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
Standardised instructions are used in an experiment so that:
- AParticipants can choose their own conditions
- BThe results become statistically significant
- CThe researcher can adapt the procedure to suit each participant's needs
- DAll participants receive the same information in the same way
2.Research methods in psychologyHard
Exercise reduces depression, and the effect occurs because exercise improves sleep, which in turn lowers depression. Sleep is best described as a:
- AMediator
- BModerator
- CExtraneous variable
- DControl variable
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 26, 48, 46, 10, 21, 38.
n = 6
s =
4.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 20. What proportion P of values is below 70?
μ = 100σ = 20x = 70
P =
5.ProbabilityMedium
In how many ways N can a committee of 2 people be chosen from a group of 13? The order does not matter.
n = 13k = 2
N =
6.ProbabilityHard
A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.7, independently. What is the probability P of exactly 3 correct answers?
n = 7p = 0.7k = 3
P =
7.Normal distributionMedium
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 246.25 and 260 g? Give P as a decimal.
μ = 250σ = 5x1 = 246.25x2 = 260
P =
8.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 10% of them lie (the 10th percentile).
μ = 178σ = 7P = 10%
x = cm
9.Confidence intervalsMedium
A sample of 15 items has mean 97.5 and standard deviation s = 12.5. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.
n = 15x̄ = 97.5s = 12.595%t(14) = 2.145
lower limit =
10.Confidence intervalsHard
The diameters of 21 machined shafts have a sample standard deviation of 7.1 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 20, 2.5% of the χ² distribution lies below 9.591 and 2.5% lies above 34.17.
n = 21s = 7.1 µmχ² = 9.591, 34.17
σ = µm
11.Hypothesis tests & confidence intervalsMedium
A sample of 64 students has a mean score of 207.3 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 200.
μ₀ = 200x̄ = 207.3s = 20n = 64
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 3? The standard deviation is 20. Find the smallest sample size n.
σ = 20E = 399%
n =
13.Linear regression & correlationMedium
The table gives 5 paired observations of x = hours studied and y = exam score. 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. Predict ŷ for x = 26.5.
n = 5x = 26.5
ŷ =
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 “rural, medium”.
row total = 56column total = 42N = 139
E =
16.Chi-square testsHard
A chi-square test of independence is carried out on a contingency table with 2 rows and 3 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 · All participants receive the same information in the same way
Why this is the answer
Standardisation removes variation in procedure as an extraneous variable and makes the study replicable.
Answer
D · All participants receive the same information in the same way
2.Research methods in psychology
Answer: A · Mediator
Why this is the answer
A mediator lies on the causal path between the predictor and outcome, explaining how or why the effect occurs.
Answer
A · Mediator
3.Descriptive statistics
Answer: s = 15.02
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 = 189/6 = 31.5
Squared deviations
Σ(x − x̄)² = (−5.5)² + 16.5² + 14.5² + (−21.5)² + (−10.5)² + 6.5² = 1128
Standard deviation
s = √(1128/5) = 15.02
Answer
s = 15.02
4.Descriptive statistics
Answer: P = 0.06681
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) = 0.06681
P = 0.06681
Answer
P = 0.06681
5.Probability
Answer: N = 78
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(13, 2) = (13·12)/(2·1) = 78
Answer
N = 78
6.Probability
Answer: P = 0.09724
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 = 3) = 35·0.73·0.34 = 35·0.343·0.0081 = 0.09724
Answer
P = 0.09724
7.Normal distribution
Answer: P = 0.7506
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 = (246.25 − 250)/5 = −0.75
z₂ = (260 − 250)/5 = 2
Area between
P = Φ(2) − Φ(−0.75) = 0.9772 − 0.2266 = 0.7506
Answer
P = 0.7506
8.Normal distribution
Answer: x = 169 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) = 0.1 → z = −1.282
Back to x
x = 178 + −1.282·7 = 169.03 cm
Answer
x = 169.03 cm
9.Confidence intervals
Answer: lower limit = 90.58
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 = 14, t = 2.145
Margin of error
E = 2.145·12.5/√15 = 6.923
Interval
97.5 ± 6.923 → [90.58, 104.4]
Answer
lower limit = 90.58
10.Confidence intervals
Answer: σ = 10.25 µm
Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.
Sum of squares
(n − 1)s² = 20·7.1² = 1008
Limits for σ²
1008/34.17 = 29.51
1008/9.591 = 105.1
Limit for σ
σ = √105.1 = 10.25 µm
Answer
σ = 10.25 µm
95% interval for σ: [5.432, 10.25] µm.
11.Hypothesis tests & confidence intervals
Answer: t = 2.92
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 = 20/√64 = 20/8 = 2.5
Test statistic
t = (207.3 − 200)/2.5 = 2.92
Answer
t = 2.92
t0.025; df = 63 = 1.998
|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).
12.Hypothesis tests & confidence intervals
Answer: n = 295
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 = (2.576·20/3)² = 294.9
Round up
n = 295
Answer
n = 295
13.Linear regression & correlation
Answer: r² = 0.9324
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 = 325/5 = 65
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 52
Syy = Σ(y − ȳ)² = 290
Coefficient of determination
r² = 52²/(10·290) = 0.9324
Answer
r² = 0.9324
About 93.24% of the variation in y is explained by the line.
14.Linear regression & correlation
Answer: ŷ = 136.1
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 = 558/5 = 111.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 489
Regression line
b = 489/90 = 5.433
a = 111.6 − 5.433·22 = −7.933
ŷ = −7.933 + 5.433·x
Prediction
ŷ = −7.933 + 5.433·26.5 = 136.1
Answer
ŷ = 136.1
15.Chi-square tests
Answer: E = 16.92
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 = 56·42/139 = 16.92
Answer
E = 16.92
16.Chi-square tests
Answer: χ²crit = 9.21
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 = (2 − 1)·(3 − 1) = 2
Table
χ²crit(df = 2, α = 0.01) = 9.210
Answer
χ²crit = 9.210
Reject H₀ if the computed χ² is larger than this value.
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