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

A student scores 130 on a test with mean 100 and standard deviation 15. What is the z-score?

  1. A1.5
  2. B2.0
  3. C0.5
  4. D3.0
2.Research methods in psychologyHard

A study shows that a questionnaire correlates strongly with an established test of the same construct. This supports:

  1. ATest–retest reliability
  2. BFace validity
  3. CConcurrent validity
  4. DEcological validity
3.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 39, 14, 31, 43, 32, 50, 52.

n = 7
s =
4.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 50 and standard deviation 20. What proportion P of values is above 70?

μ = 50σ = 20x = 70
P =
5.ProbabilityMedium
n = 14k = 5N = ?

In how many ways N can a committee of 5 people be chosen from a group of 14? The order does not matter.

n = 14k = 5
N =
6.ProbabilityHard
n = 9, p = 0.4: P(X ≤ 1) = ?k0123456789

9 components are tested, and each one fails with probability 0.4, independently. What is the probability P that at most one fails?

n = 9p = 0.4
P =
7.Normal distributionMedium
400 g407.6 gT = X₁ + … + X₁₀?

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 10 of them exceeds 407.6 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 10> 407.6 g
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 = 8, s = 9L = ?Ux̄ = 66.5E95%

A sample of 8 items has mean 66.5 and standard deviation s = 9. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 8x̄ = 66.5s = 995%t(7) = 2.365
lower limit =
10.Confidence intervalsHard
p ≈ 0.1, n = ?p̂ − Ep̂ + Ep̂E = 0.02595%

A pilot study suggests that about 0.1 of products are defective. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.025?

p ≈ 0.1E = 0.02595%
n =
11.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 72.24 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 72.24s = 8n = 81
t =
12.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 180 of 300 patients on treatment A recover, and 73 of 100 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 180n₁ = 300x₂ = 73n₂ = 100
z =
13.Linear regression & correlationMedium
xy153617431953215823702566xy1520254060

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 = 120Σy = 326Σx² = 2470Σy² = 18574Σxy = 6756
r =
14.Linear regression & correlationHard
xy2114186278371044xy5102040ŷ = 23

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 23?

n = 5ŷ = 23
x =
15.Chi-square testsMedium
lowmediumhighTotalurban20392079rural15112854Total355048133

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

row total = 54column total = 50N = 133
E =
16.Chi-square testsHard

A chi-square test of independence is carried out on a contingency table with 3 rows and 5 columns (not counting the totals). How many degrees of freedom df does the test have?

3 × 5
df =

3Solutions

1.Research methods in psychology

Answer: B · 2.0

Why this is the answer

(130 − 100) / 15 = 2. The score is two standard deviations above the mean, higher than about 97–98% of people.

Answer

B · 2.0
2.Research methods in psychology

Answer: C · Concurrent validity

Why this is the answer

Concurrent validity is shown when a new measure agrees with an established measure taken at about the same time.

Answer

C · Concurrent validity
3.Descriptive statistics

Answer: s = 13.06

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 = 261/7 = 37.29

Squared deviations

Σ(x − x̄)² = 1.714² + (−23.29)² + (−6.286)² + 5.714² + (−5.286)² + 12.71² + 14.71² = 1023

Standard deviation

s = √(1023/6) = 13.06

Answer

s = 13.06
4.Descriptive statistics

Answer: P = 0.1587

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 = 1

Standard normal table

Φ(1) = 0.8413
P = 1 − 0.8413 = 0.1587

Answer

P = 0.1587
5.Probability

Answer: N = 2002

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, 5) = (14·13·12·11·10)/(5·4·3·2·1) = 2002

Answer

N = 2002
6.Probability

Answer: P = 0.07054

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) = 0.69 = 0.01008
P(X = 1) = 9·0.4·0.68 = 0.06047

Add them

P = 0.01008 + 0.06047 = 0.07054

Answer

P = 0.07054
7.Normal distribution

Answer: P = 0.02275

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 = 400 g, σT = 1.2·√10 = 3.795 g

Standardise

z = (407.6 − 400)/3.795 = 2

Probability

P = 1 − Φ(2) = 0.02275

Answer

P = 0.02275
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: lower limit = 58.97

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 = 7, t = 2.365

Margin of error

E = 2.365·9/√8 = 7.525

Interval

66.5 ± 7.525 → [58.97, 74.03]

Answer

lower limit = 58.97
10.Confidence intervals

Answer: n = 554

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 = 1.96²·0.1·0.9/0.025² = 553.2

Round up

n = 554

Answer

n = 554
11.Hypothesis tests & confidence intervals

Answer: t = −3.105

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/√81 = 8/9 = 0.8889

Test statistic

t = (72.24 − 75)/0.8889 = −3.105

Answer

t = −3.105
t0.025; df = 80 = 1.99

|t| > 1.99, so H₀ is rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: z = −2.335

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 = 180/300 = 0.6
p̂₂ = 73/100 = 0.73
p̂ = 253/400 = 0.6325

Standard error under H₀

SE = √(0.6325·0.3675·(1/300 + 1/100)) = 0.05567

Test statistic

z = (0.6 − 0.73)/0.05567 = −2.335

Answer

z = −2.335

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

13.Linear regression & correlation

Answer: r = 0.9611

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 = 6756 − 120·326/6 = 236
Sxx = 2470 − 120²/6 = 70
Syy = 18574 − 326²/6 = 861.3

Correlation coefficient

r = 236/√(70·861.3) = 0.9611

Answer

r = 0.9611
14.Linear regression & correlation

Answer: x = 4.965

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 = 30/5 = 6
ȳ = 137/5 = 27.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 170

Regression line

b = 170/40 = 4.25
a = 27.4 − 4.25·6 = 1.9
ŷ = 1.9 + 4.25·x

Solve for x

x = (23 − 1.9)/4.25 = 4.965

Answer

x = 4.965
15.Chi-square tests

Answer: E = 20.3

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 = 54·50/133 = 20.3

Answer

E = 20.3
16.Chi-square tests

Answer: df = 8

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)·(5 − 1) = 8

Answer

df = 8

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