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
The same 25 participants complete a memory test before and after caffeine. The scores are interval and roughly normal. Which test is appropriate?
- APaired (related) samples t-test
- BMann–Whitney U test
- CChi-square test
- DIndependent samples t-test
2.Research methods in psychologyHard
A study reports t(28) = 2.50, p = .018 for two independent groups of equal size. How many participants took part?
- A30
- B28
- C29
- D56
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 18, 34, 17, 35, 27, 52, 12.
n = 7
s =
4.Descriptive statisticsHard
Exam scores are normally distributed with mean 60 and standard deviation 8. Find the z-score z of a score of 50.4.
μ = 60σ = 8x = 50.4
z =
5.ProbabilityMedium
A bag holds 8 red and 5 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
8 / 5
P =
6.ProbabilityHard
A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 3 correct answers?
n = 10p = 0.3k = 3
P =
7.Normal distributionMedium
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 70 and 115 points? Give P as a decimal.
μ = 100σ = 15x1 = 70x2 = 115
P =
8.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 40% is to the left of it. Find the cut-off x.
μ = 500 hσ = 40 h40%
x = h
9.Confidence intervalsMedium
The blood pressure of 12 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 6.2 and standard deviation 2.5. Find the upper limit of the 90% confidence interval for the mean difference.
n = 12d̄ = 6.2sd = 2.590%t(11) = 1.796
upper limit = mmHg
10.Confidence intervalsHard
A pilot study suggests that about 0.2 of products are defective. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.01?
p ≈ 0.2E = 0.0195%
n =
11.Hypothesis tests & confidence intervalsMedium
A sample of 36 students has a mean score of 47.83 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 50.
μ₀ = 50x̄ = 47.83s = 5n = 36
t =
12.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 100 against H₁: μ > 100 is done at α = 0.05 with a sample of 25; the standard deviation is known to be 5. In reality the mean is 103.2. Find the probability of a Type II error β in percent.
μ₀ = 100μ = 103.2σ = 5n = 25α = 0.05
β = %
13.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 6Σx = 42Σy = 186Σx² = 364Σy² = 6810Σxy = 1572
r =
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, 20) is made. Find its residual e = y − ŷ.
n = 6(2.5, 20)
e =
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 “urban, low”.
row total = 71column total = 33N = 118
E =
16.Chi-square testsHard
A chi-square test of independence is carried out on a contingency table with 2 rows and 4 columns (not counting the totals). Using the table, find the critical value χ²crit at significance level α = 0.05.
α = 0.05
χ²crit =
3Solutions
1.Research methods in psychology
Answer: A · Paired (related) samples t-test
Why this is the answer
A paired t-test compares means from the same participants under two conditions, using the differences within each person.
Answer
A · Paired (related) samples t-test
2.Research methods in psychology
Answer: A · 30
Why this is the answer
Degrees of freedom of 28 mean 28 + 2 = 30 participants, 15 in each group.
Answer
A · 30
3.Descriptive statistics
Answer: s = 13.78
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 = 195/7 = 27.86
Squared deviations
Σ(x − x̄)² = (−9.857)² + 6.143² + (−10.86)² + 7.143² + (−0.8571)² + 24.14² + (−15.86)² = 1139
Standard deviation
s = √(1139/6) = 13.78
Answer
s = 13.78
4.Descriptive statistics
Answer: z = −1.2
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 = (50.4 − 60)/8 = −1.2
Answer
z = −1.2
5.Probability
Answer: P = 0.359
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 = 8/13 · 7/12 = 56/156 = 0.359
Answer
P = 0.359
6.Probability
Answer: P = 0.2668
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) = 120·0.33·0.77 = 120·0.027·0.08235 = 0.2668
Answer
P = 0.2668
7.Normal distribution
Answer: P = 0.8186
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 = (70 − 100)/15 = −2
z₂ = (115 − 100)/15 = 1
Area between
P = Φ(1) − Φ(−2) = 0.8413 − 0.02275 = 0.8186
Answer
P = 0.8186
8.Normal distribution
Answer: x = 490 h
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
Φ(z) = 0.4
z from the table
z ≈ −0.25
Back to the original units
x = 500 − 0.25·40 = 490 h
Answer
x = 490 h
9.Confidence intervals
Answer: upper limit = 7.496 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 = 11, t = 1.796
Margin of error
E = 1.796·2.5/√12 = 1.296
Interval
6.2 ± 1.296 → [4.904, 7.496]
Answer
upper limit = 7.496 mmHg
10.Confidence intervals
Answer: n = 6147
Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.
Solve for n
n = 1.96²·0.2·0.8/0.01² = 6147
Round up
n = 6147
Answer
n = 6147
11.Hypothesis tests & confidence intervals
Answer: t = −2.604
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 = 5/√36 = 5/6 = 0.8333
Test statistic
t = (47.83 − 50)/0.8333 = −2.604
Answer
t = −2.604
t0.025; df = 35 = 2.03
|t| > 2.03, so H₀ is rejected at the 5% level (two-sided).
12.Hypothesis tests & confidence intervals
Answer: β = 5.997 %
Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).
Critical sample mean
SE = 5/√25 = 1
x̄c = 100 + 1.645·1 = 101.6
z under the true mean
z = (101.6 − 103.2)/1 = −1.555
Type II error
power = P(Z > −1.555) = 94%
β = 100% − 94% = 5.997%
Answer
β = 5.997%
13.Linear regression & correlation
Answer: r = 0.9988
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 = 1572 − 42·186/6 = 270
Sxx = 364 − 42²/6 = 70
Syy = 6810 − 186²/6 = 1044
Correlation coefficient
r = 270/√(70·1044) = 0.9988
Answer
r = 0.9988
14.Linear regression & correlation
Answer: e = −4.605
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 = 143/6 = 23.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −13.5
Regression line
b = −13.5/17.5 = −0.7714
a = 23.83 − (−0.7714)·3.5 = 26.53
ŷ = 26.53 − 0.7714·x
Predicted value
ŷ = 26.53 + (−0.7714)·2.5 = 24.6
Residual
e = 20 − 24.6 = −4.605
Answer
e = −4.605
The point lies below the line.
15.Chi-square tests
Answer: E = 19.86
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 = 71·33/118 = 19.86
Answer
E = 19.86
16.Chi-square tests
Answer: χ²crit = 7.815
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)·(4 − 1) = 3
Table
χ²crit(df = 3, α = 0.05) = 7.815
Answer
χ²crit = 7.815
Reject H₀ if the computed χ² is larger than this value.
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