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

Internal validity refers to:

  1. AWhether a test gives the same results over time
  2. BWhether the observed effect was really caused by the IV rather than other factors
  3. CWhether participants found the study interesting
  4. DWhether the findings generalise beyond the study to other populations, settings and times
2.Research methods in psychologyHard

A meta-analysis combines 40 studies of a therapy. Why might its overall effect size still be too large?

  1. ACombining studies always removes all bias
  2. BMeta-analyses always give smaller effects than the single studies they include
  3. CPublication bias means unpublished null studies may be missing
  4. DMeta-analyses cannot include randomised trials
3.Descriptive statisticsMedium
2530354045

Find the sample standard deviation s of the data: 41, 30, 26, 49, 23, 30.

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

A quantity is normally distributed with mean 50 and standard deviation 20. What proportion P of values is above 75?

μ = 50σ = 20x = 75
P =
5.ProbabilityMedium
red: 7blue: 7

A bag holds 7 red and 7 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

7 / 7
P =
6.ProbabilityHard
xp(x)10.2520.3530.2540.15μ

X has the distribution shown, and Y = 5X − 3. Find the standard deviation σY of Y.

x: 1, 2, 3, 4p(x): 0.25, 0.35, 0.25, 0.15Y = 5X − 3
σY =
7.Normal distributionMedium
250 kg247.6 kgT = X₁ + … + X₁₀?

The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 10 of them exceeds 247.6 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 10> 247.6 kg
P =
8.Normal distributionHard
μ = 32x = ?75%

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 75% of them lie (the 75th percentile).

μ = 32σ = 6P = 75%
x = min
9.Confidence intervalsMedium
n = 12, s = 6LUx̄ = 34E = ?99%

A sample of 12 items has mean 34 and standard deviation s = 6. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 12x̄ = 34s = 699%t(11) = 3.106
E =
10.Confidence intervalsHard
σ = 38, n = ?x̄ − Ex̄ + Ex̄E = 2.590%

The population standard deviation is 38. 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 2.5?

σ = 38E = 2.590%
n =
11.Hypothesis tests & confidence intervalsMedium
−2.0932.093t = ?df = 19

Two teaching methods are compared. Group 1: 8 students, mean score 60, standard deviation 10. Group 2: 13 students, mean score 46.2, standard deviation 8. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 8x̄₁ = 60s₁ = 10n₂ = 13x̄₂ = 46.2s₂ = 8
t =
12.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 100 against H₁: μ < 100 is done at α = 0.01 with a sample of 25; the standard deviation is known to be 7.5. In reality the mean is 96.25. Find the probability of a Type II error β in percent.

μ₀ = 100μ = 96.25σ = 7.5n = 25α = 0.01
β = %
13.Linear regression & correlationMedium
xy347453556657763871xy468506070

The table gives 6 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 = 6Σx = 33Σy = 347Σx² = 199Σy² = 20413Σxy = 1984
r =
14.Linear regression & correlationHard
xy25466981310181221xy5105101520x = 7

The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 7.

n = 6x = 7
ŷ =
15.Chi-square testsMedium
busbikecarTotalstudents374142120staff183646100Total557788220

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 × 3
χ² =
16.Chi-square testsHard

A chi-square test of independence is carried out on a contingency table with 3 rows and 5 columns (not counting the totals). How many degrees of freedom df does the test have?

3 × 5
df =

3Solutions

1.Research methods in psychology

Answer: B · Whether the observed effect was really caused by the IV rather than other factors

Why this is the answer

High internal validity means alternative explanations, such as confounds, demand characteristics or experimenter bias, have been ruled out. External validity concerns generalisation.

Answer

B · Whether the observed effect was really caused by the IV rather than other factors
2.Research methods in psychology

Answer: C · Publication bias means unpublished null studies may be missing

Why this is the answer

If null results are less often published, the pooled effect is inflated. Funnel plots and searches for unpublished data help detect and reduce this.

