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

Publication bias refers to:

  1. AResearchers publishing the same study twice
  2. BStudies with significant results being more likely to be published than null results
  3. CReviewers favouring papers by authors from famous universities over lesser-known ones
  4. DJournals publishing only qualitative research
2.Research methods in psychologyHard

In grounded theory, data collection continues until:

  1. AAt least 1,000 participants have been interviewed
  2. BTheoretical saturation is reached and new data add no new categories
  3. CA p-value below 0.05 is obtained
  4. DThe original hypothesis is confirmed
3.Descriptive statisticsMedium
20253035404550

Find the sample standard deviation s of the data: 39, 20, 34, 50, 51.

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

Exam scores are normally distributed with mean 500 and standard deviation 25. Find the z-score z of a score of 525.

μ = 500σ = 25x = 525
z =
5.ProbabilityMedium
xp(x)00.2510.320.2530.2μ = 1.4

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

x: 0, 1, 2, 3p(x): 0.25, 0.3, 0.25, 0.2
Var(X) =
6.ProbabilityHard
n = 11, p = 0.6: P(X ≤ 1) = ?k01234567891011

11 components are tested, and each one fails with probability 0.6, independently. What is the probability P that at most one fails?

n = 11p = 0.6
P =
7.Normal distributionMedium
244251.25P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 244 and 251.25 g? Give P as a decimal.

μ = 250σ = 5x1 = 244x2 = 251.25
P =
8.Normal distributionHard
0.03 mm0D − d?

Hole diameters D are normal with mean 25.03 mm and standard deviation 0.01 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 = 0.01 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
9.Confidence intervalsMedium
x̄E₁ = 3.299%, n = 36x̄E₂ = ?95%, n = 72

A 99% confidence interval for a mean, based on 36 observations and a known σ, has a margin of error of 3.2. What is the margin of error E₂ of a 95% interval based on 72 observations?

99%n₁ = 36E₁ = 3.295%n₂ = 72
E₂ =
10.Confidence intervalsHard
σ = 14, n = ?x̄ − Ex̄ + Ex̄E = 0.595%

The population standard deviation is 14. 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 0.5?

σ = 14E = 0.595%
n =
11.Hypothesis tests & confidence intervalsMedium
0.92p = ?

The mean of a population is claimed to be 100; the population standard deviation is known to be 9. A random sample of 36 observations has a mean of 101.38. Test H₀: μ = 100 against H₁: μ > 100. Find the p-value p.

μ₀ = 100σ = 9n = 36x̄ = 101.38H₁: 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 = 0.01 with a sample of 100; the standard deviation is known to be 25. In reality the mean is 90. Find the power 1 − β of the test in percent.

μ₀ = 100μ = 90σ = 25n = 100α = 0.01
1 − β = %
13.Linear regression & correlationMedium
xy1473178920111231232614529167xy15202530100150

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.

n = 6
r =
14.Linear regression & correlationHard
xy244106128181021xy5101020x = 5

The table gives 5 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 = 5.

n = 5x = 5
ŷ =
15.Chi-square testsMedium
no defectminormajorTotalday shift16242060night shift293952120Total456372180

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

A chi-square test of independence is carried out on a contingency table with 2 rows and 6 columns (not counting the totals). How many degrees of freedom df does the test have?

2 × 6
df =

3Solutions

1.Research methods in psychology

Answer: B · Studies with significant results being more likely to be published than null results

Why this is the answer

When non-significant studies stay in file drawers, the published literature overestimates effect sizes, which distorts meta-analyses.

Answer

B · Studies with significant results being more likely to be published than null results
2.Research methods in psychology

Answer: B · Theoretical saturation is reached and new data add no new categories

Why this is the answer

Grounded theory develops theory inductively, collecting and coding data together, and stops sampling when new data no longer develop the categories.

