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

Demand characteristics are:

  1. ACharacteristics of the sample
  2. BCues that make participants guess the aim and change their behaviour
  3. CThe formal demands an ethics committee makes before a study can begin
  4. DStatistical requirements for a test
2.Research methods in psychologyHard

A participant takes a drug or a sugar pill but does not know which, while the researcher does. What design is this and what bias remains possible?

  1. AOpen trial; placebo effects are fully controlled
  2. BSingle-blind; experimenter bias is still possible
  3. CSingle-blind; demand characteristics are impossible
  4. DDouble-blind; no bias is possible
3.Descriptive statisticsMedium
1520253035

Find the sample standard deviation s of the data: 21, 14, 20, 18, 39, 27.

n = 6
s =
4.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 100 and standard deviation 8. What proportion P of values is above 103.2?

μ = 100σ = 8x = 103.2
P =
5.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6?7P(six) = 1/6 on each roll

A fair die is rolled 7 times. What is the probability P of getting at least one six?

n = 7
P =
6.ProbabilityHard
xp(x)10.1520.3530.3540.15μ

X has the distribution shown, and Y = −2X − 3. Find the standard deviation σY of Y.

x: 1, 2, 3, 4p(x): 0.15, 0.35, 0.35, 0.15Y = −2X − 3
σY =
7.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.6915

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.6915. Find z to two decimals.

A = 0.6915
z =
8.Normal distributionHard
μ = 178x = ?90%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 90% of them lie (the 90th percentile).

μ = 178σ = 7P = 90%
x = cm
9.Confidence intervalsMedium
n = 20, s = 11.5LUx̄ = 101E = ?99%

A sample of 20 items has mean 101 and standard deviation s = 11.5. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 20x̄ = 101s = 11.599%t(19) = 2.861
E =
10.Confidence intervalsHard
p ≈ 0.2, n = ?p̂ − Ep̂ + Ep̂E = 0.02590%

A pilot study suggests that about 0.2 of products are defective. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.025?

p ≈ 0.2E = 0.02590%
n =
11.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 52.4 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 52.4s = 20n = 100
t =
12.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 81 measurements has mean 97.5. The population standard deviation is 15. Find the lower limit L of the 99% confidence interval for the mean.

x̄ = 97.5σ = 15n = 8199%
L =
13.Linear regression & correlationMedium
xy247352459561667xy2465060

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 5
r =
14.Linear regression & correlationHard
xy284176228301036xy510102030x = 3

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. Predict ŷ for x = 3.

n = 5x = 3
ŷ =
15.Chi-square testsMedium
no defectminormajorTotalday shift9211242night shift27233686Total364448128

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 = 86column total = 36N = 128
E =
16.Chi-square testsHard

A chi-square goodness-of-fit test checks whether counts in 8 categories follow a Poisson distribution whose mean is estimated from the same data. How many degrees of freedom df does the test have?

k = 8
df =

3Solutions

1.Research methods in psychology

Answer: B · Cues that make participants guess the aim and change their behaviour

Why this is the answer

When participants work out the hypothesis they may act to confirm or sabotage it, threatening validity. Deception and blind procedures reduce this.

Answer

B · Cues that make participants guess the aim and change their behaviour
2.Research methods in psychology

Answer: B · Single-blind; experimenter bias is still possible

Why this is the answer

In a single-blind design participants do not know their condition, which controls placebo expectations. The researcher's knowledge can still influence results, so double-blind designs are preferred.

Answer

B · Single-blind; experimenter bias is still possible
3.Descriptive statistics

Answer: s = 8.841

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 = 139/6 = 23.17

Squared deviations

Σ(x − x̄)² = (−2.167)² + (−9.167)² + (−3.167)² + (−5.167)² + 15.83² + 3.833² = 390.8

Standard deviation

s = √(390.8/5) = 8.841

Answer

s = 8.841
4.Descriptive statistics

Answer: P = 0.3446

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 = 0.4

Standard normal table

Φ(0.4) = 0.6554
P = 1 − 0.6554 = 0.3446

Answer

P = 0.3446
5.Probability

Answer: P = 0.7209

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.

Complement

P(no six) = (5/6)7 = 0.2791
P = 1 − 0.2791 = 0.7209

Answer

P = 0.7209
6.Probability

Answer: σY = 1.844

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 = 2.5, Var(X) = 0.85, σX = 0.922

Linear change

σY = |−2|·0.922 = 1.844

Answer

σY = 1.844

The mean does move: E(Y) = −2·2.5 − 3 = −8.

7.Normal distribution

Answer: z = −0.5

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 − 0.6915 = 0.3085

Read the z-table backwards

Find 0.3085 inside the table and read off z.

z = −0.50

Answer

z = −0.50
8.Normal distribution

Answer: x = 187 cm

Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.

z for the percentile

Φ(z) = 0.9 → z = 1.282

Back to x

x = 178 + 1.282·7 = 186.97 cm

Answer

x = 186.97 cm
9.Confidence intervals

Answer: E = 7.357

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 = 19, t = 2.861

Margin of error

E = 2.861·11.5/√20 = 7.357

Answer

E = 7.357
10.Confidence intervals

Answer: n = 693

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 = 1.645²·0.2·0.8/0.025² = 692.7

Round up

n = 693

Answer

n = 693
11.Hypothesis tests & confidence intervals

Answer: t = 1.2

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/√100 = 20/10 = 2

Test statistic

t = (52.4 − 50)/2 = 1.2

Answer

t = 1.2
t0.025; df = 99 = 1.984

|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: L = 93.21

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 = 2.576·15/√81 = 4.293

Interval limit

L = 97.5 − 4.293 = 93.21

Answer

L = 93.21
13.Linear regression & correlation

Answer: r = 0.9904

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 = 286/5 = 57.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 49
Syy = Σ(y − ȳ)² = 244.8

Correlation coefficient

r = 49/√(10·244.8) = 0.9904

Answer

r = 0.9904
14.Linear regression & correlation

Answer: ŷ = 12.25

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 = 113/5 = 22.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 138

Regression line

b = 138/40 = 3.45
a = 22.6 − 3.45·6 = 1.9
ŷ = 1.9 + 3.45·x

Prediction

ŷ = 1.9 + 3.45·3 = 12.25

Answer

ŷ = 12.25
15.Chi-square tests

Answer: E = 24.19

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 = 86·36/128 = 24.19

Answer

E = 24.19
16.Chi-square tests

Answer: df = 6

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 = 8 − 1 − 1 = 6

Answer

df = 6

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