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

The same 25 participants complete a memory test before and after caffeine. The scores are interval and roughly normal. Which test is appropriate?

  1. AIndependent samples t-test
  2. BChi-square test
  3. CPaired (related) samples t-test
  4. DMann–Whitney U test
2.Research methods in psychologyHard

Increasing the sample size in a study most directly:

  1. ARemoves all bias from the sample
  2. BIncreases statistical power to detect a real effect
  3. CIncreases the effect size
  4. DGuarantees a statistically significant result if any effect exists
3.Descriptive statisticsMedium
20253035404550

Find the sample standard deviation s of the data: 52, 18, 22, 46, 34.

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

Exam scores are normally distributed with mean 50 and standard deviation 5. Find the z-score z of a score of 40.

μ = 50σ = 5x = 40
z =
5.ProbabilityMedium
red: 6blue: 5

A bag holds 6 red and 5 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

6 / 5
P =
6.ProbabilityHard
xp(x)00.3510.1520.230.140.2μ

X has the distribution shown, and Y = 10X + 4. Find the standard deviation σY of Y.

x: 0, 1, 2, 3, 4p(x): 0.35, 0.15, 0.2, 0.1, 0.2Y = 10X + 4
σY =
7.Normal distributionMedium
45.577P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 45.5 and 77 points? Give P as a decimal.

μ = 65σ = 10x1 = 45.5x2 = 77
P =
8.Normal distributionHard
10 cm0X − Y?

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
n = 12, sd​ = 3.40LU = ?d̄ = −6.3E95%

The blood pressure of 12 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −6.3 and standard deviation 3.4. Find the upper limit of the 95% confidence interval for the mean difference.

n = 12d̄ = −6.3sd = 3.495%t(11) = 2.201
upper limit = mmHg
10.Confidence intervalsHard
σ = 37, n = ?x̄ − Ex̄ + Ex̄E = 0.595%

The population standard deviation is 37. 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 0.5?

σ = 37E = 0.595%
n =
11.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 48.24 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 48.24s = 8n = 100
t =
12.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 20 of 80 patients on treatment A recover, and 100 of 250 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 20n₁ = 80x₂ = 100n₂ = 250
z =
13.Linear regression & correlationMedium
xy133241345444548xy24354045

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
xy36157378499511107xy5106080100x = 4

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (4, 72) is made. Find its residual e = y − ŷ.

n = 5(4, 72)
e =
15.Chi-square testsMedium
lowmediumhighTotalurban14344593rural37201168Total515456161

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

row total = 68column total = 51N = 161
E =
16.Chi-square testsHard

A chi-square goodness-of-fit test checks whether counts in 7 categories follow a Poisson distribution whose mean is estimated from the same data. How many degrees of freedom df does the test have?

k = 7
df =

3Solutions

1.Research methods in psychology

Answer: C · Paired (related) samples t-test

Why this is the answer

A paired t-test compares means from the same participants under two conditions, using the differences within each person.

Answer

C · Paired (related) samples t-test
2.Research methods in psychology

Answer: B · Increases statistical power to detect a real effect

Why this is the answer

Larger samples reduce standard error, increasing power. They do not change the true effect size or remove systematic bias.

Answer

B · Increases statistical power to detect a real effect
3.Descriptive statistics

Answer: s = 14.72

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 = 172/5 = 34.4

Squared deviations

Σ(x − x̄)² = 17.6² + (−16.4)² + (−12.4)² + 11.6² + (−0.4)² = 867.2

Standard deviation

s = √(867.2/4) = 14.72

Answer

s = 14.72
4.Descriptive statistics

Answer: z = −2

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 = (40 − 50)/5 = −2

Answer

z = −2
5.Probability

Answer: P = 0.2727

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 = 6/11 · 5/10 = 30/110 = 0.2727

Answer

P = 0.2727
6.Probability

Answer: σY = 15.26

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 = 1.65, Var(X) = 2.328, σX = 1.526

Linear change

σY = |10|·1.526 = 15.26

Answer

σY = 15.26

The mean does move: E(Y) = 10·1.65 + 4 = 20.5.

7.Normal distribution

Answer: P = 0.8593

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 = (45.5 − 65)/10 = −1.95
z₂ = (77 − 65)/10 = 1.2

Area between

P = Φ(1.2) − Φ(−1.95) = 0.8849 − 0.02559 = 0.8593

Answer

P = 0.8593
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: upper limit = −4.14 mmHg

Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.

Critical value

df = 11, t = 2.201

Margin of error

E = 2.201·3.4/√12 = 2.16

Interval

−6.3 ± 2.16 → [−8.46, −4.14]

Answer

upper limit = −4.14 mmHg
10.Confidence intervals

Answer: n = 21037

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 = (1.96·37/0.5)² = 21037

Round up

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

n = 21037

Answer

n = 21037
11.Hypothesis tests & confidence intervals

Answer: t = −2.2

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

SE = 8/√100 = 8/10 = 0.8

Test statistic

t = (48.24 − 50)/0.8 = −2.2

Answer

t = −2.2
t0.025; df = 99 = 1.984

|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: z = −2.428

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 = 20/80 = 0.25
p̂₂ = 100/250 = 0.4
p̂ = 120/330 = 0.3636

Standard error under H₀

SE = √(0.3636·0.6364·(1/80 + 1/250)) = 0.06179

Test statistic

z = (0.25 − 0.4)/0.06179 = −2.428

Answer

z = −2.428

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

13.Linear regression & correlation

Answer: r = 0.9125

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 = 211/5 = 42.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 33
Syy = Σ(y − ȳ)² = 130.8

Correlation coefficient

r = 33/√(10·130.8) = 0.9125

Answer

r = 0.9125
14.Linear regression & correlation

Answer: e = 5.1

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 = 35/5 = 7
ȳ = 420/5 = 84

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 228

Regression line

b = 228/40 = 5.7
a = 84 − 5.7·7 = 44.1
ŷ = 44.1 + 5.7·x

Predicted value

ŷ = 44.1 + 5.7·4 = 66.9

Residual

e = 72 − 66.9 = 5.1

Answer

e = 5.1

The point lies above the line.

15.Chi-square tests

Answer: E = 21.54

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 = 68·51/161 = 21.54

Answer

E = 21.54
16.Chi-square tests

Answer: df = 5

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 = 7 − 1 − 1 = 5

Answer

df = 5

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