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

The same 25 participants complete a memory test before and after caffeine. The scores are interval and roughly normal. Which test is appropriate?

  1. APaired (related) samples t-test
  2. BMann–Whitney U test
  3. CChi-square test
  4. DIndependent samples t-test
2.Research methods in psychologyHard

In multiple regression predicting exam scores from study hours and sleep, R² = 0.36. What does this mean?

  1. AStudy hours cause 36% of exam success
  2. BThe predictors together explain 36% of the variance in exam scores
  3. C36% of students passed the exam
  4. DThe correlation between the predictors is 0.36
3.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 44, 14, 11, 54, 14.

n = 5
s =
4.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 100 and standard deviation 20. What proportion P of values is above 60?

μ = 100σ = 20x = 60
P =
5.ProbabilityMedium
xp(x)100.35200.25300.25400.15μ = 22

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

x: 10, 20, 30, 40p(x): 0.35, 0.25, 0.25, 0.15
Var(X) =
6.ProbabilityHard
€p(x)−10.7330.25490.02

A game costs €1 to play. You win €4 with probability 0.25 and €50 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €1€4: p = 0.25€50: p = 0.02
E(G) = €
7.Normal distributionMedium
32 min30 minX + Y?

X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 30 min. Round z to two decimals.

μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 30 min
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 = 30, s = 19LU = ?x̄ = 15.5E99%

A sample of 30 items has mean 15.5 and standard deviation s = 19. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.

n = 30x̄ = 15.5s = 1999%t(29) = 2.756
upper limit =
10.Confidence intervalsHard
σ = 37, n = ?x̄ − Ex̄ + Ex̄E = 395%

The population standard deviation is 37. 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 3?

σ = 37E = 395%
n =
11.Hypothesis tests & confidence intervalsMedium
1.1p = ?

The mean of a population is claimed to be 100; the population standard deviation is known to be 20. A random sample of 25 observations has a mean of 104.4. Test H₀: μ = 100 against H₁: μ > 100. Find the p-value p.

μ₀ = 100σ = 20n = 25x̄ = 104.4H₁: μ > μ₀
p =
12.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 150 of 300 patients on treatment A recover, and 90 of 150 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 150n₁ = 300x₂ = 90n₂ = 150
z =
13.Linear regression & correlationMedium
xy2314366428471056xy510304050

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 = 30Σy = 212Σx² = 220Σy² = 9366Σxy = 1394
r =
14.Linear regression & correlationHard
xy12022131741351268xy246101520x = 2.5

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. A new observation (2.5, 22) is made. Find its residual e = y − ŷ.

n = 6(2.5, 22)
e =
15.Chi-square testsMedium
improvednot improvedTotalnew drug4278120placebo4654100Total88132220

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 = 2202 × 2
χ² =
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 8 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: A · Paired (related) samples t-test

Why this is the answer

A paired t-test compares means from the same participants under two conditions, using the differences within each person.

Answer

A · Paired (related) samples t-test
2.Research methods in psychology

Answer: B · The predictors together explain 36% of the variance in exam scores

Why this is the answer

R² shows how much of the variability in the outcome is accounted for by all predictors together. It does not establish causation.

Answer

B · The predictors together explain 36% of the variance in exam scores
3.Descriptive statistics

Answer: s = 20.07

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 = 137/5 = 27.4

Squared deviations

Σ(x − x̄)² = 16.6² + (−13.4)² + (−16.4)² + 26.6² + (−13.4)² = 1611

Standard deviation

s = √(1611/4) = 20.07

Answer

s = 20.07
4.Descriptive statistics

Answer: P = 0.9772

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 = 1 − 0.02275 = 0.9772

Answer

P = 0.9772
5.Probability

Answer: Var(X) = 116

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 = 10·0.35 + 20·0.25 + 30·0.25 + 40·0.15 = 22

Mean of the squares

E(X²) = 100·0.35 + 400·0.25 + 900·0.25 + 1600·0.15 = 600

Variance

Var(X) = 600 − 22² = 116

Answer

Var(X) = 116
6.Probability

Answer: E(G) = 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: 3 (p = 0.25), 49 (p = 0.02), −1 (p = 0.73)

Expected value

E(G) = 3·0.25 + 49·0.02 − 1·0.73 = 1 €

Answer

E(G) = 1 €

A positive value means you win money on average.

7.Normal distribution

Answer: P = 0.7881

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 = 32 min, σ = √(1.5² + 2²) = 2.5 min

Standardise

z = (30 − 32)/2.5 = −0.8

Upper tail

P = 1 − Φ(−0.8) = 1 − 0.2119 = 0.7881

Answer

P = 0.7881
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: upper limit = 25.06

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 = 29, t = 2.756

Margin of error

E = 2.756·19/√30 = 9.56

Interval

15.5 ± 9.56 → [5.94, 25.06]

Answer

upper limit = 25.06
10.Confidence intervals

Answer: n = 585

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 = (1.96·37/3)² = 584.4

Round up

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

n = 585

Answer

n = 585
11.Hypothesis tests & confidence intervals

Answer: p = 0.1357

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 = (104.4 − 100)/4 = 1.1

Tail area

Φ(1.1) = 0.8643
1 − Φ(1.1) = 0.1357

p-value

p = 0.1357

Answer

p = 0.1357

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

12.Hypothesis tests & confidence intervals

Answer: z = −2.004

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.

Sample proportions

p̂₁ = 150/300 = 0.5
p̂₂ = 90/150 = 0.6
p̂ = 240/450 = 0.5333

Standard error under H₀

SE = √(0.5333·0.4667·(1/300 + 1/150)) = 0.04989

Test statistic

z = (0.5 − 0.6)/0.04989 = −2.004

Answer

z = −2.004

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

13.Linear regression & correlation

Answer: r = 0.9932

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 = 1394 − 30·212/5 = 122
Sxx = 220 − 30²/5 = 40
Syy = 9366 − 212²/5 = 377.2

Correlation coefficient

r = 122/√(40·377.2) = 0.9932

Answer

r = 0.9932
14.Linear regression & correlation

Answer: e = 4.233

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 = 91/6 = 15.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −45.5

Regression line

b = −45.5/17.5 = −2.6
a = 15.17 − (−2.6)·3.5 = 24.27
ŷ = 24.27 − 2.6·x

Predicted value

ŷ = 24.27 + (−2.6)·2.5 = 17.77

Residual

e = 22 − 17.77 = 4.233

Answer

e = 4.233

The point lies above the line.

15.Chi-square tests

Answer: χ² = 2.75

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 counts

new drug: 120·88/220 = 48, 120·132/220 = 72
placebo: 100·88/220 = 40, 100·132/220 = 60

Chi-square statistic

χ² = (42 − 48)²/48 + (78 − 72)²/72 + (46 − 40)²/40 + (54 − 60)²/60
χ² = 2.75

Answer

χ² = 2.75

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

16.Chi-square tests

Answer: χ²crit = 12.59

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 − 1 = 6

Table

χ²crit(df = 6, α = 0.05) = 12.592

Answer

χ²crit = 12.592

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

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