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

Which kind of data is collected with an open question in an interview?

  1. ARatio data
  2. BNominal quantitative data only
  3. CQualitative data
  4. DInterval data
2.Research methods in psychologyHard

A self-report scale finds that people who agree with many statements score high regardless of content. Which response bias is this, and how can it be reduced?

  1. ADemand characteristics; use a double-blind design
  2. BAcquiescence bias; include reverse-worded items
  3. CSocial desirability bias; use a larger sample
  4. DOrder effects; use counterbalancing of participants
3.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 21, 59, 24, 39, 26, 18.

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

Exam scores are normally distributed with mean 70 and standard deviation 10. Find the z-score z of a score of 76.

μ = 70σ = 10x = 76
z =
5.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6?7P(six) = 1/6 on each roll

A fair die is rolled 7 times. What is the probability P of getting at least one six?

n = 7
P =
6.ProbabilityHard
n = 10, p = 0.5: P(X ≤ 1) = ?k012345678910

10 components are tested, and each one fails with probability 0.5, independently. What is the probability P that at most one fails?

n = 10p = 0.5
P =
7.Normal distributionMedium
320 g316.6 gT = X₁ + … + X₈?

The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 8 of them is below 316.6 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 8< 316.6 g
P =
8.Normal distributionHard
157164171178185192199cmbottom 15%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded bottom 15% is to the left of it. Find the cut-off x.

μ = 178 cmσ = 7 cm15%
x = cm
9.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?90%

The battery life of two phone models is compared. Model 1: 40 phones, mean 122.5 h, known standard deviation 6 h. Model 2: 100 phones, mean 114 h, known standard deviation 9 h. Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.

n₁ = 40x̄₁ = 122.5 hσ₁ = 6 hn₂ = 100x̄₂ = 114 hσ₂ = 9 h90%
E = h
10.Confidence intervalsHard
σ = 29, n = ?x̄ − Ex̄ + Ex̄E = 2.590%

The population standard deviation is 29. 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 2.5?

σ = 29E = 2.590%
n =
11.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 87 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 87s = 20n = 16
t =
12.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 4α = 0.01βσ = 23, power 80%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 4 units with a power of 80%. The standard deviation is 23. Find the minimum sample size n.

α = 0.01power = 80%δ = 4σ = 23zα = 2.326zβ = 0.842
n =
13.Linear regression & correlationMedium
xy2484546598681076xy510506070

The table gives 5 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 30Σy = 305Σx² = 220Σy² = 19101Σxy = 1970
r =
14.Linear regression & correlationHard
xy2841261681710251225xy5101020x = 7

The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 7.

n = 6x = 7
ŷ =
15.Chi-square testsMedium
busbikecarTotalstudents18253780staff362935100Total545472180

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 = 1802 × 3
χ² =
16.Chi-square testsHard
passfailTotalplant A213960plant B293160plant C402060Total9090180dfα = 0.05α = 0.0113.8416.63525.9919.21037.81511.34549.48813.277511.07015.086612.59216.812714.06718.475815.50720.090916.91921.6661018.30723.209

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 = 1803 × 2α = 0.05
χ² =

3Solutions

1.Research methods in psychology

Answer: C · Qualitative data

Why this is the answer

Open questions produce rich qualitative data that is harder to analyse statistically than closed-question responses.

Answer

C · Qualitative data
2.Research methods in psychology

Answer: B · Acquiescence bias; include reverse-worded items

Why this is the answer

Acquiescence is a tendency to agree with items. Mixing positively and negatively worded items means that agreeing with everything no longer produces an extreme score.

Answer

B · Acquiescence bias; include reverse-worded items
3.Descriptive statistics

Answer: s = 15.43

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 = 187/6 = 31.17

Squared deviations

Σ(x − x̄)² = (−10.17)² + 27.83² + (−7.167)² + 7.833² + (−5.167)² + (−13.17)² = 1191

Standard deviation

s = √(1191/5) = 15.43

Answer

s = 15.43
4.Descriptive statistics

Answer: z = 0.6

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 = (76 − 70)/10 = 0.6

Answer

z = 0.6
5.Probability

Answer: P = 0.7209

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.

Complement

P(no six) = (5/6)7 = 0.2791
P = 1 − 0.2791 = 0.7209

Answer

P = 0.7209
6.Probability

Answer: P = 0.01074

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) = 0.510 = 0.0009766
P(X = 1) = 10·0.5·0.59 = 0.009766

Add them

P = 0.0009766 + 0.009766 = 0.01074

Answer

P = 0.01074
7.Normal distribution

Answer: P = 0.1587

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 total

μT = 320 g, σT = 1.2·√8 = 3.394 g

Standardise

z = (316.6 − 320)/3.394 = −1

Probability

P = Φ(−1) = 0.1587

Answer

P = 0.1587
8.Normal distribution

Answer: x = 170.7 cm

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

Φ(z) = 0.15

z from the table

z ≈ −1.04

Back to the original units

x = 178 − 1.04·7 = 170.7 cm

Answer

x = 170.7 cm
9.Confidence intervals

Answer: E = 2.151 h

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.

Point estimate

x̄₁ − x̄₂ = 122.5 − 114 = 8.5

Standard error

SE = √(6²/40 + 9²/100) = 1.308

Margin of error

E = 1.645·1.308 = 2.151

Answer

E = 2.151 h

The interval does not contain 0, so the means differ significantly.

10.Confidence intervals

Answer: n = 365

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 = (1.645·29/2.5)² = 364.1

Round up

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

n = 365

Answer

n = 365
11.Hypothesis tests & confidence intervals

Answer: t = 2.4

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 = 20/√16 = 20/4 = 5

Test statistic

t = (87 − 75)/5 = 2.4

Answer

t = 2.4
t0.025; df = 15 = 2.131

|t| > 2.131, so H₀ is rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: n = 332

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((2.326 + 0.842)·23/4)² = 331.8

Round up

n = 332

Answer

n = 332
13.Linear regression & correlation

Answer: r = 0.9939

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 = 1970 − 30·305/5 = 140
Sxx = 220 − 30²/5 = 40
Syy = 19101 − 305²/5 = 496

Correlation coefficient

r = 140/√(40·496) = 0.9939

Answer

r = 0.9939
14.Linear regression & correlation

Answer: ŷ = 17.17

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 = 42/6 = 7
ȳ = 103/6 = 17.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 125

Regression line

b = 125/70 = 1.786
a = 17.17 − 1.786·7 = 4.667
ŷ = 4.667 + 1.786·x

Prediction

ŷ = 4.667 + 1.786·7 = 17.17

Answer

ŷ = 17.17
15.Chi-square tests

Answer: χ² = 4.181

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: 80·54/180 = 24, 80·54/180 = 24, 80·72/180 = 32
staff: 100·54/180 = 30, 100·54/180 = 30, 100·72/180 = 40

Chi-square statistic

χ² = (18 − 24)²/24 + (25 − 24)²/24 + (37 − 32)²/32 + (36 − 30)²/30 + (29 − 30)²/30 + (35 − 40)²/40
χ² = 4.181

Answer

χ² = 4.181

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

16.Chi-square tests

Answer: χ² = 12.13

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

plant A: 30, 30
plant B: 30, 30
plant C: 30, 30

Chi-square statistic

χ² = (21 − 30)²/30 + (39 − 30)²/30 + (29 − 30)²/30 + (31 − 30)²/30 + (40 − 30)²/30 + (20 − 30)²/30
χ² = 12.13

Decision

df = 2, χ²crit = 5.991
12.13 > 5.991

Answer

χ² = 12.13

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

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