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

  1. ARun a chi-square test on the means
  2. BConclude that all four groups differ from each other
  3. CRun post hoc tests such as Tukey's HSD
  4. DReport the result as a correlation
3.Descriptive statisticsMedium
202530354045505560

Find the sample standard deviation s of the data: 60, 44, 31, 58, 19, 39.

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

A quantity is normally distributed with mean 50 and standard deviation 8. What proportion P of values is above 54.8?

μ = 50σ = 8x = 54.8
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
€p(x)−20.8440.15980.01

A game costs €2 to play. You win €6 with probability 0.15 and €100 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €2€6: p = 0.15€100: p = 0.01
E(G) = €
7.Normal distributionMedium
200 mm200.3 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 200.3 mm. Round z to two decimals.

μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 200.3 mm
P =
8.Normal distributionHard
0.03 mm0D − d?

Hole diameters D are normal with mean 25.03 mm and standard deviation 0.02 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 = 25.03 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
9.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 40 phones, mean 66 h, known standard deviation 7 h. Model 2: 80 phones, mean 62 h, known standard deviation 5 h. Find the lower limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 40x̄₁ = 66 hσ₁ = 7 hn₂ = 80x̄₂ = 62 hσ₂ = 5 h90%
lower limit = h
10.Confidence intervalsHard
σ = 27, n = ?x̄ − Ex̄ + Ex̄E = 1.595%

The population standard deviation is 27. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 1.5?

σ = 27E = 1.595%
n =
11.Hypothesis tests & confidence intervalsMedium
−0.920.92p = ?

The mean of a population is claimed to be 50; the population standard deviation is known to be 20. A random sample of 25 observations has a mean of 53.68. Test H₀: μ = 50 against H₁: μ ≠ 50. Find the p-value p.

μ₀ = 50σ = 20n = 25x̄ = 53.68H₁: μ ≠ μ₀
p =
12.Hypothesis tests & confidence intervalsHard
σ = 5, n = ?x̄ − Ex̄ + Ex̄E = 299%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 5. Find the smallest sample size n.

σ = 5E = 299%
n =
13.Linear regression & correlationMedium
xy363466566678779xy4665707580

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 = 25Σy = 352Σx² = 135Σy² = 25006Σxy = 1804
r =
14.Linear regression & correlationHard
xy14551667187520922211124121xy15202550100x = 21

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (21, 104) is made. Find its residual e = y − ŷ.

n = 6(21, 104)
e =
15.Chi-square testsMedium
no defectminormajorTotalday shift4132881night shift33111458Total744322139

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

row total = 81column total = 43N = 139
E =
16.Chi-square testsHard

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

4 × 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: C · 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

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

Answer: s = 15.77

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 = 251/6 = 41.83

Squared deviations

Σ(x − x̄)² = 18.17² + 2.167² + (−10.83)² + 16.17² + (−22.83)² + (−2.833)² = 1243

Standard deviation

s = √(1243/5) = 15.77

Answer

s = 15.77
4.Descriptive statistics

Answer: P = 0.2743

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 = 0.6

Standard normal table

Φ(0.6) = 0.7257
P = 1 − 0.7257 = 0.2743

Answer

P = 0.2743
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: E(G) = −0.1 €

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 = 0.15), 98 (p = 0.01), −2 (p = 0.84)

Expected value

E(G) = 4·0.15 + 98·0.01 − 2·0.84 = −0.1 €

Answer

E(G) = −0.1 €

A negative value means you lose money on average.

7.Normal distribution

Answer: P = 0.2743

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 = (200.3 − 200)/0.5 = 0.6

Upper tail

P = 1 − Φ(0.6) = 1 − 0.7257 = 0.2743

Answer

P = 0.2743
8.Normal distribution

Answer: P = 0.8554

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 = 25.03 − 25 = 0.03 mm, σ = 0.02828 mm

Standardise

z = (0 − 0.03)/0.02828 = −1.06

Probability

P = 1 − Φ(−1.06) = 0.8554

Answer

P = 0.8554
9.Confidence intervals

Answer: lower limit = 1.96 h

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.

Point estimate

x̄₁ − x̄₂ = 66 − 62 = 4

Standard error

SE = √(7²/40 + 5²/80) = 1.24

Margin of error

E = 1.645·1.24 = 2.04

Interval

4 ± 2.04 → [1.96, 6.04]

Answer

lower limit = 1.96 h

The interval does not contain 0, so the means differ significantly.

10.Confidence intervals

Answer: n = 1245

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 = (1.96·27/1.5)² = 1245

Round up

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

n = 1245

Answer

n = 1245
11.Hypothesis tests & confidence intervals

Answer: p = 0.3576

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.

Test statistic

SE = σ/√n = 20/√25 = 4
z = (53.68 − 50)/4 = 0.92

Tail area

Φ(0.92) = 0.8212
1 − Φ(0.92) = 0.1788

p-value

p = 2·0.1788 = 0.3576

Answer

p = 0.3576

p > 0.05, so H₀ is not rejected at the 5% level.

12.Hypothesis tests & confidence intervals

Answer: n = 42

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 = (2.576·5/2)² = 41.47

Round up

n = 42

Answer

n = 42
13.Linear regression & correlation

Answer: r = 0.9272

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 = 1804 − 25·352/5 = 44
Sxx = 135 − 25²/5 = 10
Syy = 25006 − 352²/5 = 225.2

Correlation coefficient

r = 44/√(10·225.2) = 0.9272

Answer

r = 0.9272
14.Linear regression & correlation

Answer: e = 3.481

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 = 114/6 = 19
ȳ = 521/6 = 86.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 479

Regression line

b = 479/70 = 6.843
a = 86.83 − 6.843·19 = −43.18
ŷ = −43.18 + 6.843·x

Predicted value

ŷ = −43.18 + 6.843·21 = 100.5

Residual

e = 104 − 100.5 = 3.481

Answer

e = 3.481

The point lies above the line.

15.Chi-square tests

Answer: E = 25.06

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 = 81·43/139 = 25.06

Answer

E = 25.06
16.Chi-square tests

Answer: df = 9

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 = (4 − 1)·(4 − 1) = 9

Answer

df = 9

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