48 = 2.011
|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).
12.Hypothesis tests & confidence intervals
Answer: n = 664
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).
Solve for n
n = (2.576·10/1)² = 663.6
Round up
n = 664
Answer
n = 664
13.Linear regression & correlation
Answer: r² = 0.9912
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 = 25/5 = 5
ȳ = 267/5 = 53.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 55
Syy = Σ(y − ȳ)² = 305.2
Coefficient of determination
r² = 55²/(10·305.2) = 0.9912
Answer
r² = 0.9912
About 99.12% of the variation in y is explained by the line.
14.Linear regression & correlation
Answer: e = 2.8
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 = 115/5 = 23
ȳ = 424/5 = 84.8
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 = 84.8 − 3.867·23 = −4.133
ŷ = −4.133 + 3.867·x
Predicted value
ŷ = −4.133 + 3.867·27.5 = 102.2
Residual
e = 105 − 102.2 = 2.8
Answer
e = 2.8
The point lies above the line.
15.Chi-square tests
Answer: χ² = 2.032
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
smokers: 60·48/160 = 18, 60·112/160 = 42
non-smokers: 100·48/160 = 30, 100·112/160 = 70
Chi-square statistic
χ² = (22 − 18)²/18 + (38 − 42)²/42 + (26 − 30)²/30 + (74 − 70)²/70
χ² = 2.032
Answer
χ² = 2.032
df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is not rejected.
16.Chi-square tests
Answer: χ² = 1.4
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: 20, 20
line 2: 20, 20
line 3: 20, 20
Chi-square statistic
χ² = (19 − 20)²/20 + (21 − 20)²/20 + (18 − 20)²/20 + (22 − 20)²/20 + (23 − 20)²/20 + (17 − 20)²/20
χ² = 1.4
Decision
df = 2, χ²crit = 5.991
1.4 < 5.991
Answer
χ² = 1.4
χ² is below the critical value, so H₀ is not rejected: there is no significant evidence of a relationship.
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