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

Demand characteristics are:

  1. ACues that make participants guess the aim and change their behaviour
  2. BThe formal demands an ethics committee makes before a study can begin
  3. CStatistical requirements for a test
  4. DCharacteristics of the sample
2.Research methods in psychologyHard

The "replication crisis" in psychology refers to:

  1. ACopying of questionnaires without permission
  2. BMany published findings failing to reproduce in new studies
  3. CJournals being legally required to reject all replication studies submitted
  4. DA shortage of participants for experiments
3.Descriptive statisticsMedium
3035404550

Find the sample standard deviation s of the data: 50, 47, 44, 32, 28.

n = 5
s =
4.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 170 and standard deviation 10. What proportion P of values is below 190?

μ = 170σ = 10x = 190
P =
5.ProbabilityMedium
n = 14k = 2N = ?

In how many ways N can a committee of 2 people be chosen from a group of 14? The order does not matter.

n = 14k = 2
N =
6.ProbabilityHard
xp(x)00.0510.120.3530.2540.25μ

X has the distribution shown, and Y = 0.5X + 20. Find the standard deviation σY of Y.

x: 0, 1, 2, 3, 4p(x): 0.05, 0.1, 0.35, 0.25, 0.25Y = 0.5X + 20
σY =
7.Normal distributionMedium
164.35183.6P = ?

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 164.35 and 183.6 cm? Give P as a decimal.

μ = 178σ = 7x1 = 164.35x2 = 183.6
P =
8.Normal distributionHard
0.04 mm0D − d?

Hole diameters D are normal with mean 25.04 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.04 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
9.Confidence intervalsMedium
n = 11, s = 11L = ?Ux̄ = 88E90%

A sample of 11 items has mean 88 and standard deviation s = 11. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.

n = 11x̄ = 88s = 1190%t(10) = 1.812
lower limit =
10.Confidence intervalsHard
σ = 7, n = ?x̄ − Ex̄ + Ex̄E = 295%

The population standard deviation is 7. 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 2?

σ = 7E = 295%
n =
11.Hypothesis tests & confidence intervalsMedium
−2.1312.131t = ?df = 15

16 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 9.65 and the standard deviation of the differences is 11.7. Find the paired test statistic t for H₀: μd = 0.

n = 16d̄ = 9.65sd = 11.7
t =
12.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 50 of 250 patients on treatment A recover, and 69 of 300 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 50n₁ = 250x₂ = 69n₂ = 300
z =
13.Linear regression & correlationMedium
xy132337542753960xy530405060

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 5
r =
14.Linear regression & correlationHard
xy136242345456555xy244050x = 1.5

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 1.5.

n = 5x = 1.5
ŷ =
15.Chi-square testsMedium
no defectminormajorTotalday shift1223843night shift13221247Total25452090

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

row total = 47column total = 20N = 90
E =
16.Chi-square testsHard

A chi-square goodness-of-fit test compares observed counts in 3 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?

k = 3
df =

3Solutions

1.Research methods in psychology

Answer: A · Cues that make participants guess the aim and change their behaviour

Why this is the answer

When participants work out the hypothesis they may act to confirm or sabotage it, threatening validity. Deception and blind procedures reduce this.

Answer

A · Cues that make participants guess the aim and change their behaviour
2.Research methods in psychology

Answer: B · Many published findings failing to reproduce in new studies

Why this is the answer

Large replication projects found many effects were smaller or absent, linked to publication bias, small samples and flexible analysis. Pre-registration is one response.

Answer

B · Many published findings failing to reproduce in new studies
3.Descriptive statistics

Answer: s = 9.654

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 = 201/5 = 40.2

Squared deviations

Σ(x − x̄)² = 9.8² + 6.8² + 3.8² + (−8.2)² + (−12.2)² = 372.8

Standard deviation

s = √(372.8/4) = 9.654

Answer

s = 9.654
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: N = 91

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, 2) = (14·13)/(2·1) = 91

Answer

N = 91
6.Probability

Answer: σY = 0.5585

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 = 2.55, Var(X) = 1.248, σX = 1.117

Linear change

σY = |0.5|·1.117 = 0.5585

Answer

σY = 0.5585

The mean does move: E(Y) = 0.5·2.55 + 20 = 21.27.

7.Normal distribution

Answer: P = 0.7626

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 = (164.35 − 178)/7 = −1.95
z₂ = (183.6 − 178)/7 = 0.8

Area between

P = Φ(0.8) − Φ(−1.95) = 0.7881 − 0.02559 = 0.7626

Answer

P = 0.7626
8.Normal distribution

Answer: P = 0.9868

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 = 25.04 − 25 = 0.04 mm, σ = 0.01803 mm

Standardise

z = (0 − 0.04)/0.01803 = −2.22

Probability

P = 1 − Φ(−2.22) = 0.9868

Answer

P = 0.9868
9.Confidence intervals

Answer: lower limit = 81.99

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 = 1.812

Margin of error

E = 1.812·11/√11 = 6.01

Interval

88 ± 6.01 → [81.99, 94.01]

Answer

lower limit = 81.99
10.Confidence intervals

Answer: n = 48

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 = (1.96·7/2)² = 47.06

Round up

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

n = 48

Answer

n = 48
11.Hypothesis tests & confidence intervals

Answer: t = 3.299

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 = 11.7/√16 = 2.925

Test statistic

t = 9.65/2.925 = 3.299

Answer

t = 3.299

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: z = −0.8508

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 = 50/250 = 0.2
p̂₂ = 69/300 = 0.23
p̂ = 119/550 = 0.2164

Standard error under H₀

SE = √(0.2164·0.7836·(1/250 + 1/300)) = 0.03526

Test statistic

z = (0.2 − 0.23)/0.03526 = −0.8508

Answer

z = −0.8508

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

13.Linear regression & correlation

Answer: r = 0.9883

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 = 224/5 = 44.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 144
Syy = Σ(y − ȳ)² = 530.8

Correlation coefficient

r = 144/√(40·530.8) = 0.9883

Answer

r = 0.9883
14.Linear regression & correlation

Answer: ŷ = 39

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 = 234/5 = 46.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 52

Regression line

b = 52/10 = 5.2
a = 46.8 − 5.2·3 = 31.2
ŷ = 31.2 + 5.2·x

Prediction

ŷ = 31.2 + 5.2·1.5 = 39

Answer

ŷ = 39
15.Chi-square tests

Answer: E = 10.44

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 = 47·20/90 = 10.44

Answer

E = 10.44
16.Chi-square tests

Answer: df = 2

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 = 2

Answer

df = 2

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