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 95% confidence interval for a mean of 50 is 46 to 54. What is the most accurate interpretation?

  1. A95% of the individual participants in the sample scored between 46 and 54 on the measure
  2. BThe procedure used produces intervals that contain the true mean 95% of the time
  3. CThere is a 95% chance the sample mean is 50
  4. DThe result is not significant
2.Research methods in psychologyHard

Interpretative phenomenological analysis (IPA) mainly aims to:

  1. ATest causal hypotheses with large samples
  2. BMeasure reaction times accurately
  3. CExplore how individuals make sense of their own lived experiences
  4. DCount the frequency of words in texts
3.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 56, 45, 12, 26, 21.

n = 5
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 85?

μ = 70σ = 15x = 85
P =
5.ProbabilityMedium
P(at least one six) = ??1?2?3P(six) = 1/6 on each roll

A fair die is rolled 3 times. What is the probability P of getting at least one six?

n = 3
P =
6.ProbabilityHard
€p(x)−20.7480.25980.01

A game costs €2 to play. You win €10 with probability 0.25 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€10: p = 0.25€100: p = 0.01
E(G) = €
7.Normal distributionMedium
6580.5P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 65 and 80.5 points? Give P as a decimal.

μ = 65σ = 10x1 = 65x2 = 80.5
P =
8.Normal distributionHard
0.05 mm0D − d?

Hole diameters D are normal with mean 25.05 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.05 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
9.Confidence intervalsMedium
n = 9, sd​ = 7.10L = ?Ud̄ = −2.8E95%

The blood pressure of 9 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −2.8 and standard deviation 7.1. Find the lower limit of the 95% confidence interval for the mean difference.

n = 9d̄ = −2.8sd = 7.195%t(8) = 2.306
lower limit = mmHg
10.Confidence intervalsHard
σ = 9, n = ?x̄ − Ex̄ + Ex̄E = 295%

The population standard deviation is 9. 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 2?

σ = 9E = 295%
n =
11.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 60 parts has mean length 97.5 mm, and the line's known standard deviation is 18 mm. Line 2: a sample of 40 parts has mean 92 mm, known standard deviation 14 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 60x̄₁ = 97.5 mmσ₁ = 18 mmn₂ = 40x̄₂ = 92 mmσ₂ = 14 mm
z =
12.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 16 measurements has mean 70.5. The population standard deviation is 15. Find the lower limit L of the 90% confidence interval for the mean.

x̄ = 70.5σ = 15n = 1690%
L =
13.Linear regression & correlationMedium
xy235330435539637xy246303540

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

n = 5Σx = 20Σy = 176Σx² = 90Σy² = 6240Σxy = 717
r =
14.Linear regression & correlationHard
xy121221317416513610xy246101520x = 3.5

The table gives 6 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 = 3.5.

n = 6x = 3.5
ŷ =
15.Chi-square testsMedium
busbikecarTotalstudents9224475staff3220860Total414252135

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

row total = 60column total = 52N = 135
E =
16.Chi-square testsHard
dfα = 0.05α = 0.0113.8416.63525.9919.21037.81511.34549.48813.277511.07015.086612.59216.812714.06718.475815.50720.090916.91921.6661018.30723.209

A chi-square goodness-of-fit test checks whether counts in 6 categories follow a Poisson distribution whose mean is estimated from the same data. Using the table, find the critical value χ²crit at significance level α = 0.05.

α = 0.05
χ²crit =

3Solutions

1.Research methods in psychology

Answer: B · The procedure used produces intervals that contain the true mean 95% of the time

Why this is the answer

A 95% CI gives a range of plausible values for the population parameter. Narrower intervals, from larger samples, mean more precise estimates.

Answer

B · The procedure used produces intervals that contain the true mean 95% of the time
2.Research methods in psychology

Answer: C · Explore how individuals make sense of their own lived experiences

Why this is the answer

IPA studies small, homogeneous samples in depth, examining how participants interpret significant experiences, with the researcher interpreting their sense-making.

Answer

C · Explore how individuals make sense of their own lived experiences
3.Descriptive statistics

Answer: s = 18.04

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 = 160/5 = 32

Squared deviations

Σ(x − x̄)² = 24² + 13² + (−20)² + (−6)² + (−11)² = 1302

Standard deviation

s = √(1302/4) = 18.04

Answer

s = 18.04
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) = 0.8413
P = 1 − 0.8413 = 0.1587

Answer

P = 0.1587
5.Probability

Answer: P = 0.4213

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)3 = 0.5787
P = 1 − 0.5787 = 0.4213

Answer

P = 0.4213
6.Probability

Answer: E(G) = 1.5 €

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: 8 (p = 0.25), 98 (p = 0.01), −2 (p = 0.74)

Expected value

E(G) = 8·0.25 + 98·0.01 − 2·0.74 = 1.5 €

Answer

E(G) = 1.5 €

A positive value means you win money on average.

7.Normal distribution

Answer: P = 0.4394

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 = (65 − 65)/10 = 0
z₂ = (80.5 − 65)/10 = 1.55

Area between

P = Φ(1.55) − Φ(0) = 0.9394 − 0.5 = 0.4394

Answer

P = 0.4394
8.Normal distribution

Answer: P = 0.9875

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 = 25.05 − 25 = 0.05 mm, σ = 0.02236 mm

Standardise

z = (0 − 0.05)/0.02236 = −2.24

Probability

P = 1 − Φ(−2.24) = 0.9875

Answer

P = 0.9875
9.Confidence intervals

Answer: lower limit = −8.258 mmHg

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.

Critical value

df = 8, t = 2.306

Margin of error

E = 2.306·7.1/√9 = 5.458

Interval

−2.8 ± 5.458 → [−8.258, 2.658]

Answer

lower limit = −8.258 mmHg
10.Confidence intervals

Answer: n = 78

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 = (1.96·9/2)² = 77.79

Round up

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

n = 78

Answer

n = 78
11.Hypothesis tests & confidence intervals

Answer: z = 1.714

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 = √(18²/60 + 14²/40) = √10.3 = 3.209

Test statistic

z = (97.5 − 92)/3.209 = 1.714

Answer

z = 1.714

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: L = 64.33

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 = 1.645·15/√16 = 6.169

Interval limit

L = 70.5 − 6.169 = 64.33

Answer

L = 64.33
13.Linear regression & correlation

Answer: r = 0.6142

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 = 717 − 20·176/5 = 13
Sxx = 90 − 20²/5 = 10
Syy = 6240 − 176²/5 = 44.8

Correlation coefficient

r = 13/√(10·44.8) = 0.6142

Answer

r = 0.6142
14.Linear regression & correlation

Answer: ŷ = 16.33

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 = 21/6 = 3.5
ȳ = 98/6 = 16.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −40

Regression line

b = −40/17.5 = −2.286
a = 16.33 − (−2.286)·3.5 = 24.33
ŷ = 24.33 − 2.286·x

Prediction

ŷ = 24.33 + (−2.286)·3.5 = 16.33

Answer

ŷ = 16.33
15.Chi-square tests

Answer: E = 23.11

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 = 60·52/135 = 23.11

Answer

E = 23.11
16.Chi-square tests

Answer: χ²crit = 9.488

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 − 1 = 4

Table

χ²crit(df = 4, α = 0.05) = 9.488

Answer

χ²crit = 9.488

Reject H₀ if the computed χ² is larger than this value.

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