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

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

  1. AShow whether the result is statistically significant at the chosen alpha level
  2. BReplace the need for any hypothesis
  3. CTell how many participants dropped out
  4. DShow how large an effect is, regardless of sample size
3.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 55, 47, 39, 19, 51.

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

A quantity is normally distributed with mean 50 and standard deviation 5. What proportion P of values is above 60?

μ = 50σ = 5x = 60
P =
5.ProbabilityMedium
xp(x)00.3510.320.35μ = 1

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

x: 0, 1, 2p(x): 0.35, 0.3, 0.35
σ =
6.ProbabilityHard
n = 9, p = 0.25: P(X = 1) = ?k0123456789

A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 1 correct answers?

n = 9p = 0.25k = 1
P =
7.Normal distributionMedium
750 g740.2 gX + Y?

X is normal with mean 250 g and standard deviation 4 g; Y is independent and normal with mean 500 g and standard deviation 9 g. Find the probability P that X + Y is more than 740.2 g. Round z to two decimals.

μX = 250, σX = 4μY = 500, σY = 9X + Y > 740.2 g
P =
8.Normal distributionHard
35455565758595pointstop 90%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 90% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points90%
x = points
9.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E99%

The battery life of two phone models is compared. Model 1: 100 phones, mean 75.5 h, known standard deviation 17 h. Model 2: 50 phones, mean 74.5 h, known standard deviation 6 h. Find the upper limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 100x̄₁ = 75.5 hσ₁ = 17 hn₂ = 50x̄₂ = 74.5 hσ₂ = 6 h99%
upper limit = h
10.Confidence intervalsHard
n = 10σ = ?Us = 2.395%

The diameters of 10 machined shafts have a sample standard deviation of 2.3 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 9, 2.5% of the χ² distribution lies below 2.7 and 2.5% lies above 19.023.

n = 10s = 2.3 µmχ² = 2.7, 19.023
σ = µm
11.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 78 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 78s = 5n = 16
t =
12.Hypothesis tests & confidence intervalsHard
σ = 8, n = ?x̄ − Ex̄ + Ex̄E = 199%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 8. Find the smallest sample size n.

σ = 8E = 199%
n =
13.Linear regression & correlationMedium
xy253355460577680xy246607080

The table gives 5 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
14.Linear regression & correlationHard
xy2554596728761088xy5106080ŷ = 82

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 82?

n = 5ŷ = 82
x =
15.Chi-square testsMedium
no defectminormajorTotalday shift324245119night shift37311583Total697360202

The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “night shift, no defect”.

row total = 83column total = 69N = 202
E =
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 test of independence is carried out on a contingency table with 3 rows and 3 columns (not counting the totals). Using the table, find the critical value χ²crit at significance level α = 0.01.

α = 0.01
χ²crit =

3Solutions

1.Research methods in psychology

Answer: A · 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

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

Answer: D · 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

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

Answer: s = 14.25

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 = 211/5 = 42.2

Squared deviations

Σ(x − x̄)² = 12.8² + 4.8² + (−3.2)² + (−23.2)² + 8.8² = 812.8

Standard deviation

s = √(812.8/4) = 14.25

Answer

s = 14.25
4.Descriptive statistics

Answer: P = 0.02275

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 = 2

Standard normal table

Φ(2) = 0.9772
P = 1 − 0.9772 = 0.02275

Answer

P = 0.02275
5.Probability

Answer: σ = 0.8367

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 = 0·0.35 + 1·0.3 + 2·0.35 = 1

Mean of the squares

E(X²) = 0·0.35 + 1·0.3 + 4·0.35 = 1.7

Variance

Var(X) = 1.7 − 1² = 0.7
σ = √0.7 = 0.8367

Answer

σ = 0.8367
6.Probability

Answer: P = 0.2253

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 = 1) = 9·0.251·0.758 = 9·0.25·0.1001 = 0.2253

Answer

P = 0.2253
7.Normal distribution

Answer: P = 0.8413

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 sum

μ = 750 g, σ = √(4² + 9²) = 9.849 g

Standardise

z = (740.2 − 750)/9.849 = −1

Upper tail

P = 1 − Φ(−1) = 1 − 0.1587 = 0.8413

Answer

P = 0.8413
8.Normal distribution

Answer: x = 52.2 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) = 0.1

z from the table

z ≈ −1.28

Back to the original units

x = 65 − 1.28·10 = 52.2 points

Answer

x = 52.2 points
9.Confidence intervals

Answer: upper limit = 5.894 h

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.

Point estimate

x̄₁ − x̄₂ = 75.5 − 74.5 = 1

Standard error

SE = √(17²/100 + 6²/50) = 1.9

Margin of error

E = 2.576·1.9 = 4.894

Interval

1 ± 4.894 → [−3.894, 5.894]

Answer

upper limit = 5.894 h

The interval contains 0, so the data do not show a significant difference.

10.Confidence intervals

Answer: σ = 1.582 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 = 9·2.3² = 47.61

Limits for σ²

47.61/19.023 = 2.503
47.61/2.7 = 17.63

Limit for σ

σ = √2.503 = 1.582 µm

Answer

σ = 1.582 µm

95% interval for σ: [1.582, 4.199] µm.

11.Hypothesis tests & confidence intervals

Answer: t = 2.4

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).

Standard error

SE = 5/√16 = 5/4 = 1.25

Test statistic

t = (78 − 75)/1.25 = 2.4

Answer

t = 2.4
t0.025; df = 15 = 2.131

|t| > 2.131, so H₀ is rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: n = 425

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).

Solve for n

n = (2.576·8/1)² = 424.7

Round up

n = 425

Answer

n = 425
13.Linear regression & correlation

Answer: r² = 0.9053

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 = 20/5 = 4
ȳ = 325/5 = 65

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 76
Syy = Σ(y − ȳ)² = 638

Coefficient of determination

r² = 76²/(10·638) = 0.9053

Answer

r² = 0.9053

About 90.53% of the variation in y is explained by the line.

14.Linear regression & correlation

Answer: x = 8.892

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 = 350/5 = 70

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 166

Regression line

b = 166/40 = 4.15
a = 70 − 4.15·6 = 45.1
ŷ = 45.1 + 4.15·x

Solve for x

x = (82 − 45.1)/4.15 = 8.892

Answer

x = 8.892
15.Chi-square tests

Answer: E = 28.35

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 = 83·69/202 = 28.35

Answer

E = 28.35
16.Chi-square tests

Answer: χ²crit = 13.28

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)·(3 − 1) = 4

Table

χ²crit(df = 4, α = 0.01) = 13.277

Answer

χ²crit = 13.277

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

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