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 standard error of the mean becomes smaller when:

  1. AThe variable is measured at ordinal level
  2. BThe mean increases
  3. CThe sample size decreases
  4. DThe sample size increases
2.Research methods in psychologyHard

A memory study uses a repeated measures design where everyone learns list A then list B. Recall is better for list B. What is the most likely flaw?

  1. AToo large a sample size
  2. BIndividual differences between the groups
  3. CThe use of an ordinal level of measurement
  4. DOrder effects such as practice, because conditions were not counterbalanced
3.Descriptive statisticsMedium
303540455055

Find the sample standard deviation s of the data: 58, 57, 29, 36, 56.

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

Exam scores are normally distributed with mean 500 and standard deviation 60. Find the z-score z of a score of 524.

μ = 500σ = 60x = 524
z =
5.ProbabilityMedium
xp(x)00.3510.320.35μ = 1

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

x: 0, 1, 2p(x): 0.35, 0.3, 0.35
σ =
6.ProbabilityHard
n = 5, p = 0.25: P(X ≤ 1) = ?k012345

5 components are tested, and each one fails with probability 0.25, independently. What is the probability P that at most one fails?

n = 5p = 0.25
P =
7.Normal distributionMedium
200 g202.7 gT = X₁ + … + X₅?

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 5 of them exceeds 202.7 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 5> 202.7 g
P =
8.Normal distributionHard
35455565758595pointstop 30%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 30% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points30%
x = points
9.Confidence intervalsMedium
n = 15, s = 4.5L = ?Ux̄ = 117E95%

A sample of 15 items has mean 117 and standard deviation s = 4.5. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 15x̄ = 117s = 4.595%t(14) = 2.145
lower limit =
10.Confidence intervalsHard
p ≈ 0.6, n = ?p̂ − Ep̂ + Ep̂E = 0.0199%

A pilot study suggests that about 0.6 of products are defective. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.01?

p ≈ 0.6E = 0.0199%
n =
11.Hypothesis tests & confidence intervalsMedium
−2.3652.365t = ?df = 7

8 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 −7.05 and the standard deviation of the differences is 9.5. Find the paired test statistic t for H₀: μd = 0.

n = 8d̄ = −7.05sd = 9.5
t =
12.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 63 of 250 patients on treatment A recover, and 56 of 150 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 63n₁ = 250x₂ = 56n₂ = 150
z =
13.Linear regression & correlationMedium
xy284861181610191222xy510101520

The table gives 6 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 42Σy = 84Σx² = 364Σy² = 1350Σxy = 696
r =
14.Linear regression & correlationHard
xy252350458565671xy246506070x = 2.5

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 (2.5, 56) is made. Find its residual e = y − ŷ.

n = 5(2.5, 56)
e =
15.Chi-square testsMedium
no defectminormajorTotalday shift18432081night shift24104276Total425362157

The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “night shift, no defect”.

row total = 76column total = 42N = 157
E =
16.Chi-square testsHard
yesnoTotalage 18–30384280age 31–50231740age 51+79180Total14060200dfα = 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 = 2003 × 2α = 0.05
χ² =

3Solutions

1.Research methods in psychology

Answer: D · The sample size increases

Why this is the answer

The standard error estimates how much sample means vary around the population mean. Larger samples give more precise estimates.

Answer

D · The sample size increases
2.Research methods in psychology

Answer: D · Order effects such as practice, because conditions were not counterbalanced

Why this is the answer

When all participants experience conditions in the same order, improvement may reflect practice rather than the list. Counterbalancing (ABBA or half AB, half BA) controls this.

Answer

D · Order effects such as practice, because conditions were not counterbalanced
3.Descriptive statistics

Answer: s = 13.66

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 = 236/5 = 47.2

Squared deviations

Σ(x − x̄)² = 10.8² + 9.8² + (−18.2)² + (−11.2)² + 8.8² = 746.8

Standard deviation

s = √(746.8/4) = 13.66

Answer

s = 13.66
4.Descriptive statistics

Answer: z = 0.4

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 = (524 − 500)/60 = 0.4

Answer

z = 0.4
5.Probability

Answer: σ = 0.8367

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 + 1·0.3 + 2·0.35 = 1

Mean of the squares

E(X²) = 0·0.35 + 1·0.3 + 4·0.35 = 1.7

Variance

Var(X) = 1.7 − 1² = 0.7
σ = √0.7 = 0.8367

Answer

σ = 0.8367
6.Probability

Answer: P = 0.6328

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) = 0.755 = 0.2373
P(X = 1) = 5·0.25·0.754 = 0.3955

Add them

P = 0.2373 + 0.3955 = 0.6328

Answer

P = 0.6328
7.Normal distribution

Answer: P = 0.1562

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 = 200 g, σT = 1.2·√5 = 2.683 g

Standardise

z = (202.7 − 200)/2.683 = 1.01

Probability

P = 1 − Φ(1.01) = 0.1562

Answer

P = 0.1562
8.Normal distribution

Answer: x = 70.2 points

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) = 0.7

z from the table

z ≈ 0.52

Back to the original units

x = 65 + 0.52·10 = 70.2 points

Answer

x = 70.2 points
9.Confidence intervals

Answer: lower limit = 114.5

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 = 14, t = 2.145

Margin of error

E = 2.145·4.5/√15 = 2.492

Interval

117 ± 2.492 → [114.5, 119.5]

Answer

lower limit = 114.5
10.Confidence intervals

Answer: n = 15926

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 2.576²·0.6·0.4/0.01² = 15926

Round up

n = 15926

Answer

n = 15926
11.Hypothesis tests & confidence intervals

Answer: t = −2.099

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 = 9.5/√8 = 3.359

Test statistic

t = −7.05/3.359 = −2.099

Answer

t = −2.099

With df = 7, the two-sided 5% critical value is 2.365; |t| is smaller, so H₀ is not rejected.

12.Hypothesis tests & confidence intervals

Answer: z = −2.57

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 = 63/250 = 0.252
p̂₂ = 56/150 = 0.3733
p̂ = 119/400 = 0.2975

Standard error under H₀

SE = √(0.2975·0.7025·(1/250 + 1/150)) = 0.04722

Test statistic

z = (0.252 − 0.3733)/0.04722 = −2.57

Answer

z = −2.57

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

13.Linear regression & correlation

Answer: r = 0.9786

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 = 696 − 42·84/6 = 108
Sxx = 364 − 42²/6 = 70
Syy = 1350 − 84²/6 = 174

Correlation coefficient

r = 108/√(70·174) = 0.9786

Answer

r = 0.9786
14.Linear regression & correlation

Answer: e = 4.75

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 = 20/5 = 4
ȳ = 296/5 = 59.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 53

Regression line

b = 53/10 = 5.3
a = 59.2 − 5.3·4 = 38
ŷ = 38 + 5.3·x

Predicted value

ŷ = 38 + 5.3·2.5 = 51.25

Residual

e = 56 − 51.25 = 4.75

Answer

e = 4.75

The point lies above the line.

15.Chi-square tests

Answer: E = 20.33

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 = 76·42/157 = 20.33

Answer

E = 20.33
16.Chi-square tests

Answer: χ² = 53.75

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

age 18–30: 56, 24
age 31–50: 28, 12
age 51+: 56, 24

Chi-square statistic

χ² = (38 − 56)²/56 + (42 − 24)²/24 + (23 − 28)²/28 + (17 − 12)²/12 + (79 − 56)²/56 + (1 − 24)²/24
χ² = 53.75

Decision

df = 2, χ²crit = 5.991
53.75 > 5.991

Answer

χ² = 53.75

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

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