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 researcher studies the effect of gender on memory scores. Why is this a quasi-experiment rather than a true experiment?
- AThe IV is a pre-existing characteristic, so participants cannot be randomly allocated to it
- BThe DV is not measured
- CThe study takes place in a laboratory
- DThere are only two conditions, and a true experiment needs at least three levels of the IV
2.Research methods in psychologyHard
A researcher measures the effect of noise on concentration. Participants in the noise condition happened to be tested in the morning and the quiet condition in the afternoon. What is the problem?
- ATime of day is a confounding variable
- BThe data are at nominal level
- CTime of day is the dependent variable
- DThe study has too many conditions
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 15, 23, 52, 19, 15, 53, 25.
n = 7
s =
4.Descriptive statisticsHard
Exam scores are normally distributed with mean 60 and standard deviation 12. Find the z-score z of a score of 36.
μ = 60σ = 12x = 36
z =
5.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 10, 20, 30, 40p(x): 0.2, 0.2, 0.25, 0.35
σ =
6.ProbabilityHard
X has the distribution shown, and Y = 5X + 20. Find the standard deviation σY of Y.
x: 0, 2, 4, 6p(x): 0.35, 0.25, 0.2, 0.2Y = 5X + 20
σY =
7.Normal distributionMedium
X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 114.6 s. Round z to two decimals.
μX = 60, σX = 5μY = 45, σY = 4X + Y > 114.6 s
P =
8.Normal distributionHard
Hole diameters D are normal with mean 25.02 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 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 = 0.015 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
9.Confidence intervalsMedium
A sample of 6 items has mean 24.5 and standard deviation s = 6. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.
n = 6x̄ = 24.5s = 690%t(5) = 2.015
lower limit =
10.Confidence intervalsHard
The population standard deviation is 38. 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 4?
σ = 38E = 495%
n =
11.Hypothesis tests & confidence intervalsMedium
A sample of 64 students has a mean score of 79.5 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 79.5s = 15n = 64
t =
12.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.01 must detect a shift of the mean of 4 units with a power of 95%. The standard deviation is 28. Find the minimum sample size n.
α = 0.01power = 95%δ = 4σ = 28zα = 2.326zβ = 1.645
n =
13.Linear regression & correlationMedium
The table gives 6 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 6Σx = 21Σy = 132Σx² = 91Σy² = 2990Σxy = 429
r =
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. For which value x does the line predict ŷ = 39?
n = 6ŷ = 39
x =
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 = 2002 × 3
χ² =
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: A · The IV is a pre-existing characteristic, so participants cannot be randomly allocated to it
Why this is the answer
In a quasi-experiment the IV is a naturally occurring difference between people. Without random allocation, other differences between groups may confound the results.
Answer
A · The IV is a pre-existing characteristic, so participants cannot be randomly allocated to it
2.Research methods in psychology
Answer: A · Time of day is a confounding variable
Why this is the answer
Because noise and testing time vary together, any difference could be due to either. Random allocation of testing times or holding time constant would remove the confound.
Answer
A · Time of day is a confounding variable
3.Descriptive statistics
Answer: s = 16.58
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 = 202/7 = 28.86
Squared deviations
Σ(x − x̄)² = (−13.86)² + (−5.857)² + 23.14² + (−9.857)² + (−13.86)² + 24.14² + (−3.857)² = 1649
Standard deviation
s = √(1649/6) = 16.58
Answer
s = 16.58
4.Descriptive statistics
Answer: z = −2
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 = (36 − 60)/12 = −2
Answer
z = −2
5.Probability
Answer: σ = 11.35
Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.
Mean
μ = 10·0.2 + 20·0.2 + 30·0.25 + 40·0.35 = 27.5
Mean of the squares
E(X²) = 100·0.2 + 400·0.2 + 900·0.25 + 1600·0.35 = 885
Variance
Var(X) = 885 − 27.5² = 128.8
σ = √128.8 = 11.35
Answer
σ = 11.35
6.Probability
Answer: σY = 11.35
Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.
Spread of X
μ = 2.5, Var(X) = 5.15, σX = 2.269
Linear change
σY = |5|·2.269 = 11.35
Answer
σY = 11.35
The mean does move: E(Y) = 5·2.5 + 20 = 32.5.
7.Normal distribution
Answer: P = 0.06681
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 sum
μ = 105 s, σ = √(5² + 4²) = 6.403 s
Standardise
z = (114.6 − 105)/6.403 = 1.5
Upper tail
P = 1 − Φ(1.5) = 1 − 0.9332 = 0.06681
Answer
P = 0.06681
8.Normal distribution
Answer: P = 0.7881
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 = 0.025 mm
Standardise
z = (0 − 0.02)/0.025 = −0.8
Probability
P = 1 − Φ(−0.8) = 0.7881
Answer
P = 0.7881
9.Confidence intervals
Answer: lower limit = 19.56
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 = 5, t = 2.015
Margin of error
E = 2.015·6/√6 = 4.936
Interval
24.5 ± 4.936 → [19.56, 29.44]
Answer
lower limit = 19.56
10.Confidence intervals
Answer: n = 347
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 = (1.96·38/4)² = 346.7
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 347
Answer
n = 347
11.Hypothesis tests & confidence intervals
Answer: t = 2.4
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 = 15/√64 = 15/8 = 1.875
Test statistic
t = (79.5 − 75)/1.875 = 2.4
Answer
t = 2.4
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 = 773
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)·28/4)² = 772.7
Round up
n = 773
Answer
n = 773
13.Linear regression & correlation
Answer: r = −0.8506
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.
Centred sums
Sxy = 429 − 21·132/6 = −33
Sxx = 91 − 21²/6 = 17.5
Syy = 2990 − 132²/6 = 86
Correlation coefficient
r = −33/√(17.5·86) = −0.8506
Answer
r = −0.8506
14.Linear regression & correlation
Answer: x = 9.109
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 = 185/6 = 30.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 271
Regression line
b = 271/70 = 3.871
a = 30.83 − 3.871·7 = 3.733
ŷ = 3.733 + 3.871·x
Solve for x
x = (39 − 3.733)/3.871 = 9.109
Answer
x = 9.109
15.Chi-square tests
Answer: χ² = 13.12
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
urban: 100·100/200 = 50, 100·50/200 = 25, 100·50/200 = 25
rural: 100·100/200 = 50, 100·50/200 = 25, 100·50/200 = 25
Chi-square statistic
χ² = (44 − 50)²/50 + (20 − 25)²/25 + (36 − 25)²/25 + (56 − 50)²/50 + (30 − 25)²/25 + (14 − 25)²/25
χ² = 13.12
Answer
χ² = 13.12
df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is rejected.
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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