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)·p^k(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: S_xx = Σ(x − x̄)² = Σx² − (Σx)²/n, S_xy = Σxy − ΣxΣy/n
Slope: b = S_xy/S_xx
Intercept: a = ȳ − b·x̄
Correlation: r = S_xy/√(S_xxS_yy)
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 > χ²_crit

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

The "replication crisis" in psychology refers to:

AA shortage of participants for experiments
BCopying of questionnaires without permission
CMany published findings failing to reproduce in new studies
DJournals being legally required to reject all replication studies submitted

3.Descriptive statisticsMedium

Find the sample standard deviation s of the data: 10, 59, 23, 20, 14, 20, 47.

n = 7

s =

4.Descriptive statisticsHard

A quantity is normally distributed with mean 100 and standard deviation 20. What proportion P of values is below 60?

μ = 100σ = 20x = 60

P =

5.ProbabilityMedium

A bag holds 4 red and 8 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

4 / 8

P =

6.ProbabilityHard

A test has 7 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 = 7p = 0.3k = 3

P =

7.Normal distributionMedium

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.7190. Find z to two decimals.

A = 0.7190

z =

8.Normal distributionHard

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 99% of them lie (the 99th percentile).

μ = 250σ = 5P = 99%

x = g

9.Confidence intervalsMedium

A 90% confidence interval for a mean, based on 36 observations and a known σ, has a margin of error of 5. What is the margin of error E₂ of a 99% interval from the same data?

90%n₁ = 36E₁ = 599%

E₂ =

10.Confidence intervalsHard

The population standard deviation is 14. 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 0.5?

σ = 14E = 0.590%

n =

11.Hypothesis tests & confidence intervalsMedium

The mean of a population is claimed to be 50; the population standard deviation is known to be 10. A random sample of 100 observations has a mean of 52.22. Test H₀: μ = 50 against H₁: μ > 50. Find the p-value p.

μ₀ = 50σ = 10n = 100x̄ = 52.22H₁: μ > μ₀

p =

12.Hypothesis tests & confidence intervalsHard

In a trial, 25 of 100 patients on treatment A recover, and 24 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 25n₁ = 100x₂ = 24n₂ = 120

z =

13.Linear regression & correlationMedium

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 132Σy = 720Σx² = 2974Σy² = 88228Σxy = 16192

r =

14.Linear regression & correlationHard

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (21.5, 81) is made. Find its residual e = y − ŷ.

n = 5(21.5, 81)

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 = 802 × 3

χ² =

16.Chi-square testsHard

The table shows observed counts. Test at α = 0.05 whether the two variables are independent (no continuity correction). Compute the test statistic χ², then compare it with the critical value from the table.

N = 2003 × 2α = 0.05

χ² =

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: C · 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

C · Many published findings failing to reproduce in new studies

3.Descriptive statistics

Answer: s = 18.23

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 = (−17.57)² + 31.43² + (−4.571)² + (−7.571)² + (−13.57)² + (−7.571)² + 19.43² = 1994

Standard deviation

s = √(1994/6) = 18.23

Answer

s = 18.23

4.Descriptive statistics

Answer: P = 0.02275

Standardise

z = −2

Standard normal table

Φ(−2) = 0.02275

P = 0.02275

Answer

P = 0.02275

5.Probability

Answer: P = 0.09091

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 = 4/12 · 3/11 = 12/132 = 0.09091

Answer

P = 0.09091

6.Probability

Answer: P = 0.2269

Binomial formula

P(X = 3) = 35·0.3³·0.7⁴ = 35·0.027·0.2401 = 0.2269

Answer

P = 0.2269

7.Normal distribution

Answer: z = −0.58

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 − 0.7190 = 0.2810

Read the z-table backwards

Find 0.2810 inside the table and read off z.

z = −0.58

Answer

z = −0.58

8.Normal distribution

Answer: x = 261.6 g

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) = 0.99 → z = 2.326

Back to x

x = 250 + 2.326·5 = 261.63 g

Answer

x = 261.63 g

9.Confidence intervals

Answer: E₂ = 7.83

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.

Ratio of margins

E₂/E₁ = (2.576/1.645)·√(36/36) = 1.566

New margin

E₂ = 5·1.566 = 7.83

Answer

E₂ = 7.83

10.Confidence intervals

Answer: n = 2122

Solve for n

n = (1.645·14/0.5)² = 2122

Round up

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

n = 2122

Answer

n = 2122

11.Hypothesis tests & confidence intervals

Answer: p = 0.01321

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.

Test statistic

SE = σ/√n = 10/√100 = 1

z = (52.22 − 50)/1 = 2.22

Tail area

Φ(2.22) = 0.9868

1 − Φ(2.22) = 0.0132

p-value

p = 0.01321

Answer

p = 0.01321

p ≤ 0.05, so H₀ is rejected at the 5% level.

12.Hypothesis tests & confidence intervals

Answer: z = 0.8875

Sample proportions

p̂₁ = 25/100 = 0.25

p̂₂ = 24/120 = 0.2

p̂ = 49/220 = 0.2227

Standard error under H₀

SE = √(0.2227·0.7773·(1/100 + 1/120)) = 0.05634

Test statistic

z = (0.25 − 0.2)/0.05634 = 0.8875

Answer

z = 0.8875

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

13.Linear regression & correlation

Answer: r = 0.984

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

S_xy = 16192 − 132·720/6 = 352

S_xx = 2974 − 132²/6 = 70

S_yy = 88228 − 720²/6 = 1828

Correlation coefficient

r = 352/√(70·1828) = 0.984

Answer

r = 0.984

14.Linear regression & correlation

Answer: e = −2.2

Means

x̄ = 100/5 = 20

ȳ = 381/5 = 76.2

Sums of squares and products

S_xx = Σ(x − x̄)² = 90

S_xy = Σ(x − x̄)(y − ȳ) = 420

Regression line

b = 420/90 = 4.667

a = 76.2 − 4.667·20 = −17.13

ŷ = −17.13 + 4.667·x

Predicted value

ŷ = −17.13 + 4.667·21.5 = 83.2

Residual

e = 81 − 83.2 = −2.2

Answer

e = −2.2

The point lies below the line.

15.Chi-square tests

Answer: χ² = 2.443

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

students: 40·20/80 = 10, 40·28/80 = 14, 40·32/80 = 16

staff: 40·20/80 = 10, 40·28/80 = 14, 40·32/80 = 16

Chi-square statistic

χ² = (7 − 10)²/10 + (15 − 14)²/14 + (18 − 16)²/16 + (13 − 10)²/10 + (13 − 14)²/14 + (14 − 16)²/16

χ² = 2.443

Answer

χ² = 2.443

df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is not rejected.

16.Chi-square tests

Answer: χ² = 16.67

Expected counts

plant A: 60, 40

plant B: 24, 16

plant C: 36, 24

Chi-square statistic

χ² = (74 − 60)²/60 + (26 − 40)²/40 + (17 − 24)²/24 + (23 − 16)²/16 + (29 − 36)²/36 + (31 − 24)²/24

χ² = 16.67

Decision

df = 2, χ²_crit = 5.991

16.67 > 5.991

Answer

χ² = 16.67

χ² is above the critical value, so H₀ is rejected: the variables appear to be related.

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