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

Which kind of data is collected with an open question in an interview?

  1. ARatio data
  2. BInterval data
  3. CNominal quantitative data only
  4. DQualitative data
2.Research methods in psychologyHard

A study of 10 participants finds a large but non-significant effect. What is the most appropriate conclusion?

  1. AThe study may have lacked power, so the result is inconclusive
  2. BThe effect must be a Type I error
  3. CThe effect size must be miscalculated
  4. DThe study proves the null hypothesis is true, since the effect was not significant
3.Descriptive statisticsMedium
25303540455055

Find the sample standard deviation s of the data: 47, 49, 40, 30, 58, 29, 22.

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

A quantity is normally distributed with mean 170 and standard deviation 12. What proportion P of values is below 158?

μ = 170σ = 12x = 158
P =
5.ProbabilityMedium
xp(x)10.2520.3530.3540.05μ = 2.2

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

x: 1, 2, 3, 4p(x): 0.25, 0.35, 0.35, 0.05
σ =
6.ProbabilityHard
n = 6, p = 0.25: P(X = 1) = ?k0123456

A test has 6 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 1 correct answers?

n = 6p = 0.25k = 1
P =
7.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.9385

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9385. Find z to two decimals.

A = 0.9385
z =
8.Normal distributionHard
0.02 mm0D − d?

Hole diameters D are normal with mean 25.02 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.02 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
9.Confidence intervalsMedium
n = 25, s = 2.5LUx̄ = 48.5E = ?90%

A sample of 25 items has mean 48.5 and standard deviation s = 2.5. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.

n = 25x̄ = 48.5s = 2.590%t(24) = 1.711
E =
10.Confidence intervalsHard
σ = 29, n = ?x̄ − Ex̄ + Ex̄E = 490%

The population standard deviation is 29. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 4?

σ = 29E = 490%
n =
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 122 mm, and the line's known standard deviation is 12 mm. Line 2: a sample of 60 parts has mean 126.7 mm, known standard deviation 3 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 50x̄₁ = 122 mmσ₁ = 12 mmn₂ = 60x̄₂ = 126.7 mmσ₂ = 3 mm
z =
12.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 49 measurements has mean 196.7. The population standard deviation is 10. Find the upper limit U of the 90% confidence interval for the mean.

x̄ = 196.7σ = 10n = 4990%
U =
13.Linear regression & correlationMedium
xy1644196022642586288931109xy1520253050100

The table gives 6 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 = 6Σx = 141Σy = 452Σx² = 3471Σy² = 36830Σxy = 11273
r =
14.Linear regression & correlationHard
xy122223321420516xy2420x = 1.5

The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 1.5.

n = 5x = 1.5
ŷ =
15.Chi-square testsMedium
yesnoTotalmen164460women142640Total3070100

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 = 1002 × 2
χ² =
16.Chi-square testsHard

A chi-square test of independence is carried out on a contingency table with 2 rows and 4 columns (not counting the totals). How many degrees of freedom df does the test have?

2 × 4
df =

3Solutions

1.Research methods in psychology

Answer: D · Qualitative data

Why this is the answer

Open questions produce rich qualitative data that is harder to analyse statistically than closed-question responses.

Answer

D · Qualitative data
2.Research methods in psychology

Answer: A · The study may have lacked power, so the result is inconclusive

Why this is the answer

With small samples, even real effects often fail to reach significance (Type II error). Researchers should report the effect size and confidence interval and replicate with a larger sample.

Answer

A · The study may have lacked power, so the result is inconclusive
3.Descriptive statistics

Answer: s = 12.88

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 = 275/7 = 39.29

Squared deviations

Σ(x − x̄)² = 7.714² + 9.714² + 0.7143² + (−9.286)² + 18.71² + (−10.29)² + (−17.29)² = 995.4

Standard deviation

s = √(995.4/6) = 12.88

Answer

s = 12.88
4.Descriptive statistics

Answer: P = 0.1587

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

Standard normal table

Φ(−1) = 0.1587
P = 0.1587

Answer

P = 0.1587
5.Probability

Answer: σ = 0.8718

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 = 1·0.25 + 2·0.35 + 3·0.35 + 4·0.05 = 2.2

Mean of the squares

E(X²) = 1·0.25 + 4·0.35 + 9·0.35 + 16·0.05 = 5.6

Variance

Var(X) = 5.6 − 2.2² = 0.76
σ = √0.76 = 0.8718

Answer

σ = 0.8718
6.Probability

Answer: P = 0.356

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) = 6·0.251·0.755 = 6·0.25·0.2373 = 0.356

Answer

P = 0.356
7.Normal distribution

Answer: z = 1.87

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 + 0.9385)/2 = 0.9693

Read the z-table backwards

Find 0.9693 inside the table and read off z.

z = 1.87

Answer

z = 1.87
8.Normal distribution

Answer: P = 0.8133

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 = 25.02 − 25 = 0.02 mm, σ = 0.02236 mm

Standardise

z = (0 − 0.02)/0.02236 = −0.89

Probability

P = 1 − Φ(−0.89) = 0.8133

Answer

P = 0.8133
9.Confidence intervals

Answer: E = 0.8555

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 = 24, t = 1.711

Margin of error

E = 1.711·2.5/√25 = 0.8555

Answer

E = 0.8555
10.Confidence intervals

Answer: n = 143

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 = (1.645·29/4)² = 142.2

Round up

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

n = 143

Answer

n = 143
11.Hypothesis tests & confidence intervals

Answer: z = −2.7

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 = √(12²/50 + 3²/60) = √3.03 = 1.741

Test statistic

z = (122 − 126.7)/1.741 = −2.7

Answer

z = −2.7

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: U = 199

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 = 1.645·10/√49 = 2.35

Interval limit

U = 196.7 + 2.35 = 199

Answer

U = 199
13.Linear regression & correlation

Answer: r = 0.9839

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 = 11273 − 141·452/6 = 651
Sxx = 3471 − 141²/6 = 157.5
Syy = 36830 − 452²/6 = 2779

Correlation coefficient

r = 651/√(157.5·2779) = 0.9839

Answer

r = 0.9839
14.Linear regression & correlation

Answer: ŷ = 22.65

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 = 15/5 = 3
ȳ = 102/5 = 20.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −15

Regression line

b = −15/10 = −1.5
a = 20.4 − (−1.5)·3 = 24.9
ŷ = 24.9 − 1.5·x

Prediction

ŷ = 24.9 + (−1.5)·1.5 = 22.65

Answer

ŷ = 22.65
15.Chi-square tests

Answer: χ² = 0.7937

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

men: 60·30/100 = 18, 60·70/100 = 42
women: 40·30/100 = 12, 40·70/100 = 28

Chi-square statistic

χ² = (16 − 18)²/18 + (44 − 42)²/42 + (14 − 12)²/12 + (26 − 28)²/28
χ² = 0.7937

Answer

χ² = 0.7937

df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is not rejected.

16.Chi-square tests

Answer: df = 3

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 = (2 − 1)·(4 − 1) = 3

Answer

df = 3

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