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

Operationalisation means:

  1. ACarrying out the research in a real operational setting, such as a hospital
  2. BDefining variables in a precise, measurable way
  3. CChoosing a statistical test
  4. DGiving participants instructions
2.Research methods in psychologyHard

Effect size measures such as Cohen's d are reported because they:

  1. ATell how many participants dropped out
  2. BReplace the need for any hypothesis
  3. CShow how large an effect is, regardless of sample size
  4. DShow whether the result is statistically significant at the chosen alpha level
3.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 31, 37, 15, 26, 11, 57, 28.

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

Exam scores are normally distributed with mean 60 and standard deviation 20. Find the z-score z of a score of 100.

μ = 60σ = 20x = 100
z =
5.ProbabilityMedium
xp(x)10.3520.1530.140.350.1μ = 2.65

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

x: 1, 2, 3, 4, 5p(x): 0.35, 0.15, 0.1, 0.3, 0.1
Var(X) =
6.ProbabilityHard
n = 9, p = 0.3: P(X ≤ 1) = ?k0123456789

9 components are tested, and each one fails with probability 0.3, independently. What is the probability P that at most one fails?

n = 9p = 0.3
P =
7.Normal distributionMedium
96.25112.75P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 96.25 and 112.75 points? Give P as a decimal.

μ = 100σ = 15x1 = 96.25x2 = 112.75
P =
8.Normal distributionHard
35455565758595pointstop 15%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 15% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points15%
x = points
9.Confidence intervalsMedium
n = 6, sd​ = 6.20L = ?Ud̄ = −5.6E99%

The blood pressure of 6 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −5.6 and standard deviation 6.2. Find the lower limit of the 99% confidence interval for the mean difference.

n = 6d̄ = −5.6sd = 6.299%t(5) = 4.032
lower limit = mmHg
10.Confidence intervalsHard
p ≈ 0.6, n = ?p̂ − Ep̂ + Ep̂E = 0.02599%

A pilot study suggests that about 0.6 of products are defective. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.025?

p ≈ 0.6E = 0.02599%
n =
11.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 100 parts has mean length 131 mm, and the line's known standard deviation is 16 mm. Line 2: a sample of 100 parts has mean 128.2 mm, known standard deviation 7 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 100x̄₁ = 131 mmσ₁ = 16 mmn₂ = 100x̄₂ = 128.2 mmσ₂ = 7 mm
z =
12.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 80 against H₁: μ < 80 is done at α = 0.10 with a sample of 36; the standard deviation is known to be 12. In reality the mean is 74. Find the power 1 − β of the test in percent.

μ₀ = 80μ = 74σ = 12n = 36α = 0.10
1 − β = %
13.Linear regression & correlationMedium
xy24486138181019xy5105101520

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.

n = 5
r =
14.Linear regression & correlationHard
xy231339443548660763xy24630405060x = 5.5

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 5.5.

n = 6x = 5.5
ŷ =
15.Chi-square testsMedium
no defectminormajorTotalday shift18131849night shift9342871Total274746120

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

row total = 71column total = 46N = 120
E =
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 · Defining variables in a precise, measurable way

Why this is the answer

Operationalising variables turns abstract concepts into specific measurements so that research can be replicated.

Answer

B · Defining variables in a precise, measurable way
2.Research methods in psychology

Answer: C · Show how large an effect is, regardless of sample size

Why this is the answer

With large samples tiny effects can be significant. Effect sizes express the magnitude of a difference, e.g. d = 0.2 small, 0.5 medium, 0.8 large.

Answer

C · Show how large an effect is, regardless of sample size
3.Descriptive statistics

Answer: s = 15.17

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 = 205/7 = 29.29

Squared deviations

Σ(x − x̄)² = 1.714² + 7.714² + (−14.29)² + (−3.286)² + (−18.29)² + 27.71² + (−1.286)² = 1381

Standard deviation

s = √(1381/6) = 15.17

Answer

s = 15.17
4.Descriptive statistics

Answer: z = 2

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 = (100 − 60)/20 = 2

Answer

z = 2
5.Probability

Answer: Var(X) = 2.128

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 = 1·0.35 + 2·0.15 + 3·0.1 + 4·0.3 + 5·0.1 = 2.65

Mean of the squares

E(X²) = 1·0.35 + 4·0.15 + 9·0.1 + 16·0.3 + 25·0.1 = 9.15

Variance

Var(X) = 9.15 − 2.65² = 2.128

Answer

Var(X) = 2.128
6.Probability

Answer: P = 0.196

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.

Zero and one failure

P(X = 0) = 0.79 = 0.04035
P(X = 1) = 9·0.3·0.78 = 0.1556

Add them

P = 0.04035 + 0.1556 = 0.196

Answer

P = 0.196
7.Normal distribution

Answer: P = 0.401

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 = (96.25 − 100)/15 = −0.25
z₂ = (112.75 − 100)/15 = 0.85

Area between

P = Φ(0.85) − Φ(−0.25) = 0.8023 − 0.4013 = 0.401

Answer

P = 0.401
8.Normal distribution

Answer: x = 75.4 points

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) = 0.85

z from the table

z ≈ 1.04

Back to the original units

x = 65 + 1.04·10 = 75.4 points

Answer

x = 75.4 points
9.Confidence intervals

Answer: lower limit = −15.81 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 = 5, t = 4.032

Margin of error

E = 4.032·6.2/√6 = 10.21

Interval

−5.6 ± 10.21 → [−15.81, 4.606]

Answer

lower limit = −15.81 mmHg
10.Confidence intervals

Answer: n = 2549

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 2.576²·0.6·0.4/0.025² = 2548

Round up

n = 2549

Answer

n = 2549
11.Hypothesis tests & confidence intervals

Answer: z = 1.603

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 = √(16²/100 + 7²/100) = √3.05 = 1.746

Test statistic

z = (131 − 128.2)/1.746 = 1.603

Answer

z = 1.603

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

12.Hypothesis tests & confidence intervals

Answer: 1 − β = 95.71 %

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 = 12/√36 = 2
x̄c = 80 − 1.282·2 = 77.44

z under the true mean

z = (77.44 − 74)/2 = 1.718

Power

power = P(Z < 1.718) = 95.71%
1 − β = 95.71%

Answer

1 − β = 95.71%
13.Linear regression & correlation

Answer: r = 0.9841

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 = 30/5 = 6
ȳ = 62/5 = 12.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 80
Syy = Σ(y − ȳ)² = 165.2

Correlation coefficient

r = 80/√(40·165.2) = 0.9841

Answer

r = 0.9841
14.Linear regression & correlation

Answer: ŷ = 53.85

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 = 27/6 = 4.5
ȳ = 284/6 = 47.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 114

Regression line

b = 114/17.5 = 6.514
a = 47.33 − 6.514·4.5 = 18.02
ŷ = 18.02 + 6.514·x

Prediction

ŷ = 18.02 + 6.514·5.5 = 53.85

Answer

ŷ = 53.85
15.Chi-square tests

Answer: E = 27.22

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 = 71·46/120 = 27.22

Answer

E = 27.22
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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