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 study finds that students who sleep more have higher grades. Which design change would best allow a causal conclusion?

  1. ARandomly assign students to different sleep schedules
  2. BSurvey more students
  3. CMeasure grades more accurately
  4. DAsk students to report their sleep twice and average the two reports
3.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 54, 26, 34, 25, 13.

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

Exam scores are normally distributed with mean 50 and standard deviation 10. Find the z-score z of a score of 42.

μ = 50σ = 10x = 42
z =
5.ProbabilityMedium
xp(x)00.210.320.330.0540.15μ = 1.65

For the distribution shown, find the standard deviation σ of X.

x: 0, 1, 2, 3, 4p(x): 0.2, 0.3, 0.3, 0.05, 0.15
σ =
6.ProbabilityHard
€p(x)−10.8310.1590.02

A game costs €1 to play. You win €2 with probability 0.15 and €10 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €1€2: p = 0.15€10: p = 0.02
E(G) = €
7.Normal distributionMedium
70107.5P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 70 and 107.5 points? Give P as a decimal.

μ = 100σ = 15x1 = 70x2 = 107.5
P =
8.Normal distributionHard
μ = 500x = ?99%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 99% of them lie (the 99th percentile).

μ = 500σ = 40P = 99%
x = h
9.Confidence intervalsMedium
n = 25, s = 5L = ?Ux̄ = 85E95%

A sample of 25 items has mean 85 and standard deviation s = 5. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 25x̄ = 85s = 595%t(24) = 2.064
lower limit =
10.Confidence intervalsHard
n = 15σ = ?Us = 9.995%

The diameters of 15 machined shafts have a sample standard deviation of 9.9 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 14, 2.5% of the χ² distribution lies below 5.629 and 2.5% lies above 26.119.

n = 15s = 9.9 µmχ² = 5.629, 26.119
σ = µm
11.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 30 parts has mean length 118 mm, and the line's known standard deviation is 5 mm. Line 2: a sample of 50 parts has mean 124.2 mm, known standard deviation 15 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 30x̄₁ = 118 mmσ₁ = 5 mmn₂ = 50x̄₂ = 124.2 mmσ₂ = 15 mm
z =
12.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 49 measurements has mean 69.6. The population standard deviation is 20. Find the upper limit U of the 99% confidence interval for the mean.

x̄ = 69.6σ = 20n = 4999%
U =
13.Linear regression & correlationMedium
xy233343445554662766xy2464060

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 = 27Σy = 303Σx² = 139Σy² = 16079Σxy = 1479
r =
14.Linear regression & correlationHard
xy14336257478599311110xy51050100x = 2

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 2.

n = 6x = 2
ŷ =
15.Chi-square testsMedium
no defectminormajorTotalday shift15233674night shift13333783Total285673157

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

row total = 74column total = 56N = 157
E =
16.Chi-square testsHard

A chi-square goodness-of-fit test compares observed counts in 6 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?

k = 6
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: A · Randomly assign students to different sleep schedules

Why this is the answer

Only manipulating sleep with random allocation controls third variables and allows causal inference, within ethical limits.

Answer

A · Randomly assign students to different sleep schedules
3.Descriptive statistics

Answer: s = 15.18

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 = 152/5 = 30.4

Squared deviations

Σ(x − x̄)² = 23.6² + (−4.4)² + 3.6² + (−5.4)² + (−17.4)² = 921.2

Standard deviation

s = √(921.2/4) = 15.18

Answer

s = 15.18
4.Descriptive statistics

Answer: z = −0.8

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 = (42 − 50)/10 = −0.8

Answer

z = −0.8
5.Probability

Answer: σ = 1.276

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 = 0·0.2 + 1·0.3 + 2·0.3 + 3·0.05 + 4·0.15 = 1.65

Mean of the squares

E(X²) = 0·0.2 + 1·0.3 + 4·0.3 + 9·0.05 + 16·0.15 = 4.35

Variance

Var(X) = 4.35 − 1.65² = 1.627
σ = √1.627 = 1.276

Answer

σ = 1.276
6.Probability

Answer: E(G) = −0.5 €

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: 1 (p = 0.15), 9 (p = 0.02), −1 (p = 0.83)

Expected value

E(G) = 1·0.15 + 9·0.02 − 1·0.83 = −0.5 €

Answer

E(G) = −0.5 €

A negative value means you lose money on average.

7.Normal distribution

Answer: P = 0.6687

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 = (70 − 100)/15 = −2
z₂ = (107.5 − 100)/15 = 0.5

Area between

P = Φ(0.5) − Φ(−2) = 0.6915 − 0.02275 = 0.6687

Answer

P = 0.6687
8.Normal distribution

Answer: x = 593.1 h

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) = 0.99 → z = 2.326

Back to x

x = 500 + 2.326·40 = 593.05 h

Answer

x = 593.05 h
9.Confidence intervals

Answer: lower limit = 82.94

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 = 24, t = 2.064

Margin of error

E = 2.064·5/√25 = 2.064

Interval

85 ± 2.064 → [82.94, 87.06]

Answer

lower limit = 82.94
10.Confidence intervals

Answer: σ = 7.248 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 14·9.9² = 1372

Limits for σ²

1372/26.119 = 52.53
1372/5.629 = 243.8

Limit for σ

σ = √52.53 = 7.248 µm

Answer

σ = 7.248 µm

95% interval for σ: [7.248, 15.61] µm.

11.Hypothesis tests & confidence intervals

Answer: z = −2.685

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 = √(5²/30 + 15²/50) = √5.333 = 2.309

Test statistic

z = (118 − 124.2)/2.309 = −2.685

Answer

z = −2.685

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

12.Hypothesis tests & confidence intervals

Answer: U = 76.96

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 = 2.576·20/√49 = 7.36

Interval limit

U = 69.6 + 7.36 = 76.96

Answer

U = 76.96
13.Linear regression & correlation

Answer: r = 0.9902

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 = 1479 − 27·303/6 = 115.5
Sxx = 139 − 27²/6 = 17.5
Syy = 16079 − 303²/6 = 777.5

Correlation coefficient

r = 115.5/√(17.5·777.5) = 0.9902

Answer

r = 0.9902
14.Linear regression & correlation

Answer: ŷ = 52.75

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 = 36/6 = 6
ȳ = 467/6 = 77.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 439

Regression line

b = 439/70 = 6.271
a = 77.83 − 6.271·6 = 40.2
ŷ = 40.2 + 6.271·x

Prediction

ŷ = 40.2 + 6.271·2 = 52.75

Answer

ŷ = 52.75
15.Chi-square tests

Answer: E = 26.39

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 = 74·56/157 = 26.39

Answer

E = 26.39
16.Chi-square tests

Answer: df = 5

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 = 5

Answer

df = 5

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