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 wants to test whether there is an association between gender (male/female) and preferred film genre (four categories). Which test is appropriate?

  1. APearson's correlation
  2. BChi-square test of independence
  3. CPaired t-test
  4. DOne-way ANOVA
2.Research methods in psychologyHard

Why do many researchers argue that p-values should be reported together with effect sizes and confidence intervals?

  1. AEffect sizes make sample size irrelevant to significance
  2. BA p-value does not show how large or precise the effect is
  3. CConfidence intervals are only needed in qualitative research
  4. DP-values show the probability the hypothesis is true
3.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 38, 23, 13, 35, 44, 46, 50.

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

A quantity is normally distributed with mean 50 and standard deviation 15. What proportion P of values is above 80?

μ = 50σ = 15x = 80
P =
5.ProbabilityMedium
red: 8blue: 4

A bag holds 8 red and 4 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

8 / 4
P =
6.ProbabilityHard
xp(x)10.320.330.1540.25μ

X has the distribution shown, and Y = 0.5X + 1. Find the standard deviation σY of Y.

x: 1, 2, 3, 4p(x): 0.3, 0.3, 0.15, 0.25Y = 0.5X + 1
σY =
7.Normal distributionMedium
400 g403.8 gT = X₁ + … + X₁₀?

The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 10 of them is below 403.8 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 10< 403.8 g
P =
8.Normal distributionHard
157164171178185192199cmtop 70%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded top 70% is to the right of it. Find the cut-off x.

μ = 178 cmσ = 7 cm70%
x = cm
9.Confidence intervalsMedium
x̄E₁ = 9.795%, n = 100x̄E₂ = ?99%, n = 100

A 95% confidence interval for a mean, based on 100 observations and a known σ, has a margin of error of 9.7. What is the margin of error E₂ of a 99% interval from the same data?

95%n₁ = 100E₁ = 9.799%
E₂ =
10.Confidence intervalsHard
σ = 21, n = ?x̄ − Ex̄ + Ex̄E = 1.599%

The population standard deviation is 21. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 1.5?

σ = 21E = 1.599%
n =
11.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 73.4 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 73.4s = 8n = 81
t =
12.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 100 measurements has mean 502. The population standard deviation is 20. Find the upper limit U of the 99% confidence interval for the mean.

x̄ = 502σ = 20n = 10099%
U =
13.Linear regression & correlationMedium
xy123219319417516615xy2461520

The table gives 6 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.

n = 6
r =
14.Linear regression & correlationHard
xy2104196278351044xy5102040x = 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
busbikecarTotalstudents172756100staff429740Total215663140

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 = 1402 × 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 11 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 = 0.01.

α = 0.01
χ²crit =

3Solutions

1.Research methods in psychology

Answer: B · Chi-square test of independence

Why this is the answer

The chi-square test compares observed with expected frequencies in a contingency table to test whether two categorical variables are associated.

Answer

B · Chi-square test of independence
2.Research methods in psychology

Answer: B · A p-value does not show how large or precise the effect is

Why this is the answer

A p-value depends heavily on sample size and only indicates how surprising the data are under the null. Effect sizes show magnitude and intervals show precision.

Answer

B · A p-value does not show how large or precise the effect is
3.Descriptive statistics

Answer: s = 13.3

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 = 249/7 = 35.57

Squared deviations

Σ(x − x̄)² = 2.429² + (−12.57)² + (−22.57)² + (−0.5714)² + 8.429² + 10.43² + 14.43² = 1062

Standard deviation

s = √(1062/6) = 13.3

Answer

s = 13.3
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: P = 0.4242

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.

First and second draw

P = 8/12 · 7/11 = 56/132 = 0.4242

Answer

P = 0.4242
6.Probability

Answer: σY = 0.5761

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.

Spread of X

μ = 2.35, Var(X) = 1.328, σX = 1.152

Linear change

σY = |0.5|·1.152 = 0.5761

Answer

σY = 0.5761

The mean does move: E(Y) = 0.5·2.35 + 1 = 2.175.

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 total

μT = 400 g, σT = 1.2·√10 = 3.795 g

Standardise

z = (403.8 − 400)/3.795 = 1

Probability

P = Φ(1) = 0.8413

Answer

P = 0.8413
8.Normal distribution

Answer: x = 174.4 cm

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) = 0.3

z from the table

z ≈ −0.52

Back to the original units

x = 178 − 0.52·7 = 174.4 cm

Answer

x = 174.4 cm
9.Confidence intervals

Answer: E₂ = 12.75

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)·√(100/100) = 1.314

New margin

E₂ = 9.7·1.314 = 12.75

Answer

E₂ = 12.75
10.Confidence intervals

Answer: n = 1301

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 = (2.576·21/1.5)² = 1301

Round up

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

n = 1301

Answer

n = 1301
11.Hypothesis tests & confidence intervals

Answer: t = −1.8

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 = 8/√81 = 8/9 = 0.8889

Test statistic

t = (73.4 − 75)/0.8889 = −1.8

Answer

t = −1.8
t0.025; df = 80 = 1.99

|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: U = 507.2

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/√100 = 5.152

Interval limit

U = 502 + 5.152 = 507.2

Answer

U = 507.2
13.Linear regression & correlation

Answer: r = −0.9539

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 = 21/6 = 3.5
ȳ = 109/6 = 18.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −25.5
Syy = Σ(y − ȳ)² = 40.83

Correlation coefficient

r = −25.5/√(17.5·40.83) = −0.9539

Answer

r = −0.9539
14.Linear regression & correlation

Answer: ŷ = 22.8

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 = 135/5 = 27

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 168

Regression line

b = 168/40 = 4.2
a = 27 − 4.2·6 = 1.8
ŷ = 1.8 + 4.2·x

Prediction

ŷ = 1.8 + 4.2·5 = 22.8

Answer

ŷ = 22.8
15.Chi-square tests

Answer: χ² = 25.13

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

students: 100·21/140 = 15, 100·56/140 = 40, 100·63/140 = 45
staff: 40·21/140 = 6, 40·56/140 = 16, 40·63/140 = 18

Chi-square statistic

χ² = (17 − 15)²/15 + (27 − 40)²/40 + (56 − 45)²/45 + (4 − 6)²/6 + (29 − 16)²/16 + (7 − 18)²/18
χ² = 25.13

Answer

χ² = 25.13

df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is rejected.

16.Chi-square tests

Answer: χ²crit = 23.21

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 = 11 − 1 = 10

Table

χ²crit(df = 10, α = 0.01) = 23.209

Answer

χ²crit = 23.209

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

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