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 researcher sends a questionnaire to all students and then asks respondents to recruit friends who also use a rare sleep app. Which sampling method is used for the second step?

  1. AStratified sampling
  2. BRandom sampling
  3. CSystematic sampling
  4. DSnowball sampling
2.Research methods in psychologyHard

Effect size measures such as Cohen's d are reported because they:

  1. AShow how large an effect is, regardless of sample size
  2. BTell how many participants dropped out
  3. CReplace the need for any hypothesis
  4. DShow whether the result is statistically significant at the chosen alpha level
3.Descriptive statisticsMedium
2025303540

Find the sample standard deviation s of the data: 40, 35, 28, 41, 17.

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

A quantity is normally distributed with mean 60 and standard deviation 10. What proportion P of values is below 75?

μ = 60σ = 10x = 75
P =
5.ProbabilityMedium
xp(x)00.1520.2540.2560.35μ = 3.6

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

x: 0, 2, 4, 6p(x): 0.15, 0.25, 0.25, 0.35
σ =
6.ProbabilityHard
xp(x)10.0520.3530.3540.25μ

X has the distribution shown, and Y = 0.5X + 10. Find the standard deviation σY of Y.

x: 1, 2, 3, 4p(x): 0.05, 0.35, 0.35, 0.25Y = 0.5X + 10
σY =
7.Normal distributionMedium
200 mm200.4 mmX + Y?

X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 200.4 mm. Round z to two decimals.

μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 200.4 mm
P =
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
x̄E₁ = 7.699%, n = 64x̄E₂ = ?99%, n = 128

A 99% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 7.6. What is the margin of error E₂ if the same study uses 128 observations instead?

99%n₁ = 64E₁ = 7.6n₂ = 128
E₂ =
10.Confidence intervalsHard
μ₁ − μ₂, t(20) = 2.0860LU = ?x̄₁ − x̄₂E95%

Two fertilisers are tested. Plot group 1: 16 plots, mean yield 85 kg, standard deviation 6 kg. Plot group 2: 6 plots, mean 77.5 kg, standard deviation 9 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 16x̄₁ = 85s₁ = 6n₂ = 6x̄₂ = 77.5s₂ = 995%t(20) = 2.086
upper limit = kg
11.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 50 parts has mean length 47 mm, and the line's known standard deviation is 19 mm. Line 2: a sample of 80 parts has mean 41 mm, known standard deviation 15 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 50x̄₁ = 47 mmσ₁ = 19 mmn₂ = 80x̄₂ = 41 mmσ₂ = 15 mm
z =
12.Hypothesis tests & confidence intervalsHard
σ = 8, n = ?x̄ − Ex̄ + Ex̄E = 395%

How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 3? The standard deviation is 8. Find the smallest sample size n.

σ = 8E = 395%
n =
13.Linear regression & correlationMedium
xy237342440550657755xy2464050

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 27Σy = 281Σx² = 139Σy² = 13507Σxy = 1337
r =
14.Linear regression & correlationHard
xy155359572786997xy56080100x = 2

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 2.

n = 5x = 2
ŷ =
15.Chi-square testsMedium
lowmediumhighTotalurban8233061rural40242993Total484759154

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

row total = 93column total = 59N = 154
E =
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 6 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 = 0.01.

α = 0.01
χ²crit =

3Solutions

1.Research methods in psychology

Answer: D · Snowball sampling

Why this is the answer

Snowball sampling is useful for hard-to-reach groups, but the sample is likely to share characteristics and is not representative.

Answer

D · Snowball sampling
2.Research methods in psychology

Answer: A · Show how large an effect is, regardless of sample size

Why this is the answer

With large samples tiny effects can be significant. Effect sizes express the magnitude of a difference, e.g. d = 0.2 small, 0.5 medium, 0.8 large.

Answer

A · Show how large an effect is, regardless of sample size
3.Descriptive statistics

Answer: s = 9.935

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 = 161/5 = 32.2

Squared deviations

Σ(x − x̄)² = 7.8² + 2.8² + (−4.2)² + 8.8² + (−15.2)² = 394.8

Standard deviation

s = √(394.8/4) = 9.935

Answer

s = 9.935
4.Descriptive statistics

Answer: P = 0.9332

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 = 1.5

Standard normal table

Φ(1.5) = 0.9332
P = 0.9332

Answer

P = 0.9332
5.Probability

Answer: σ = 2.154

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 = 0·0.15 + 2·0.25 + 4·0.25 + 6·0.35 = 3.6

Mean of the squares

E(X²) = 0·0.15 + 4·0.25 + 16·0.25 + 36·0.35 = 17.6

Variance

Var(X) = 17.6 − 3.6² = 4.64
σ = √4.64 = 2.154

Answer

σ = 2.154
6.Probability

Answer: σY = 0.4359

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 = 2.8, Var(X) = 0.76, σX = 0.8718

Linear change

σY = |0.5|·0.8718 = 0.4359

Answer

σY = 0.4359

The mean does move: E(Y) = 0.5·2.8 + 10 = 11.4.

7.Normal distribution

Answer: P = 0.2119

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 sum

μ = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm

Standardise

z = (200.4 − 200)/0.5 = 0.8

Upper tail

P = 1 − Φ(0.8) = 1 − 0.7881 = 0.2119

Answer

P = 0.2119
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: E₂ = 5.374

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.

Ratio of margins

E₂/E₁ = (2.576/2.576)·√(64/128) = 0.7071

New margin

E₂ = 7.6·0.7071 = 5.374

Answer

E₂ = 5.374
10.Confidence intervals

Answer: upper limit = 14.36 kg

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.

Pooled standard deviation

sp² = (15·6² + 5·9²)/20 = 47.25
sp = 6.874

Margin of error

E = 2.086·6.874·√(1/16 + 1/6) = 6.864

Interval

7.5 ± 6.864 → [0.6358, 14.36]

Answer

upper limit = 14.36 kg
11.Hypothesis tests & confidence intervals

Answer: z = 1.894

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.

Standard error of the difference

SE = √(19²/50 + 15²/80) = √10.03 = 3.167

Test statistic

z = (47 − 41)/3.167 = 1.894

Answer

z = 1.894

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: n = 28

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).

Solve for n

n = (1.96·8/3)² = 27.32

Round up

n = 28

Answer

n = 28
13.Linear regression & correlation

Answer: r = 0.9306

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 = 1337 − 27·281/6 = 72.5
Sxx = 139 − 27²/6 = 17.5
Syy = 13507 − 281²/6 = 346.8

Correlation coefficient

r = 72.5/√(17.5·346.8) = 0.9306

Answer

r = 0.9306
14.Linear regression & correlation

Answer: ŷ = 57.15

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 = 25/5 = 5
ȳ = 369/5 = 73.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 222

Regression line

b = 222/40 = 5.55
a = 73.8 − 5.55·5 = 46.05
ŷ = 46.05 + 5.55·x

Prediction

ŷ = 46.05 + 5.55·2 = 57.15

Answer

ŷ = 57.15
15.Chi-square tests

Answer: E = 35.63

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 = 93·59/154 = 35.63

Answer

E = 35.63
16.Chi-square tests

Answer: χ²crit = 15.09

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 = 6 − 1 = 5

Table

χ²crit(df = 5, α = 0.01) = 15.086

Answer

χ²crit = 15.086

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

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