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 researcher compares children's aggression before and after television arrives in a remote town. The researcher did not control the arrival of television. This is a:

  1. ACorrelational survey of television viewing
  2. BNatural experiment
  3. CField experiment
  4. DLaboratory experiment
2.Research methods in psychologyHard

A researcher wants to test whether a new teaching method causes better grades, but cannot randomise students. Which design gives the strongest causal evidence in this situation?

  1. AMeasuring only the class using the new method after teaching
  2. BAsking students whether they think the method worked
  3. CComparing matched classes measured before and after, with a comparison group
  4. DCorrelating grades with how much students like the teacher
3.Descriptive statisticsMedium
20253035404550

Find the sample standard deviation s of the data: 19, 22, 23, 48, 25, 51, 30.

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

Exam scores are normally distributed with mean 60 and standard deviation 5. Find the z-score z of a score of 50.

μ = 60σ = 5x = 50
z =
5.ProbabilityMedium
xp(x)10.0520.330.3540.3μ = 2.9

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

x: 1, 2, 3, 4p(x): 0.05, 0.3, 0.35, 0.3
Var(X) =
6.ProbabilityHard
n = 8, p = 0.5: P(X = 5) = ?k012345678

A test has 8 multiple-choice questions. A student who guesses gets each one right with probability 0.5, independently. What is the probability P of exactly 5 correct answers?

n = 8p = 0.5k = 5
P =
7.Normal distributionMedium
125 kg124.4 kgT = X₁ + … + X₅?

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 5 of them is below 124.4 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 5< 124.4 kg
P =
8.Normal distributionHard
0.03 mm0D − d?

Hole diameters D are normal with mean 25.03 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.03 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
9.Confidence intervalsMedium
x̄E₁ = 3.595%, n = 25x̄E₂ = ?99%, n = 25

A 95% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 3.5. What is the margin of error E₂ of a 99% interval from the same data?

95%n₁ = 25E₁ = 3.599%
E₂ =
10.Confidence intervalsHard
σ = 11, n = ?x̄ − Ex̄ + Ex̄E = 390%

The population standard deviation is 11. 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 = 11E = 390%
n =
11.Hypothesis tests & confidence intervalsMedium
1.62p = ?

The mean of a population is claimed to be 250; the population standard deviation is known to be 30. A random sample of 144 observations has a mean of 254.05. Test H₀: μ = 250 against H₁: μ > 250. Find the p-value p.

μ₀ = 250σ = 30n = 144x̄ = 254.05H₁: μ > μ₀
p =
12.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 100 against H₁: μ < 100 is done at α = 0.10 with a sample of 36; the standard deviation is known to be 18. In reality the mean is 96.1. Find the power 1 − β of the test in percent.

μ₀ = 100μ = 96.1σ = 18n = 36α = 0.10
1 − β = %
13.Linear regression & correlationMedium
xy2941662382810351244xy5102040

The table gives 6 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 42Σy = 155Σx² = 364Σy² = 4811Σxy = 1322
r =
14.Linear regression & correlationHard
xy125224320417516xy242025ŷ = 22

The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 22?

n = 5ŷ = 22
x =
15.Chi-square testsMedium
no defectminormajorTotalday shift18293380night shift182557100Total365490180

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 = 1802 × 3
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 compares observed counts in 9 categories with fully specified expected proportions (no parameters are estimated from the 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 · Natural experiment

Why this is the answer

In a natural experiment the IV varies because of events outside the researcher's control. Ecological validity is high, but random allocation is not possible.

Answer

B · Natural experiment
2.Research methods in psychology

Answer: C · Comparing matched classes measured before and after, with a comparison group

Why this is the answer

A non-equivalent groups pretest-posttest design controls for initial differences and for change over time, though it is weaker than a randomised experiment.

