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

A p-value of 0.03 means:

  1. AThe result will replicate 97% of the time
  2. BIf the null hypothesis were true, data this extreme would occur about 3% of the time
  3. CGiven the observed data, there is only a 3% probability that the null hypothesis is true
  4. DThe effect is large
2.Research methods in psychologyHard

In qualitative research, reflexivity means that the researcher:

  1. AReflects on how their own background and assumptions may shape the analysis
  2. BConverts themes into numerical data
  3. CUses a random sample to remove bias
  4. DRepeats the study to check reliability
3.Descriptive statisticsMedium
101520253035404550

Find the sample standard deviation s of the data: 38, 50, 52, 10, 42, 14, 12.

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

A quantity is normally distributed with mean 70 and standard deviation 15. What proportion P of values is above 88.75?

μ = 70σ = 15x = 88.75
P =
5.ProbabilityMedium
n = 7k = 4N = ?

In how many ways N can a committee of 4 people be chosen from a group of 7? The order does not matter.

n = 7k = 4
N =
6.ProbabilityHard
€p(x)−100.88300.11900.02

A game costs €10 to play. You win €40 with probability 0.1 and €200 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €10€40: p = 0.1€200: p = 0.02
E(G) = €
7.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.9586

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9586. Find z to two decimals.

A = 0.9586
z =
8.Normal distributionHard
380420460500540580620hbottom 75%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 75% is to the left of it. Find the cut-off x.

μ = 500 hσ = 40 h75%
x = h
9.Confidence intervalsMedium
n = 20, s = 13.5LUx̄ = 10.5E = ?95%

A sample of 20 items has mean 10.5 and standard deviation s = 13.5. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.

n = 20x̄ = 10.5s = 13.595%t(19) = 2.093
E =
10.Confidence intervalsHard
σ = 26, n = ?x̄ − Ex̄ + Ex̄E = 390%

The population standard deviation is 26. 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 3?

σ = 26E = 390%
n =
11.Hypothesis tests & confidence intervalsMedium
1.7p = ?

The mean of a population is claimed to be 100; the population standard deviation is known to be 15. A random sample of 100 observations has a mean of 102.55. Test H₀: μ = 100 against H₁: μ > 100. Find the p-value p.

μ₀ = 100σ = 15n = 100x̄ = 102.55H₁: μ > μ₀
p =
12.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 25 measurements has mean 197.6. The population standard deviation is 10. Find the lower limit L of the 90% confidence interval for the mean.

x̄ = 197.6σ = 10n = 2590%
L =
13.Linear regression & correlationMedium
xy368468585691795xy46708090

The table gives 5 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
14.Linear regression & correlationHard
xy35255275897011791382xy51050607080x = 10

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 10.

n = 6x = 10
ŷ =
15.Chi-square testsMedium
busbikecarTotalstudents8201240staff12252360Total204535100

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 × 3
χ² =
16.Chi-square testsHard

A chi-square goodness-of-fit test compares observed counts in 8 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?

k = 8
df =

3Solutions

1.Research methods in psychology

Answer: B · If the null hypothesis were true, data this extreme would occur about 3% of the time

Why this is the answer

The p-value is the probability of results at least as extreme as those observed if the null were true. It does not give the probability that a hypothesis is true or the size of the effect.

Answer

B · If the null hypothesis were true, data this extreme would occur about 3% of the time
2.Research methods in psychology

Answer: A · Reflects on how their own background and assumptions may shape the analysis

Why this is the answer

Reflexivity acknowledges the researcher's influence on data collection and interpretation, and is a key quality criterion in qualitative work.

Answer

A · Reflects on how their own background and assumptions may shape the analysis
3.Descriptive statistics

Answer: s = 18.54

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 = 218/7 = 31.14

Squared deviations

Σ(x − x̄)² = 6.857² + 18.86² + 20.86² + (−21.14)² + 10.86² + (−17.14)² + (−19.14)² = 2063

Standard deviation

s = √(2063/6) = 18.54

Answer

s = 18.54
4.Descriptive statistics

Answer: P = 0.1056

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 = 1.25

Standard normal table

Φ(1.25) = 0.8944
P = 1 − 0.8944 = 0.1056

Answer

P = 0.1056
5.Probability

Answer: N = 35

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.

Combinations

C(7, 4) = (7·6·5·4)/(4·3·2·1) = 35

Answer

N = 35
6.Probability

Answer: E(G) = −2 €

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: 30 (p = 0.1), 190 (p = 0.02), −10 (p = 0.88)

Expected value

E(G) = 30·0.1 + 190·0.02 − 10·0.88 = −2 €

Answer

E(G) = −2 €

A negative value means you lose money on average.

7.Normal distribution

Answer: z = 2.04

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 + 0.9586)/2 = 0.9793

Read the z-table backwards

Find 0.9793 inside the table and read off z.

z = 2.04

Answer

z = 2.04
8.Normal distribution

Answer: x = 527 h

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

Φ(z) = 0.75

z from the table

z ≈ 0.67

Back to the original units

x = 500 + 0.67·40 = 527 h

Answer

x = 527 h
9.Confidence intervals

Answer: E = 6.318

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 = 2.093

Margin of error

E = 2.093·13.5/√20 = 6.318

Answer

E = 6.318
10.Confidence intervals

Answer: n = 204

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 = (1.645·26/3)² = 203.3

Round up

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

n = 204

Answer

n = 204
11.Hypothesis tests & confidence intervals

Answer: p = 0.04457

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 = 15/√100 = 1.5
z = (102.55 − 100)/1.5 = 1.7

Tail area

Φ(1.7) = 0.9554
1 − Φ(1.7) = 0.0446

p-value

p = 0.04457

Answer

p = 0.04457

p ≤ 0.05, so H₀ is rejected at the 5% level.

12.Hypothesis tests & confidence intervals

Answer: L = 194.3

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/√25 = 3.29

Interval limit

L = 197.6 − 3.29 = 194.3

Answer

L = 194.3
13.Linear regression & correlation

Answer: r² = 0.9133

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 = 407/5 = 81.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 77
Syy = Σ(y − ȳ)² = 649.2

Coefficient of determination

r² = 77²/(10·649.2) = 0.9133

Answer

r² = 0.9133

About 91.33% of the variation in y is explained by the line.

14.Linear regression & correlation

Answer: ŷ = 72.44

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 = 48/6 = 8
ȳ = 393/6 = 65.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 243

Regression line

b = 243/70 = 3.471
a = 65.5 − 3.471·8 = 37.73
ŷ = 37.73 + 3.471·x

Prediction

ŷ = 37.73 + 3.471·10 = 72.44

Answer

ŷ = 72.44
15.Chi-square tests

Answer: χ² = 0.8466

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

students: 40·20/100 = 8, 40·45/100 = 18, 40·35/100 = 14
staff: 60·20/100 = 12, 60·45/100 = 27, 60·35/100 = 21

Chi-square statistic

χ² = (8 − 8)²/8 + (20 − 18)²/18 + (12 − 14)²/14 + (12 − 12)²/12 + (25 − 27)²/27 + (23 − 21)²/21
χ² = 0.8466

Answer

χ² = 0.8466

df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is not rejected.

16.Chi-square tests

Answer: df = 7

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 = 7

Answer

df = 7

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