Answer

C · Publication bias means unpublished null studies may be missing
3.Descriptive statistics

Answer: s = 9.867

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 = 199/6 = 33.17

Squared deviations

Σ(x − x̄)² = 7.833² + (−3.167)² + (−7.167)² + 15.83² + (−10.17)² + (−3.167)² = 486.8

Standard deviation

s = √(486.8/5) = 9.867

Answer

s = 9.867
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: P = 0.2308

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.

First and second draw

P = 7/14 · 6/13 = 42/182 = 0.2308

Answer

P = 0.2308
6.Probability

Answer: σY = 5.025

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.

Spread of X

μ = 2.3, Var(X) = 1.01, σX = 1.005

Linear change

σY = |5|·1.005 = 5.025

Answer

σY = 5.025

The mean does move: E(Y) = 5·2.3 − 3 = 8.5.

7.Normal distribution

Answer: P = 0.9357

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 total

μT = 250 kg, σT = 0.5·√10 = 1.581 kg

Standardise

z = (247.6 − 250)/1.581 = −1.52

Probability

P = 1 − Φ(−1.52) = 0.9357

Answer

P = 0.9357
8.Normal distribution

Answer: x = 36.05 min

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 for the percentile

Φ(z) = 0.75 → z = 0.6745

Back to x

x = 32 + 0.6745·6 = 36.05 min

Answer

x = 36.05 min
9.Confidence intervals

Answer: E = 5.38

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 = 11, t = 3.106

Margin of error

E = 3.106·6/√12 = 5.38

Answer

E = 5.38
10.Confidence intervals

Answer: n = 626

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 = (1.645·38/2.5)² = 625.2

Round up

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

n = 626

Answer

n = 626
11.Hypothesis tests & confidence intervals

Answer: t = 3.494

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.

Pooled variance

sp² = (7·10² + 12·8²)/19 = 77.26
sp = 8.79

Standard error

SE = 8.79·√(1/8 + 1/13) = 3.95

Test statistic

t = (60 − 46.2)/3.95 = 3.494

Answer

t = 3.494

With df = 19, the two-sided 5% critical value is 2.093; |t| is larger, so H₀ is rejected.

12.Hypothesis tests & confidence intervals

Answer: β = 43.09 %

Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).

Critical sample mean

SE = 7.5/√25 = 1.5
x̄c = 100 − 2.326·1.5 = 96.51

z under the true mean

z = (96.51 − 96.25)/1.5 = 0.174

Type II error

power = P(Z < 0.174) = 56.91%
β = 100% − 56.91% = 43.09%

Answer

β = 43.09%
13.Linear regression & correlation

Answer: r = 0.9719

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 = 1984 − 33·347/6 = 75.5
Sxx = 199 − 33²/6 = 17.5
Syy = 20413 − 347²/6 = 344.8

Correlation coefficient

r = 75.5/√(17.5·344.8) = 0.9719

Answer

r = 0.9719
14.Linear regression & correlation

Answer: ŷ = 12

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 = 42/6 = 7
ȳ = 72/6 = 12

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 120

Regression line

b = 120/70 = 1.714
a = 12 − 1.714·7 = 0
ŷ = 0 + 1.714·x

Prediction

ŷ = 0 + 1.714·7 = 12

Answer

ŷ = 12
15.Chi-square tests

Answer: χ² = 5.296

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: 120·55/220 = 30, 120·77/220 = 42, 120·88/220 = 48
staff: 100·55/220 = 25, 100·77/220 = 35, 100·88/220 = 40

Chi-square statistic

χ² = (37 − 30)²/30 + (41 − 42)²/42 + (42 − 48)²/48 + (18 − 25)²/25 + (36 − 35)²/35 + (46 − 40)²/40
χ² = 5.296

Answer

χ² = 5.296

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

16.Chi-square tests

Answer: df = 8

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 = (3 − 1)·(5 − 1) = 8

Answer

df = 8

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