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:
- AUses the wrong statistical test
- BFails to reject a null hypothesis that is false
- CCollects too much data
- DRejects a null hypothesis that is actually true
2.Research methods in psychologyHard
The "replication crisis" in psychology refers to:
- AA shortage of participants for experiments
- BMany published findings failing to reproduce in new studies
- CJournals being legally required to reject all replication studies submitted
- DCopying of questionnaires without permission
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 32, 47, 57, 35, 15, 28, 58.
n = 7
s =
4.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 12. What proportion P of values is above 107.2?
μ = 100σ = 12x = 107.2
P =
5.ProbabilityMedium
A bag holds 8 red and 8 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
8 / 8
P =
6.ProbabilityHard
11 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?
n = 11p = 0.2
P =
7.Normal distributionMedium
The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 5 of them is below 202.7 g. Round z to two decimals.
μ = 40 gσ = 1.2 gn = 5< 202.7 g
P =
8.Normal distributionHard
Hole diameters D are normal with mean 25.02 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.02 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
9.Confidence intervalsMedium
The blood pressure of 20 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 5.8 and standard deviation 0.9. Find the upper limit of the 99% confidence interval for the mean difference.
n = 20d̄ = 5.8sd = 0.999%t(19) = 2.861
upper limit = mmHg
10.Confidence intervalsHard
The population standard deviation is 34. 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 1?
σ = 34E = 190%
n =
11.Hypothesis tests & confidence intervalsMedium
A sample of 16 students has a mean score of 77.1 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 77.1s = 4n = 16
t =
12.Hypothesis tests & confidence intervalsHard
A sample of 81 measurements has mean 75.5. The population standard deviation is 12. Find the lower limit L of the 95% confidence interval for the mean.
x̄ = 75.5σ = 12n = 8195%
L =
13.Linear regression & correlationMedium
The table gives 5 paired observations of x = car age in years and y = price in k€. 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 = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (9, 12) is made. Find its residual e = y − ŷ.
n = 6(9, 12)
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 = 2202 × 3
χ² =
16.Chi-square testsHard
A chi-square test of independence is carried out on a contingency table with 4 rows and 4 columns (not counting the totals). 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: D · 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
D · Rejects a null hypothesis that is actually true
2.Research methods in psychology
Answer: B · Many published findings failing to reproduce in new studies
Why this is the answer
Large replication projects found many effects were smaller or absent, linked to publication bias, small samples and flexible analysis. Pre-registration is one response.
Answer
B · Many published findings failing to reproduce in new studies
3.Descriptive statistics
Answer: s = 15.87
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 = 272/7 = 38.86
Squared deviations
Σ(x − x̄)² = (−6.857)² + 8.143² + 18.14² + (−3.857)² + (−23.86)² + (−10.86)² + 19.14² = 1511
Standard deviation
s = √(1511/6) = 15.87
Answer
s = 15.87
4.Descriptive statistics
Answer: P = 0.2743
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 = 0.6
Standard normal table
Φ(0.6) = 0.7257
P = 1 − 0.7257 = 0.2743
Answer
P = 0.2743
5.Probability
Answer: P = 0.2333
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 = 8/16 · 7/15 = 56/240 = 0.2333
Answer
P = 0.2333
6.Probability
Answer: P = 0.3221
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) = 0.811 = 0.0859
P(X = 1) = 11·0.2·0.810 = 0.2362
Add them
P = 0.0859 + 0.2362 = 0.3221
Answer
P = 0.3221
7.Normal distribution
Answer: P = 0.8438
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 = 200 g, σT = 1.2·√5 = 2.683 g
Standardise
z = (202.7 − 200)/2.683 = 1.01
Probability
P = Φ(1.01) = 0.8438
Answer
P = 0.8438
8.Normal distribution
Answer: P = 0.8133
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 difference
μ = 25.02 − 25 = 0.02 mm, σ = 0.02236 mm
Standardise
z = (0 − 0.02)/0.02236 = −0.89
Probability
P = 1 − Φ(−0.89) = 0.8133
Answer
P = 0.8133
9.Confidence intervals
Answer: upper limit = 6.376 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 = 19, t = 2.861
Margin of error
E = 2.861·0.9/√20 = 0.5758
Interval
5.8 ± 0.5758 → [5.224, 6.376]
Answer
upper limit = 6.376 mmHg
10.Confidence intervals
Answer: n = 3129
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 = (1.645·34/1)² = 3128
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 3129
Answer
n = 3129
11.Hypothesis tests & confidence intervals
Answer: t = 2.1
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 = 4/√16 = 4/4 = 1
Test statistic
t = (77.1 − 75)/1 = 2.1
Answer
t = 2.1
t0.025; df = 15 = 2.131
|t| < 2.131, so H₀ is not rejected at the 5% level (two-sided).
12.Hypothesis tests & confidence intervals
Answer: L = 72.89
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 = 1.96·12/√81 = 2.613
Interval limit
L = 75.5 − 2.613 = 72.89
Answer
L = 72.89
13.Linear regression & correlation
Answer: r² = 0.8585
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 = 15/5 = 3
ȳ = 74/5 = 14.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −25
Syy = Σ(y − ȳ)² = 72.8
Coefficient of determination
r² = −25²/(10·72.8) = 0.8585
Answer
r² = 0.8585
About 85.85% of the variation in y is explained by the line.
14.Linear regression & correlation
Answer: e = −1.776
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 = 65/6 = 10.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 103
Regression line
b = 103/70 = 1.471
a = 10.83 − 1.471·7 = 0.5333
ŷ = 0.5333 + 1.471·x
Predicted value
ŷ = 0.5333 + 1.471·9 = 13.78
Residual
e = 12 − 13.78 = −1.776
Answer
e = −1.776
The point lies below the line.
15.Chi-square tests
Answer: χ² = 24.87
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
students: 100·66/220 = 30, 100·66/220 = 30, 100·88/220 = 40
staff: 120·66/220 = 36, 120·66/220 = 36, 120·88/220 = 48
Chi-square statistic
χ² = (40 − 30)²/30 + (38 − 30)²/30 + (22 − 40)²/40 + (26 − 36)²/36 + (28 − 36)²/36 + (66 − 48)²/48
χ² = 24.87
Answer
χ² = 24.87
df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is rejected.
16.Chi-square tests
Answer: χ²crit = 21.67
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 = (4 − 1)·(4 − 1) = 9
Table
χ²crit(df = 9, α = 0.01) = 21.666
Answer
χ²crit = 21.666
Reject H₀ if the computed χ² is larger than this value.
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