Answer

C · Comparing matched classes measured before and after, with a comparison group
3.Descriptive statistics

Answer: s = 13.01

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 = (−12.14)² + (−9.143)² + (−8.143)² + 16.86² + (−6.143)² + 19.86² + (−1.143)² = 1015

Standard deviation

s = √(1015/6) = 13.01

Answer

s = 13.01
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 = (50 − 60)/5 = −2

Answer

z = −2
5.Probability

Answer: Var(X) = 0.79

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 = 1·0.05 + 2·0.3 + 3·0.35 + 4·0.3 = 2.9

Mean of the squares

E(X²) = 1·0.05 + 4·0.3 + 9·0.35 + 16·0.3 = 9.2

Variance

Var(X) = 9.2 − 2.9² = 0.79

Answer

Var(X) = 0.79
6.Probability

Answer: P = 0.2188

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.

Binomial formula

P(X = 5) = 56·0.55·0.53 = 56·0.03125·0.125 = 0.2188

Answer

P = 0.2188
7.Normal distribution

Answer: P = 0.2946

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 = 125 kg, σT = 0.5·√5 = 1.118 kg

Standardise

z = (124.4 − 125)/1.118 = −0.54

Probability

P = Φ(−0.54) = 0.2946

Answer

P = 0.2946
8.Normal distribution

Answer: P = 0.8849

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 = 25.03 − 25 = 0.03 mm, σ = 0.025 mm

Standardise

z = (0 − 0.03)/0.025 = −1.2

Probability

P = 1 − Φ(−1.2) = 0.8849

Answer

P = 0.8849
9.Confidence intervals

Answer: E₂ = 4.6

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.

Ratio of margins

E₂/E₁ = (2.576/1.96)·√(25/25) = 1.314

New margin

E₂ = 3.5·1.314 = 4.6

Answer

E₂ = 4.6
10.Confidence intervals

Answer: n = 37

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 = (1.645·11/3)² = 36.38

Round up

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

n = 37

Answer

n = 37
11.Hypothesis tests & confidence intervals

Answer: p = 0.05262

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 = 30/√144 = 2.5
z = (254.05 − 250)/2.5 = 1.62

Tail area

Φ(1.62) = 0.9474
1 − Φ(1.62) = 0.0526

p-value

p = 0.05262

Answer

p = 0.05262

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

12.Hypothesis tests & confidence intervals

Answer: 1 − β = 50.72 %

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 = 18/√36 = 3
x̄c = 100 − 1.282·3 = 96.15

z under the true mean

z = (96.15 − 96.1)/3 = 0.018

Power

power = P(Z < 0.018) = 50.72%
1 − β = 50.72%

Answer

1 − β = 50.72%
13.Linear regression & correlation

Answer: r = 0.9973

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 = 1322 − 42·155/6 = 237
Sxx = 364 − 42²/6 = 70
Syy = 4811 − 155²/6 = 806.8

Correlation coefficient

r = 237/√(70·806.8) = 0.9973

Answer

r = 0.9973
14.Linear regression & correlation

Answer: x = 2.36

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 = 15/5 = 3
ȳ = 102/5 = 20.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −25

Regression line

b = −25/10 = −2.5
a = 20.4 − (−2.5)·3 = 27.9
ŷ = 27.9 − 2.5·x

Solve for x

x = (22 − 27.9)/(−2.5) = 2.36

Answer

x = 2.36
15.Chi-square tests

Answer: χ² = 4.53

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

day shift: 80·36/180 = 16, 80·54/180 = 24, 80·90/180 = 40
night shift: 100·36/180 = 20, 100·54/180 = 30, 100·90/180 = 50

Chi-square statistic

χ² = (18 − 16)²/16 + (29 − 24)²/24 + (33 − 40)²/40 + (18 − 20)²/20 + (25 − 30)²/30 + (57 − 50)²/50
χ² = 4.53

Answer

χ² = 4.53

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

16.Chi-square tests

Answer: χ²crit = 15.51

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 = 9 − 1 = 8

Table

χ²crit(df = 8, α = 0.05) = 15.507

Answer

χ²crit = 15.507

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

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