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 student scores 130 on a test with mean 100 and standard deviation 15. What is the z-score?
- A3.0
- B0.5
- C1.5
- D2.0
2.Research methods in psychologyHard
Two observers record aggressive acts in a playground and their counts correlate at r = 0.45. What should the researcher do?
- ASwitch to a one-tailed hypothesis
- BRefine the behaviour categories and train observers, then recheck agreement
- CAccept the data because any positive correlation is enough
- DUse only the higher of the two counts
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 37, 43, 17, 54, 57, 16.
n = 6
s =
4.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 5. What proportion P of values is below 50?
μ = 60σ = 5x = 50
P =
5.ProbabilityMedium
In how many ways N can a committee of 2 people be chosen from a group of 14? The order does not matter.
n = 14k = 2
N =
6.ProbabilityHard
A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 5 correct answers?
n = 7p = 0.1k = 5
P =
7.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.8962. Find z to two decimals.
A = 0.8962
z =
8.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 20% of them lie (the 20th percentile).
μ = 100σ = 15P = 20%
x = points
9.Confidence intervalsMedium
The blood pressure of 20 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 6.1 and standard deviation 4.9. Find the margin of error E of the 99% confidence interval for the mean difference.
n = 20d̄ = 6.1sd = 4.999%t(19) = 2.861
E = mmHg
10.Confidence intervalsHard
The population standard deviation is 15. 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 = 15E = 1.599%
n =
11.Hypothesis tests & confidence intervalsMedium
A sample of 16 students has a mean score of 96.5 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 100.
μ₀ = 100x̄ = 96.5s = 10n = 16
t =
12.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.01 must detect a shift of the mean of 3 units with a power of 95%. The standard deviation is 16. Find the minimum sample size n.
α = 0.01power = 95%δ = 3σ = 16zα = 2.326zβ = 1.645
n =
13.Linear regression & correlationMedium
The table gives 5 paired observations of x = load in kN and y = extension in mm. 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 = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (10, 102) is made. Find its residual e = y − ŷ.
n = 6(10, 102)
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
A chi-square test of independence is carried out on a contingency table with 2 rows and 6 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 · 2.0
Why this is the answer
(130 − 100) / 15 = 2. The score is two standard deviations above the mean, higher than about 97–98% of people.
Answer
D · 2.0
2.Research methods in psychology
Answer: B · Refine the behaviour categories and train observers, then recheck agreement
Why this is the answer
Low inter-observer reliability means the categories are applied inconsistently. Clearer operational definitions and practice sessions should raise agreement before main data collection.
Answer
B · Refine the behaviour categories and train observers, then recheck agreement
3.Descriptive statistics
Answer: s = 17.69
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 = 224/6 = 37.33
Squared deviations
Σ(x − x̄)² = (−0.3333)² + 5.667² + (−20.33)² + 16.67² + 19.67² + (−21.33)² = 1565
Standard deviation
s = √(1565/5) = 17.69
Answer
s = 17.69
4.Descriptive statistics
Answer: P = 0.02275
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 = −2
Standard normal table
Φ(−2) = 0.02275
P = 0.02275
Answer
P = 0.02275
5.Probability
Answer: N = 91
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.
Combinations
C(14, 2) = (14·13)/(2·1) = 91
Answer
N = 91
6.Probability
Answer: P = 0.0001701
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.
Binomial formula
P(X = 5) = 21·0.15·0.92 = 21·1 × 10⁻⁵·0.81 = 0.0001701
Answer
P = 0.0001701
7.Normal distribution
Answer: z = −1.26
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 of z
Φ(z) = 1 − 0.8962 = 0.1038
Read the z-table backwards
Find 0.1038 inside the table and read off z.
z = −1.26
Answer
z = −1.26
8.Normal distribution
Answer: x = 87.38 points
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 = 100 + −0.8416·15 = 87.38 points
Answer
x = 87.38 points
9.Confidence intervals
Answer: E = 3.135 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 = 2.861·4.9/√20 = 3.135
Answer
E = 3.135 mmHg
10.Confidence intervals
Answer: n = 664
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 = (2.576·15/1.5)² = 663.6
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 664
Answer
n = 664
11.Hypothesis tests & confidence intervals
Answer: t = −1.4
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).
Standard error
SE = 10/√16 = 10/4 = 2.5
Test statistic
t = (96.5 − 100)/2.5 = −1.4
Answer
t = −1.4
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: n = 449
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 = ((2.326 + 1.645)·16/3)² = 448.5
Round up
n = 449
Answer
n = 449
13.Linear regression & correlation
Answer: r² = 0.992
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 = 30/5 = 6
ȳ = 130/5 = 26
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 148
Syy = Σ(y − ȳ)² = 552
Coefficient of determination
r² = 148²/(40·552) = 0.992
Answer
r² = 0.992
About 99.2% of the variation in y is explained by the line.
14.Linear regression & correlation
Answer: e = 2.305
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 = 36/6 = 6
ȳ = 448/6 = 74.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 438
Regression line
b = 438/70 = 6.257
a = 74.67 − 6.257·6 = 37.12
ŷ = 37.12 + 6.257·x
Predicted value
ŷ = 37.12 + 6.257·10 = 99.7
Residual
e = 102 − 99.7 = 2.305
Answer
e = 2.305
The point lies above the line.
15.Chi-square tests
Answer: χ² = 0.9375
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
men: 80·96/160 = 48, 80·64/160 = 32
women: 80·96/160 = 48, 80·64/160 = 32
Chi-square statistic
χ² = (45 − 48)²/48 + (35 − 32)²/32 + (51 − 48)²/48 + (29 − 32)²/32
χ² = 0.9375
Answer
χ² = 0.9375
df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is not rejected.
16.Chi-square tests
Answer: χ²crit = 15.09
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)·(6 − 1) = 5
Table
χ²crit(df = 5, α = 0.01) = 15.086
Answer
χ²crit = 15.086
Reject H₀ if the computed χ² is larger than this value.
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