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

In a positively skewed distribution, such as income:

  1. AThe mean is greater than the median
  2. BThe mean and median are always equal
  3. CThe mean is less than the median
  4. DThere are no extreme scores
2.Research methods in psychologyHard

A one-way ANOVA comparing four groups is significant. What should the researcher do next to find which groups differ?

  1. ARun post hoc tests such as Tukey's HSD
  2. BRun a chi-square test on the means
  3. CConclude that all four groups differ from each other
  4. DReport the result as a correlation
3.Descriptive statisticsMedium
20253035404550

Find the sample standard deviation s of the data: 53, 17, 47, 46, 53.

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

Exam scores are normally distributed with mean 70 and standard deviation 20. Find the z-score z of a score of 78.

μ = 70σ = 20x = 78
z =
5.ProbabilityMedium
red: 3blue: 6

A bag holds 3 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

3 / 6
P =
6.ProbabilityHard
€p(x)−20.7840.2180.02

A game costs €2 to play. You win €6 with probability 0.2 and €20 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €2€6: p = 0.2€20: p = 0.02
E(G) = €
7.Normal distributionMedium
640 g635.2 gT = X₁ + … + X₁₆?

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 16 of them is below 635.2 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 16< 635.2 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
x̄E₁ = 8.595%, n = 64x̄E₂ = ?90%, n = 256

A 95% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 8.5. What is the margin of error E₂ of a 90% interval based on 256 observations?

95%n₁ = 64E₁ = 8.590%n₂ = 256
E₂ =
10.Confidence intervalsHard
n = 10Lσ = ?s = 7.195%

The diameters of 10 machined shafts have a sample standard deviation of 7.1 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 9, 2.5% of the χ² distribution lies below 2.7 and 2.5% lies above 19.023.

n = 10s = 7.1 µmχ² = 2.7, 19.023
σ = µm
11.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 197.9 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 197.9s = 8n = 100
t =
12.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 4α = 0.05βσ = 27, power 80%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 4 units with a power of 80%. The standard deviation is 27. Find the minimum sample size n.

α = 0.05power = 80%δ = 4σ = 27zα = 1.645zβ = 0.842
n =
13.Linear regression & correlationMedium
xy26347367989310102xy5106080100

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 = 410Σx² = 220Σy² = 34592Σxy = 2656
r =
14.Linear regression & correlationHard
xy146353558771973xy5506070ŷ = 71

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 71?

n = 5ŷ = 71
x =
15.Chi-square testsMedium
busbikecarTotalstudents1791642staff29181360Total462729102

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

row total = 42column total = 46N = 102
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 checks whether counts in 7 categories follow a Poisson distribution whose mean is estimated from the same 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: A · The mean is greater than the median

Why this is the answer

A few very high values in the right tail raise the mean above the median, which is why the median is often reported for skewed data.

Answer

A · The mean is greater than the median
2.Research methods in psychology

Answer: A · Run post hoc tests such as Tukey's HSD

Why this is the answer

A significant F shows that not all means are equal. Post hoc tests or planned contrasts identify which pairs differ while controlling the error rate.

Answer

A · Run post hoc tests such as Tukey's HSD
3.Descriptive statistics

Answer: s = 15.01

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 = 216/5 = 43.2

Squared deviations

Σ(x − x̄)² = 9.8² + (−26.2)² + 3.8² + 2.8² + 9.8² = 900.8

Standard deviation

s = √(900.8/4) = 15.01

Answer

s = 15.01
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 = (78 − 70)/20 = 0.4

Answer

z = 0.4
5.Probability

Answer: P = 0.08333

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 = 3/9 · 2/8 = 6/72 = 0.08333

Answer

P = 0.08333
6.Probability

Answer: E(G) = −0.4 €

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.

Net outcomes

G: 4 (p = 0.2), 18 (p = 0.02), −2 (p = 0.78)

Expected value

E(G) = 4·0.2 + 18·0.02 − 2·0.78 = −0.4 €

Answer

E(G) = −0.4 €

A negative value means you lose money on average.

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 = 640 g, σT = 1.2·√16 = 4.8 g

Standardise

z = (635.2 − 640)/4.8 = −1

Probability

P = Φ(−1) = 0.1587

Answer

P = 0.1587
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: E₂ = 3.567

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 = (1.645/1.96)·√(64/256) = 0.4196

New margin

E₂ = 8.5·0.4196 = 3.567

Answer

E₂ = 3.567
10.Confidence intervals

Answer: σ = 12.96 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 9·7.1² = 453.7

Limits for σ²

453.7/19.023 = 23.85
453.7/2.7 = 168

Limit for σ

σ = √168 = 12.96 µm

Answer

σ = 12.96 µm

95% interval for σ: [4.884, 12.96] µm.

11.Hypothesis tests & confidence intervals

Answer: t = −2.625

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/√100 = 8/10 = 0.8

Test statistic

t = (197.9 − 200)/0.8 = −2.625

Answer

t = −2.625
t0.025; df = 99 = 1.984

|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: n = 282

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 = ((1.645 + 0.842)·27/4)² = 281.8

Round up

n = 282

Answer

n = 282
13.Linear regression & correlation

Answer: r = 0.994

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 = 2656 − 30·410/5 = 196
Sxx = 220 − 30²/5 = 40
Syy = 34592 − 410²/5 = 972

Correlation coefficient

r = 196/√(40·972) = 0.994

Answer

r = 0.994
14.Linear regression & correlation

Answer: x = 8

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 = 301/5 = 60.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 144

Regression line

b = 144/40 = 3.6
a = 60.2 − 3.6·5 = 42.2
ŷ = 42.2 + 3.6·x

Solve for x

x = (71 − 42.2)/3.6 = 8

Answer

x = 8
15.Chi-square tests

Answer: E = 18.94

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 = 42·46/102 = 18.94

Answer

E = 18.94
16.Chi-square tests

Answer: χ²crit = 11.07

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 = 7 − 1 − 1 = 5

Table

χ²crit(df = 5, α = 0.05) = 11.070

Answer

χ²crit = 11.070

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

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