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
In a correlation, r = 0.40. What proportion of variance in one variable is shared with the other?
- A16%
- B4%
- C40%
- D60%
2.Research methods in psychologyHard
A researcher records data at nine time points and tests at the 5% level each time without correction. What is the problem?
- AThe data become more reliable each time, so the final test is the most accurate
- BThere is no problem with this approach
- CThe effect size falls with each test
- DThe chance of at least one false positive rises well above 5%
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 12, 30, 12, 41, 40, 15, 59.
n = 7
s =
4.Descriptive statisticsHard
Exam scores are normally distributed with mean 50 and standard deviation 15. Find the z-score z of a score of 59.
μ = 50σ = 15x = 59
z =
5.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2p(x): 0.35, 0.35, 0.3
Var(X) =
6.ProbabilityHard
X has the distribution shown, and Y = −2X + 10. Find the standard deviation σY of Y.
x: 0, 1, 2, 3p(x): 0.3, 0.35, 0.05, 0.3Y = −2X + 10
σY =
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 9 of them is below 367.2 g. Round z to two decimals.
μ = 40 gσ = 1.2 gn = 9< 367.2 g
P =
8.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 20% of them lie (the 20th percentile).
μ = 65σ = 10P = 20%
x = points
9.Confidence intervalsMedium
A 95% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 5.4. What is the margin of error E₂ of a 90% interval based on 225 observations?
95%n₁ = 25E₁ = 5.490%n₂ = 225
E₂ =
10.Confidence intervalsHard
The population standard deviation is 29. 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 1.5?
σ = 29E = 1.595%
n =
11.Hypothesis tests & confidence intervalsMedium
Two teaching methods are compared. Group 1: 5 students, mean score 82.5, standard deviation 10.5. Group 2: 12 students, mean score 98.7, standard deviation 11. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).
n₁ = 5x̄₁ = 82.5s₁ = 10.5n₂ = 12x̄₂ = 98.7s₂ = 11
t =
12.Hypothesis tests & confidence intervalsHard
A sample of 64 measurements has mean 118.9. The population standard deviation is 10. Find the lower limit L of the 99% confidence interval for the mean.
x̄ = 118.9σ = 10n = 6499%
L =
13.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.
n = 6
r² =
14.Linear regression & correlationHard
The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. A new observation (2.5, 24) is made. Find its residual e = y − ŷ.
n = 6(2.5, 24)
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 = 1202 × 3
χ² =
16.Chi-square testsHard
A chi-square test of independence is carried out on a contingency table with 2 rows and 2 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 · 16%
Why this is the answer
The coefficient of determination is r², here 0.40² = 0.16, so 16% of the variance is shared.
Answer
A · 16%
2.Research methods in psychology
Answer: D · The chance of at least one false positive rises well above 5%
Why this is the answer
With nine independent tests the chance of at least one Type I error is about 1 − 0.95⁹ ≈ 37%. Corrections such as Bonferroni address this.
Answer
D · The chance of at least one false positive rises well above 5%
3.Descriptive statistics
Answer: s = 17.96
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 = 209/7 = 29.86
Squared deviations
Σ(x − x̄)² = (−17.86)² + 0.1429² + (−17.86)² + 11.14² + 10.14² + (−14.86)² + 29.14² = 1935
Standard deviation
s = √(1935/6) = 17.96
Answer
s = 17.96
4.Descriptive statistics
Answer: z = 0.6
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 = (59 − 50)/15 = 0.6
Answer
z = 0.6
5.Probability
Answer: Var(X) = 0.6475
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 = 0·0.35 + 1·0.35 + 2·0.3 = 0.95
Mean of the squares
E(X²) = 0·0.35 + 1·0.35 + 4·0.3 = 1.55
Variance
Var(X) = 1.55 − 0.95² = 0.6475
Answer
Var(X) = 0.6475
6.Probability
Answer: σY = 2.39
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 = 1.35, Var(X) = 1.428, σX = 1.195
Linear change
σY = |−2|·1.195 = 2.39
Answer
σY = 2.39
The mean does move: E(Y) = −2·1.35 + 10 = 7.3.
7.Normal distribution
Answer: P = 0.9772
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 = 360 g, σT = 1.2·√9 = 3.6 g
Standardise
z = (367.2 − 360)/3.6 = 2
Probability
P = Φ(2) = 0.9772
Answer
P = 0.9772
8.Normal distribution
Answer: x = 56.58 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 = 65 + −0.8416·10 = 56.58 points
Answer
x = 56.58 points
9.Confidence intervals
Answer: E₂ = 1.511
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.
Ratio of margins
E₂/E₁ = (1.645/1.96)·√(25/225) = 0.2798
New margin
E₂ = 5.4·0.2798 = 1.511
Answer
E₂ = 1.511
10.Confidence intervals
Answer: n = 1436
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 = (1.96·29/1.5)² = 1436
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 1436
Answer
n = 1436
11.Hypothesis tests & confidence intervals
Answer: t = −2.8
Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.
Pooled variance
sp² = (4·10.5² + 11·11²)/15 = 118.1
sp = 10.87
Standard error
SE = 10.87·√(1/5 + 1/12) = 5.785
Test statistic
t = (82.5 − 98.7)/5.785 = −2.8
Answer
t = −2.8
With df = 15, the two-sided 5% critical value is 2.131; |t| is larger, so H₀ is rejected.
12.Hypothesis tests & confidence intervals
Answer: L = 115.7
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 = 2.576·10/√64 = 3.22
Interval limit
L = 118.9 − 3.22 = 115.7
Answer
L = 115.7
13.Linear regression & correlation
Answer: r² = 0.9898
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 = 155/6 = 25.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 249
Syy = Σ(y − ȳ)² = 894.8
Coefficient of determination
r² = 249²/(70·894.8) = 0.9898
Answer
r² = 0.9898
About 98.98% of the variation in y is explained by the line.
14.Linear regression & correlation
Answer: e = 3.181
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 = 21/6 = 3.5
ȳ = 116/6 = 19.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −26
Regression line
b = −26/17.5 = −1.486
a = 19.33 − (−1.486)·3.5 = 24.53
ŷ = 24.53 − 1.486·x
Predicted value
ŷ = 24.53 + (−1.486)·2.5 = 20.82
Residual
e = 24 − 20.82 = 3.181
Answer
e = 3.181
The point lies above the line.
15.Chi-square tests
Answer: χ² = 5.219
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: 80·36/120 = 24, 80·36/120 = 24, 80·48/120 = 32
staff: 40·36/120 = 12, 40·36/120 = 12, 40·48/120 = 16
Chi-square statistic
χ² = (20 − 24)²/24 + (29 − 24)²/24 + (31 − 32)²/32 + (16 − 12)²/12 + (7 − 12)²/12 + (17 − 16)²/16
χ² = 5.219
Answer
χ² = 5.219
df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is not rejected.
16.Chi-square tests
Answer: χ²crit = 6.635
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)·(2 − 1) = 1
Table
χ²crit(df = 1, α = 0.01) = 6.635
Answer
χ²crit = 6.635
Reject H₀ if the computed χ² is larger than this value.
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