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

Experimenter bias is reduced most directly by:

  1. AUsing only volunteers
  2. BUsing a larger sample
  3. CUsing a researcher who does not know the hypothesis or conditions to collect data
  4. DUsing a one-tailed hypothesis
2.Research methods in psychologyHard

A memory study uses a repeated measures design where everyone learns list A then list B. Recall is better for list B. What is the most likely flaw?

  1. AToo large a sample size
  2. BThe use of an ordinal level of measurement
  3. CIndividual differences between the groups
  4. DOrder effects such as practice, because conditions were not counterbalanced
3.Descriptive statisticsMedium
15202530354045505560

Find the sample standard deviation s of the data: 52, 31, 11, 19, 60, 26, 12.

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

A quantity is normally distributed with mean 100 and standard deviation 10. What proportion P of values is below 80?

μ = 100σ = 10x = 80
P =
5.ProbabilityMedium
xp(x)00.3510.3520.3μ = 0.95

For the distribution shown, find the variance Var(X) of X.

x: 0, 1, 2p(x): 0.35, 0.35, 0.3
Var(X) =
6.ProbabilityHard
n = 5, p = 0.1: P(X ≤ 1) = ?k012345

5 components are tested, and each one fails with probability 0.1, independently. What is the probability P that at most one fails?

n = 5p = 0.1
P =
7.Normal distributionMedium
176.25188.85P = ?

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 176.25 and 188.85 cm? Give P as a decimal.

μ = 178σ = 7x1 = 176.25x2 = 188.85
P =
8.Normal distributionHard
μ = 32x = ?25%

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 25% of them lie (the 25th percentile).

μ = 32σ = 6P = 25%
x = min
9.Confidence intervalsMedium
n = 21, s = 10LU = ?x̄ = 89.5E95%

A sample of 21 items has mean 89.5 and standard deviation s = 10. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.

n = 21x̄ = 89.5s = 1095%t(20) = 2.086
upper limit =
10.Confidence intervalsHard
n = 30Lσ = ?s = 0.595%

The diameters of 30 machined shafts have a sample standard deviation of 0.5 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 29, 2.5% of the χ² distribution lies below 16.047 and 2.5% lies above 45.722.

n = 30s = 0.5 µmχ² = 16.047, 45.722
σ = µm
11.Hypothesis tests & confidence intervalsMedium
−2.3652.365t = ?df = 7

8 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is −12.87 and the standard deviation of the differences is 9.1. Find the paired test statistic t for H₀: μd = 0.

n = 8d̄ = −12.87sd = 9.1
t =
12.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 16 measurements has mean 121.1. The population standard deviation is 5. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 121.1σ = 5n = 1695%
L =
13.Linear regression & correlationMedium
xy18802193241092713630139xy20253080100120140

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 = 120Σy = 557Σx² = 2970Σy² = 64747Σxy = 13851
r =
14.Linear regression & correlationHard
xy353461564678782xy46607080ŷ = 56

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 56?

n = 5ŷ = 56
x =
15.Chi-square testsMedium
busbikecarTotalstudents12194374staff17443091Total296373165

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

row total = 91column total = 63N = 165
E =
16.Chi-square testsHard

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). How many degrees of freedom df does the test have?

k = 6
df =

3Solutions

1.Research methods in psychology

Answer: C · Using a researcher who does not know the hypothesis or conditions to collect data

Why this is the answer

Experimenters who know the hypothesis may unintentionally influence participants or recordings. Single- or double-blind procedures and standardised instructions reduce this.

Answer

C · Using a researcher who does not know the hypothesis or conditions to collect data
2.Research methods in psychology

Answer: D · Order effects such as practice, because conditions were not counterbalanced

Why this is the answer

When all participants experience conditions in the same order, improvement may reflect practice rather than the list. Counterbalancing (ABBA or half AB, half BA) controls this.

Answer

D · Order effects such as practice, because conditions were not counterbalanced
3.Descriptive statistics

Answer: s = 19.18

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 = 211/7 = 30.14

Squared deviations

Σ(x − x̄)² = 21.86² + 0.8571² + (−19.14)² + (−11.14)² + 29.86² + (−4.143)² + (−18.14)² = 2207

Standard deviation

s = √(2207/6) = 19.18

Answer

s = 19.18
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) = 0.02275
P = 0.02275

Answer

P = 0.02275
5.Probability

Answer: Var(X) = 0.6475

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 = 0·0.35 + 1·0.35 + 2·0.3 = 0.95

Mean of the squares

E(X²) = 0·0.35 + 1·0.35 + 4·0.3 = 1.55

Variance

Var(X) = 1.55 − 0.95² = 0.6475

Answer

Var(X) = 0.6475
6.Probability

Answer: P = 0.9185

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) = 0.95 = 0.5905
P(X = 1) = 5·0.1·0.94 = 0.3281

Add them

P = 0.5905 + 0.3281 = 0.9185

Answer

P = 0.9185
7.Normal distribution

Answer: P = 0.5381

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-scores

z₁ = (176.25 − 178)/7 = −0.25
z₂ = (188.85 − 178)/7 = 1.55

Area between

P = Φ(1.55) − Φ(−0.25) = 0.9394 − 0.4013 = 0.5381

Answer

P = 0.5381
8.Normal distribution

Answer: x = 27.95 min

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) = 0.25 → z = −0.6745

Back to x

x = 32 + −0.6745·6 = 27.95 min

Answer

x = 27.95 min
9.Confidence intervals

Answer: upper limit = 94.05

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 = 20, t = 2.086

Margin of error

E = 2.086·10/√21 = 4.552

Interval

89.5 ± 4.552 → [84.95, 94.05]

Answer

upper limit = 94.05
10.Confidence intervals

Answer: σ = 0.6722 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 = 29·0.5² = 7.25

Limits for σ²

7.25/45.722 = 0.1586
7.25/16.047 = 0.4518

Limit for σ

σ = √0.4518 = 0.6722 µm

Answer

σ = 0.6722 µm

95% interval for σ: [0.3982, 0.6722] µm.

11.Hypothesis tests & confidence intervals

Answer: t = −4

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 of the mean difference

SE = 9.1/√8 = 3.217

Test statistic

t = −12.87/3.217 = −4

Answer

t = −4

With df = 7, the two-sided 5% critical value is 2.365; |t| is larger, so H₀ is rejected.

12.Hypothesis tests & confidence intervals

Answer: L = 118.6

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 = 1.96·5/√16 = 2.45

Interval limit

L = 121.1 − 2.45 = 118.6

Answer

L = 118.6
13.Linear regression & correlation

Answer: r = 0.9803

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 = 13851 − 120·557/5 = 483
Sxx = 2970 − 120²/5 = 90
Syy = 64747 − 557²/5 = 2697

Correlation coefficient

r = 483/√(90·2697) = 0.9803

Answer

r = 0.9803
14.Linear regression & correlation

Answer: x = 3.453

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 = 338/5 = 67.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 75

Regression line

b = 75/10 = 7.5
a = 67.6 − 7.5·5 = 30.1
ŷ = 30.1 + 7.5·x

Solve for x

x = (56 − 30.1)/7.5 = 3.453

Answer

x = 3.453
15.Chi-square tests

Answer: E = 34.75

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 = 91·63/165 = 34.75

Answer

E = 34.75
16.Chi-square tests

Answer: df = 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.

Degrees of freedom

df = 6 − 1 = 5

Answer

df = 5

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