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 study takes place in a laboratory
- BThe DV is not measured
- CThere are only two conditions, and a true experiment needs at least three levels of the IV
- DThe IV is a pre-existing characteristic, so participants cannot be randomly allocated to it
2.Research methods in psychologyHard
Which non-parametric test compares two related samples, such as before-and-after scores on an ordinal scale?
- AChi-square test of independence
- BMann–Whitney U test
- CIndependent t-test
- DWilcoxon signed-rank test
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 53, 54, 32, 37, 46.
n = 5
s =
4.Descriptive statisticsHard
Exam scores are normally distributed with mean 60 and standard deviation 10. Find the z-score z of a score of 52.
μ = 60σ = 10x = 52
z =
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
6 components are tested, and each one fails with probability 0.25, independently. What is the probability P that at most one fails?
n = 6p = 0.25
P =
7.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.2843. Find z to two decimals.
A = 0.2843
z =
8.Normal distributionHard
The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
9.Confidence intervalsMedium
A sample of 11 items has mean 79 and standard deviation s = 7.5. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.
n = 11x̄ = 79s = 7.595%t(10) = 2.228
lower limit =
10.Confidence intervalsHard
The population standard deviation is 27. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 5?
σ = 27E = 590%
n =
11.Hypothesis tests & confidence intervalsMedium
6 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is −3.37 and the standard deviation of the differences is 5.5. Find the paired test statistic t for H₀: μd = 0.
n = 6d̄ = −3.37sd = 5.5
t =
12.Hypothesis tests & confidence intervalsHard
A sample of 25 measurements has mean 496.7. The population standard deviation is 15. Find the lower limit L of the 95% confidence interval for the mean.
x̄ = 496.7σ = 15n = 2595%
L =
13.Linear regression & correlationMedium
The table gives 6 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.
n = 6
r =
14.Linear regression & correlationHard
The table gives 5 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 = 26?
n = 5ŷ = 26
x =
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 “day shift, minor”.
row total = 42column total = 52N = 122
E =
16.Chi-square testsHard
A chi-square goodness-of-fit test compares observed counts in 9 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?
k = 9
df =
3Solutions
1.Research methods in psychology
Answer: D · 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
D · The IV is a pre-existing characteristic, so participants cannot be randomly allocated to it
2.Research methods in psychology
Answer: D · Wilcoxon signed-rank test
Why this is the answer
The Wilcoxon signed-rank test is for paired ordinal data; Mann–Whitney U is its counterpart for independent groups.
Answer
D · Wilcoxon signed-rank test
3.Descriptive statistics
Answer: s = 9.711
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 = 222/5 = 44.4
Squared deviations
Σ(x − x̄)² = 8.6² + 9.6² + (−12.4)² + (−7.4)² + 1.6² = 377.2
Standard deviation
s = √(377.2/4) = 9.711
Answer
s = 9.711
4.Descriptive statistics
Answer: z = −0.8
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 = (52 − 60)/10 = −0.8
Answer
z = −0.8
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 = 0.5339
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.
Zero and one failure
P(X = 0) = 0.756 = 0.178
P(X = 1) = 6·0.25·0.755 = 0.356
Add them
P = 0.178 + 0.356 = 0.5339
Answer
P = 0.5339
7.Normal distribution
Answer: z = 0.57
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 − 0.2843 = 0.7157
Read the z-table backwards
Find 0.7157 inside the table and read off z.
z = 0.57
Answer
z = 0.57
8.Normal distribution
Answer: P = 0.7389
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 = 180 − 170 = 10 cm, σ = 15.62 cm
Standardise
z = (0 − 10)/15.62 = −0.64
Probability
P = 1 − Φ(−0.64) = 0.7389
Answer
P = 0.7389
9.Confidence intervals
Answer: lower limit = 73.96
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 = 10, t = 2.228
Margin of error
E = 2.228·7.5/√11 = 5.038
Interval
79 ± 5.038 → [73.96, 84.04]
Answer
lower limit = 73.96
10.Confidence intervals
Answer: n = 79
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 = (1.645·27/5)² = 78.91
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 79
Answer
n = 79
11.Hypothesis tests & confidence intervals
Answer: t = −1.501
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 of the mean difference
SE = 5.5/√6 = 2.245
Test statistic
t = −3.37/2.245 = −1.501
Answer
t = −1.501
With df = 5, the two-sided 5% critical value is 2.571; |t| is smaller, so H₀ is not rejected.
12.Hypothesis tests & confidence intervals
Answer: L = 490.8
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 = 1.96·15/√25 = 5.88
Interval limit
L = 496.7 − 5.88 = 490.8
Answer
L = 490.8
13.Linear regression & correlation
Answer: r = −0.9978
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 = 111/6 = 18.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −47.5
Syy = Σ(y − ȳ)² = 129.5
Correlation coefficient
r = −47.5/√(17.5·129.5) = −0.9978
Answer
r = −0.9978
14.Linear regression & correlation
Answer: x = 6.968
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 = 115/5 = 23
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 124
Regression line
b = 124/40 = 3.1
a = 23 − 3.1·6 = 4.4
ŷ = 4.4 + 3.1·x
Solve for x
x = (26 − 4.4)/3.1 = 6.968
Answer
x = 6.968
15.Chi-square tests
Answer: E = 17.9
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 = 42·52/122 = 17.9
Answer
E = 17.9
16.Chi-square tests
Answer: df = 8
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 = 9 − 1 = 8
Answer
df = 8
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