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
Experimenter bias is reduced most directly by:
- AUsing a larger sample
- BUsing only volunteers
- CUsing a researcher who does not know the hypothesis or conditions to collect data
- DUsing a one-tailed hypothesis
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?
- A56
- B28
- C30
- D29
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 43, 37, 19, 22, 41, 16, 15.
n = 7
s =
4.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 5. What proportion P of values is below 55?
μ = 60σ = 5x = 55
P =
5.ProbabilityMedium
In how many ways N can a committee of 5 people be chosen from a group of 14? The order does not matter.
n = 14k = 5
N =
6.ProbabilityHard
A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 1 correct answers?
n = 9p = 0.4k = 1
P =
7.Normal distributionMedium
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 22.4 and 35.3 min? Give P as a decimal.
μ = 32σ = 6x1 = 22.4x2 = 35.3
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 75% of them lie (the 75th percentile).
μ = 178σ = 7P = 75%
x = cm
9.Confidence intervalsMedium
A sample of 25 items has mean 34 and standard deviation s = 17.5. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.
n = 25x̄ = 34s = 17.599%t(24) = 2.797
upper limit =
10.Confidence intervalsHard
The population standard deviation is 26. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 1.5?
σ = 26E = 1.599%
n =
11.Hypothesis tests & confidence intervalsMedium
13 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 4.79 and the standard deviation of the differences is 9.6. Find the paired test statistic t for H₀: μd = 0.
n = 13d̄ = 4.79sd = 9.6
t =
12.Hypothesis tests & confidence intervalsHard
A sample of 36 measurements has mean 72. The population standard deviation is 15. Find the upper limit U of the 95% confidence interval for the mean.
x̄ = 72σ = 15n = 3695%
U =
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 = 130Σx² = 91Σy² = 2974Σxy = 407
r =
14.Linear regression & correlationHard
The table gives 5 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 (4.5, 23) is made. Find its residual e = y − ŷ.
n = 5(4.5, 23)
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 = 1402 × 2
χ² =
16.Chi-square testsHard
A chi-square test of independence is carried out on a contingency table with 3 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: C · Using a researcher who does not know the hypothesis or conditions to collect data
Why this is the answer
Experimenters who know the hypothesis may unintentionally influence participants or recordings. Single- or double-blind procedures and standardised instructions reduce this.
Answer
C · Using a researcher who does not know the hypothesis or conditions to collect data
2.Research methods in psychology
Answer: C · 30
Why this is the answer
Degrees of freedom of 28 mean 28 + 2 = 30 participants, 15 in each group.
Answer
C · 30
3.Descriptive statistics
Answer: s = 12.27
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 = 193/7 = 27.57
Squared deviations
Σ(x − x̄)² = 15.43² + 9.429² + (−8.571)² + (−5.571)² + 13.43² + (−11.57)² + (−12.57)² = 903.7
Standard deviation
s = √(903.7/6) = 12.27
Answer
s = 12.27
4.Descriptive statistics
Answer: P = 0.1587
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
Standard normal table
Φ(−1) = 0.1587
P = 0.1587
Answer
P = 0.1587
5.Probability
Answer: N = 2002
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(14, 5) = (14·13·12·11·10)/(5·4·3·2·1) = 2002
Answer
N = 2002
6.Probability
Answer: P = 0.06047
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 = 1) = 9·0.41·0.68 = 9·0.4·0.0168 = 0.06047
Answer
P = 0.06047
7.Normal distribution
Answer: P = 0.654
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 = (22.4 − 32)/6 = −1.6
z₂ = (35.3 − 32)/6 = 0.55
Area between
P = Φ(0.55) − Φ(−1.6) = 0.7088 − 0.0548 = 0.654
Answer
P = 0.654
8.Normal distribution
Answer: x = 182.7 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) = 0.75 → z = 0.6745
Back to x
x = 178 + 0.6745·7 = 182.72 cm
Answer
x = 182.72 cm
9.Confidence intervals
Answer: upper limit = 43.79
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.
Critical value
df = n − 1 = 24, t = 2.797
Margin of error
E = 2.797·17.5/√25 = 9.79
Interval
34 ± 9.79 → [24.21, 43.79]
Answer
upper limit = 43.79
10.Confidence intervals
Answer: n = 1994
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 = (2.576·26/1.5)² = 1994
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 1994
Answer
n = 1994
11.Hypothesis tests & confidence intervals
Answer: t = 1.799
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 = 9.6/√13 = 2.663
Test statistic
t = 4.79/2.663 = 1.799
Answer
t = 1.799
With df = 12, the two-sided 5% critical value is 2.179; |t| is smaller, so H₀ is not rejected.
12.Hypothesis tests & confidence intervals
Answer: U = 76.9
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/√36 = 4.9
Interval limit
U = 72 + 4.9 = 76.9
Answer
U = 76.9
13.Linear regression & correlation
Answer: r = −0.9148
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 = 407 − 21·130/6 = −48
Sxx = 91 − 21²/6 = 17.5
Syy = 2974 − 130²/6 = 157.3
Correlation coefficient
r = −48/√(17.5·157.3) = −0.9148
Answer
r = −0.9148
14.Linear regression & correlation
Answer: e = 3.6
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 = 15/5 = 3
ȳ = 112/5 = 22.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −20
Regression line
b = −20/10 = −2
a = 22.4 − (−2)·3 = 28.4
ŷ = 28.4 − 2·x
Predicted value
ŷ = 28.4 + (−2)·4.5 = 19.4
Residual
e = 23 − 19.4 = 3.6
Answer
e = 3.6
The point lies above the line.
15.Chi-square tests
Answer: χ² = 0.1667
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
smokers: 100·42/140 = 30, 100·98/140 = 70
non-smokers: 40·42/140 = 12, 40·98/140 = 28
Chi-square statistic
χ² = (31 − 30)²/30 + (69 − 70)²/70 + (11 − 12)²/12 + (29 − 28)²/28
χ² = 0.1667
Answer
χ² = 0.1667
df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is not rejected.
16.Chi-square tests
Answer: χ²crit = 12.59
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 = (3 − 1)·(4 − 1) = 6
Table
χ²crit(df = 6, α = 0.05) = 12.592
Answer
χ²crit = 12.592
Reject H₀ if the computed χ² is larger than this value.
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