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.

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1Formulas and methods

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

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

  1. Convert the value to a z-score: z = (x − μ)/σ.
  2. Look up Φ(z), the area to the left of z.
  3. For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
  4. 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

  1. Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
  2. Find the critical value for the confidence level.
  3. Margin of error: E = critical value × standard deviation/√n.
  4. 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

  1. State H₀ and H₁ and decide whether the test is one- or two-sided.
  2. Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
  3. Compare with the critical value, or find the p-value.
  4. 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 p-value of 0.03 means:

  1. AThe effect is large
  2. BIf the null hypothesis were true, data this extreme would occur about 3% of the time
  3. CThe result will replicate 97% of the time
  4. DGiven the observed data, there is only a 3% probability that the null hypothesis is true
2.Research methods in psychologyHard

In thematic analysis, the analyst mainly:

  1. ATests a numerical hypothesis
  2. BCalculates the mean score of each theme to compare them statistically
  3. CCounts word frequencies only
  4. DIdentifies recurring patterns of meaning across qualitative data
3.Descriptive statisticsMedium
15202530

Find the sample standard deviation s of the data: 28, 34, 23, 24, 15, 11.

n = 6
s =
4.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 60 and standard deviation 5. What proportion P of values is below 70?

μ = 60σ = 5x = 70
P =
5.ProbabilityMedium
xp(x)00.3510.220.230.25μ = 1.35

For the distribution shown, find the standard deviation σ of X.

x: 0, 1, 2, 3p(x): 0.35, 0.2, 0.2, 0.25
σ =
6.ProbabilityHard
n = 12, p = 0.2: P(X ≤ 1) = ?k0123456789101112

12 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?

n = 12p = 0.2
P =
7.Normal distributionMedium
200 mm199.3 mmX + Y?

X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 199.3 mm. Round z to two decimals.

μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 199.3 mm
P =
8.Normal distributionHard
μ = 500x = ?97.5%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 97.5% of them lie (the 97.5th percentile).

μ = 500σ = 40P = 97.5%
x = h
9.Confidence intervalsMedium
n = 21, s = 1LUx̄ = 42E = ?95%

A sample of 21 items has mean 42 and standard deviation s = 1. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.

n = 21x̄ = 42s = 195%t(20) = 2.086
E =
10.Confidence intervalsHard
σ = 15, n = ?x̄ − Ex̄ + Ex̄E = 1.599%

The population standard deviation is 15. 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 = 15E = 1.599%
n =
11.Hypothesis tests & confidence intervalsMedium
−2.1452.145t = ?df = 14

Two teaching methods are compared. Group 1: 6 students, mean score 105.5, standard deviation 6. Group 2: 10 students, mean score 112.1, standard deviation 3.5. 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 = 6x̄₁ = 105.5s₁ = 6n₂ = 10x̄₂ = 112.1s₂ = 3.5
t =
12.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 38 of 250 patients on treatment A recover, and 20 of 80 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 38n₁ = 250x₂ = 20n₂ = 80
z =
13.Linear regression & correlationMedium
xy331432533634735xy463234

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
14.Linear regression & correlationHard
xy153360578790911011121xy51050100x = 8

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 8.

n = 6x = 8
ŷ =
15.Chi-square testsMedium
busbikecarTotalstudents23212367staff254334102Total486457169

The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “staff, bus”.

row total = 102column total = 48N = 169
E =
16.Chi-square testsHard

A chi-square goodness-of-fit test compares observed counts in 8 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?

k = 8
df =

3Solutions

1.Research methods in psychology

Answer: B · If the null hypothesis were true, data this extreme would occur about 3% of the time

Why this is the answer

The p-value is the probability of results at least as extreme as those observed if the null were true. It does not give the probability that a hypothesis is true or the size of the effect.

Answer

B · If the null hypothesis were true, data this extreme would occur about 3% of the time
2.Research methods in psychology

Answer: D · Identifies recurring patterns of meaning across qualitative data

Why this is the answer

Thematic analysis involves familiarisation, coding, generating, reviewing and naming themes and writing up, focusing on meaning rather than statistics.

