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

In thematic analysis, the analyst mainly:

  1. ACounts word frequencies only
  2. BTests a numerical hypothesis
  3. CIdentifies recurring patterns of meaning across qualitative data
  4. DCalculates the mean score of each theme to compare them statistically
3.Descriptive statisticsMedium
1015202530354045505560

Find the sample standard deviation s of the data: 10, 47, 60, 21, 41.

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

A quantity is normally distributed with mean 170 and standard deviation 15. What proportion P of values is above 200?

μ = 170σ = 15x = 200
P =
5.ProbabilityMedium
red: 5blue: 3

A bag holds 5 red and 3 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

5 / 3
P =
6.ProbabilityHard
n = 12, p = 0.5: P(X = 1) = ?k0123456789101112

A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.5, independently. What is the probability P of exactly 1 correct answers?

n = 12p = 0.5k = 1
P =
7.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.8444

Drag z so that the shaded area between −z and z under the standard normal curve is 0.8444. Find z to two decimals.

A = 0.8444
z =
8.Normal distributionHard
4 s0X − Y?

Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
9.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?95%

The battery life of two phone models is compared. Model 1: 50 phones, mean 77 h, known standard deviation 5 h. Model 2: 40 phones, mean 62 h, known standard deviation 5 h. Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.

n₁ = 50x̄₁ = 77 hσ₁ = 5 hn₂ = 40x̄₂ = 62 hσ₂ = 5 h95%
E = h
10.Confidence intervalsHard
σ = 24, n = ?x̄ − Ex̄ + Ex̄E = 2.599%

The population standard deviation is 24. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 2.5?

σ = 24E = 2.599%
n =
11.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 115.1 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 115.1s = 20n = 81
t =
12.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 49 measurements has mean 77.1. The population standard deviation is 8. Find the upper limit U of the 95% confidence interval for the mean.

x̄ = 77.1σ = 8n = 4995%
U =
13.Linear regression & correlationMedium
xy17331937213723522555xy20254050

The table gives 5 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 = 5Σx = 105Σy = 214Σx² = 2245Σy² = 9556Σxy = 4612
r =
14.Linear regression & correlationHard
xy18612175249027973012133135xy202530100x = 22.5

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 22.5.

n = 6x = 22.5
ŷ =
15.Chi-square testsMedium
no defectminormajorTotalday shift234453120night shift215524100Total449977220

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 = 2202 × 3
χ² =
16.Chi-square testsHard
dfα = 0.05α = 0.0113.8416.63525.9919.21037.81511.34549.48813.277511.07015.086612.59216.812714.06718.475815.50720.090916.91921.6661018.30723.209

A chi-square goodness-of-fit test compares observed counts in 11 categories with fully specified expected proportions (no parameters are estimated from the data). Using the table, find the critical value χ²crit at significance level α = 0.05.

α = 0.05
χ²crit =

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: C · Identifies recurring patterns of meaning across qualitative data

Why this is the answer

Thematic analysis involves familiarisation, coding, generating, reviewing and naming themes and writing up, focusing on meaning rather than statistics.

Answer

C · Identifies recurring patterns of meaning across qualitative data
3.Descriptive statistics

Answer: s = 20.14

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 = 179/5 = 35.8

Squared deviations

Σ(x − x̄)² = (−25.8)² + 11.2² + 24.2² + (−14.8)² + 5.2² = 1623

Standard deviation

s = √(1623/4) = 20.14

Answer

s = 20.14
4.Descriptive statistics

Answer: P = 0.02275

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 = 2

Standard normal table

Φ(2) = 0.9772
P = 1 − 0.9772 = 0.02275

Answer

P = 0.02275
5.Probability

Answer: P = 0.3571

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 = 5/8 · 4/7 = 20/56 = 0.3571

Answer

P = 0.3571
6.Probability

Answer: P = 0.00293

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.

Binomial formula

P(X = 1) = 12·0.51·0.511 = 12·0.5·0.0004883 = 0.00293

Answer

P = 0.00293
7.Normal distribution

Answer: z = 1.42

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 + 0.8444)/2 = 0.9222

Read the z-table backwards

Find 0.9222 inside the table and read off z.

z = 1.42

Answer

z = 1.42
8.Normal distribution

Answer: P = 0.7881

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 = 52 − 48 = 4 s, σ = 5 s

Standardise

z = (0 − 4)/5 = −0.8

Probability

P = 1 − Φ(−0.8) = 0.7881

Answer

P = 0.7881
9.Confidence intervals

Answer: E = 2.079 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 = 77 − 62 = 15

Standard error

SE = √(5²/50 + 5²/40) = 1.061

Margin of error

E = 1.96·1.061 = 2.079

Answer

E = 2.079 h

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

10.Confidence intervals

Answer: n = 612

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 = (2.576·24/2.5)² = 611.6

Round up

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

n = 612

Answer

n = 612
11.Hypothesis tests & confidence intervals

Answer: t = −2.205

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/√81 = 20/9 = 2.222

Test statistic

t = (115.1 − 120)/2.222 = −2.205

Answer

t = −2.205
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: U = 79.34

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).

Margin of error

E = 1.96·8/√49 = 2.24

Interval limit

U = 77.1 + 2.24 = 79.34

Answer

U = 79.34
13.Linear regression & correlation

Answer: r = 0.9366

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 = 4612 − 105·214/5 = 118
Sxx = 2245 − 105²/5 = 40
Syy = 9556 − 214²/5 = 396.8

Correlation coefficient

r = 118/√(40·396.8) = 0.9366

Answer

r = 0.9366
14.Linear regression & correlation

Answer: ŷ = 81.79

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 = 153/6 = 25.5
ȳ = 579/6 = 96.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 772.5

Regression line

b = 772.5/157.5 = 4.905
a = 96.5 − 4.905·25.5 = −28.57
ŷ = −28.57 + 4.905·x

Prediction

ŷ = −28.57 + 4.905·22.5 = 81.79

Answer

ŷ = 81.79
15.Chi-square tests

Answer: χ² = 10.5

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.

Expected counts

day shift: 120·44/220 = 24, 120·99/220 = 54, 120·77/220 = 42
night shift: 100·44/220 = 20, 100·99/220 = 45, 100·77/220 = 35

Chi-square statistic

χ² = (23 − 24)²/24 + (44 − 54)²/54 + (53 − 42)²/42 + (21 − 20)²/20 + (55 − 45)²/45 + (24 − 35)²/35
χ² = 10.5

Answer

χ² = 10.5

df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is rejected.

16.Chi-square tests

Answer: χ²crit = 18.31

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 = 11 − 1 = 10

Table

χ²crit(df = 10, α = 0.05) = 18.307

Answer

χ²crit = 18.307

Reject H₀ if the computed χ² is larger than this value.

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