Answer

B · Theoretical saturation is reached and new data add no new categories
3.Descriptive statistics

Answer: s = 12.76

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 = 194/5 = 38.8

Squared deviations

Σ(x − x̄)² = 0.2² + (−18.8)² + (−4.8)² + 11.2² + 12.2² = 650.8

Standard deviation

s = √(650.8/4) = 12.76

Answer

s = 12.76
4.Descriptive statistics

Answer: z = 1

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 = (525 − 500)/25 = 1

Answer

z = 1
5.Probability

Answer: Var(X) = 1.14

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 = 0·0.25 + 1·0.3 + 2·0.25 + 3·0.2 = 1.4

Mean of the squares

E(X²) = 0·0.25 + 1·0.3 + 4·0.25 + 9·0.2 = 3.1

Variance

Var(X) = 3.1 − 1.4² = 1.14

Answer

Var(X) = 1.14
6.Probability

Answer: P = 0.000734

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) = 0.411 = 4.194 × 10⁻⁵
P(X = 1) = 11·0.6·0.410 = 0.0006921

Add them

P = 4.194 × 10⁻⁵ + 0.0006921 = 0.000734

Answer

P = 0.000734
7.Normal distribution

Answer: P = 0.4836

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 = (244 − 250)/5 = −1.2
z₂ = (251.25 − 250)/5 = 0.25

Area between

P = Φ(0.25) − Φ(−1.2) = 0.5987 − 0.1151 = 0.4836

Answer

P = 0.4836
8.Normal distribution

Answer: P = 0.9515

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 = 0.01803 mm

Standardise

z = (0 − 0.03)/0.01803 = −1.66

Probability

P = 1 − Φ(−1.66) = 0.9515

Answer

P = 0.9515
9.Confidence intervals

Answer: E₂ = 1.722

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 = (1.96/2.576)·√(36/72) = 0.538

New margin

E₂ = 3.2·0.538 = 1.722

Answer

E₂ = 1.722
10.Confidence intervals

Answer: n = 3012

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 = (1.96·14/0.5)² = 3012

Round up

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

n = 3012

Answer

n = 3012
11.Hypothesis tests & confidence intervals

Answer: p = 0.1788

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 = 9/√36 = 1.5
z = (101.38 − 100)/1.5 = 0.92

Tail area

Φ(0.92) = 0.8212
1 − Φ(0.92) = 0.1788

p-value

p = 0.1788

Answer

p = 0.1788

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

12.Hypothesis tests & confidence intervals

Answer: 1 − β = 95.29 %

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 = 25/√100 = 2.5
x̄c = 100 − 2.326·2.5 = 94.19

z under the true mean

z = (94.19 − 90)/2.5 = 1.674

Power

power = P(Z < 1.674) = 95.29%
1 − β = 95.29%

Answer

1 − β = 95.29%
13.Linear regression & correlation

Answer: r = 0.9972

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 = 129/6 = 21.5
ȳ = 708/6 = 118

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 975
Syy = Σ(y − ȳ)² = 6070

Correlation coefficient

r = 975/√(157.5·6070) = 0.9972

Answer

r = 0.9972
14.Linear regression & correlation

Answer: ŷ = 10.9

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 = 65/5 = 13

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 84

Regression line

b = 84/40 = 2.1
a = 13 − 2.1·6 = 0.4
ŷ = 0.4 + 2.1·x

Prediction

ŷ = 0.4 + 2.1·5 = 10.9

Answer

ŷ = 10.9
15.Chi-square tests

Answer: χ² = 1.743

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: 60·45/180 = 15, 60·63/180 = 21, 60·72/180 = 24
night shift: 120·45/180 = 30, 120·63/180 = 42, 120·72/180 = 48

Chi-square statistic

χ² = (16 − 15)²/15 + (24 − 21)²/21 + (20 − 24)²/24 + (29 − 30)²/30 + (39 − 42)²/42 + (52 − 48)²/48
χ² = 1.743

Answer

χ² = 1.743

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

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 = (2 − 1)·(6 − 1) = 5

Answer

df = 5

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