Answer

D · Identifies recurring patterns of meaning across qualitative data
3.Descriptive statistics

Answer: s = 8.408

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 = 135/6 = 22.5

Squared deviations

Σ(x − x̄)² = 5.5² + 11.5² + 0.5² + 1.5² + (−7.5)² + (−11.5)² = 353.5

Standard deviation

s = √(353.5/5) = 8.408

Answer

s = 8.408
4.Descriptive statistics

Answer: P = 0.9772

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 = 2

Standard normal table

Φ(2) = 0.9772
P = 0.9772

Answer

P = 0.9772
5.Probability

Answer: σ = 1.195

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 + 1·0.2 + 2·0.2 + 3·0.25 = 1.35

Mean of the squares

E(X²) = 0·0.35 + 1·0.2 + 4·0.2 + 9·0.25 = 3.25

Variance

Var(X) = 3.25 − 1.35² = 1.427
σ = √1.427 = 1.195

Answer

σ = 1.195
6.Probability

Answer: P = 0.2749

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) = 0.812 = 0.06872
P(X = 1) = 12·0.2·0.811 = 0.2062

Add them

P = 0.06872 + 0.2062 = 0.2749

Answer

P = 0.2749
7.Normal distribution

Answer: P = 0.9192

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 = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm

Standardise

z = (199.3 − 200)/0.5 = −1.4

Upper tail

P = 1 − Φ(−1.4) = 1 − 0.08076 = 0.9192

Answer

P = 0.9192
8.Normal distribution

Answer: x = 578.4 h

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) = 0.975 → z = 1.96

Back to x

x = 500 + 1.96·40 = 578.4 h

Answer

x = 578.4 h
9.Confidence intervals

Answer: E = 0.4552

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 = 20, t = 2.086

Margin of error

E = 2.086·1/√21 = 0.4552

Answer

E = 0.4552
10.Confidence intervals

Answer: n = 664

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 = (2.576·15/1.5)² = 663.6

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 664

Answer

n = 664
11.Hypothesis tests & confidence intervals

Answer: t = −2.807

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 = (5·6² + 9·3.5²)/14 = 20.73
sp = 4.553

Standard error

SE = 4.553·√(1/6 + 1/10) = 2.351

Test statistic

t = (105.5 − 112.1)/2.351 = −2.807

Answer

t = −2.807

With df = 14, the two-sided 5% critical value is 2.145; |t| is larger, so H₀ is rejected.

12.Hypothesis tests & confidence intervals

Answer: z = −2.004

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.

Sample proportions

p̂₁ = 38/250 = 0.152
p̂₂ = 20/80 = 0.25
p̂ = 58/330 = 0.1758

Standard error under H₀

SE = √(0.1758·0.8242·(1/250 + 1/80)) = 0.04889

Test statistic

z = (0.152 − 0.25)/0.04889 = −2.004

Answer

z = −2.004

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

13.Linear regression & correlation

Answer: r² = 1

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 = 25/5 = 5
ȳ = 165/5 = 33

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 10
Syy = Σ(y − ȳ)² = 10

Coefficient of determination

r² = 10²/(10·10) = 1

Answer

r² = 1

About 100% of the variation in y is explained by the line.

14.Linear regression & correlation

Answer: ŷ = 99.68

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 = 36/6 = 6
ȳ = 512/6 = 85.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 502

Regression line

b = 502/70 = 7.171
a = 85.33 − 7.171·6 = 42.3
ŷ = 42.3 + 7.171·x

Prediction

ŷ = 42.3 + 7.171·8 = 99.68

Answer

ŷ = 99.68
15.Chi-square tests

Answer: E = 28.97

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 = 102·48/169 = 28.97

Answer

E = 28.97
16.Chi-square tests

Answer: df = 7

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 = 8 − 1 = 7

Answer

df = 7

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