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 sends a questionnaire to all students and then asks respondents to recruit friends who also use a rare sleep app. Which sampling method is used for the second step?

  1. ASystematic sampling
  2. BStratified sampling
  3. CRandom sampling
  4. DSnowball sampling
2.Research methods in psychologyHard

A researcher wants to show that a new anxiety scale measures something different from depression. Low correlations with a depression scale would support:

  1. ADiscriminant validity
  2. BFace validity
  3. CConvergent validity
  4. DPredictive validity
3.Descriptive statisticsMedium
3035404550

Find the sample standard deviation s of the data: 52, 28, 40, 30, 33.

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

A quantity is normally distributed with mean 170 and standard deviation 8. What proportion P of values is below 154?

μ = 170σ = 8x = 154
P =
5.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6?7?8P(six) = 1/6 on each roll

A fair die is rolled 8 times. What is the probability P of getting at least one six?

n = 8
P =
6.ProbabilityHard
xp(x)10.1520.2530.340.3μ

X has the distribution shown, and Y = 10X + 4. Find the standard deviation σY of Y.

x: 1, 2, 3, 4p(x): 0.15, 0.25, 0.3, 0.3Y = 10X + 4
σY =
7.Normal distributionMedium
32 min37 minX + Y?

X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 37 min. Round z to two decimals.

μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 37 min
P =
8.Normal distributionHard
10 cm0X − Y?

The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
9.Confidence intervalsMedium
x̄E₁ = 3.595%, n = 25x̄E₂ = ?90%, 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 90% interval from the same data?

95%n₁ = 25E₁ = 3.590%
E₂ =
10.Confidence intervalsHard
μ₁ − μ₂, t(20) = 1.7250LUx̄₁ − x̄₂E = ?90%

Two fertilisers are tested. Plot group 1: 8 plots, mean yield 13 kg, standard deviation 8.5 kg. Plot group 2: 14 plots, mean 7 kg, standard deviation 9.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.

n₁ = 8x̄₁ = 13s₁ = 8.5n₂ = 14x̄₂ = 7s₂ = 9.590%t(20) = 1.725
E = kg
11.Hypothesis tests & confidence intervalsMedium
−2.1452.145t = ?df = 14

Two teaching methods are compared. Group 1: 8 students, mean score 96.5, standard deviation 5.5. Group 2: 8 students, mean score 102, standard deviation 5.5. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 8x̄₁ = 96.5s₁ = 5.5n₂ = 8x̄₂ = 102s₂ = 5.5
t =
12.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 100 measurements has mean 119.6. The population standard deviation is 10. Find the lower limit L of the 90% confidence interval for the mean.

x̄ = 119.6σ = 10n = 10090%
L =
13.Linear regression & correlationMedium
xy126224321419515610xy24610152025

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
xy153356567773975xy56070x = 4

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 4.

n = 5x = 4
ŷ =
15.Chi-square testsMedium
busbikecarTotalstudents403141112staff41192484Total815065196

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

row total = 84column total = 81N = 196
E =
16.Chi-square testsHard
passfailTotalplant A221840plant B122840plant C223860Total5684140dfα = 0.05α = 0.0113.8416.63525.9919.21037.81511.34549.48813.277511.07015.086612.59216.812714.06718.475815.50720.090916.91921.6661018.30723.209

The table shows observed counts. Test at α = 0.05 whether the two variables are independent (no continuity correction). Compute the test statistic χ², then compare it with the critical value from the table.

N = 1403 × 2α = 0.05
χ² =

3Solutions

1.Research methods in psychology

Answer: D · Snowball sampling

Why this is the answer

Snowball sampling is useful for hard-to-reach groups, but the sample is likely to share characteristics and is not representative.

Answer

D · Snowball sampling
2.Research methods in psychology

Answer: A · Discriminant validity

Why this is the answer

Discriminant (divergent) validity is shown when a measure correlates weakly with measures of different constructs. Convergent validity is high correlation with measures of the same construct.

Answer

A · Discriminant validity
3.Descriptive statistics

Answer: s = 9.737

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 = 183/5 = 36.6

Squared deviations

Σ(x − x̄)² = 15.4² + (−8.6)² + 3.4² + (−6.6)² + (−3.6)² = 379.2

Standard deviation

s = √(379.2/4) = 9.737

Answer

s = 9.737
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) = 0.02275
P = 0.02275

Answer

P = 0.02275
5.Probability

Answer: P = 0.7674

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.

Complement

P(no six) = (5/6)8 = 0.2326
P = 1 − 0.2326 = 0.7674

Answer

P = 0.7674
6.Probability

Answer: σY = 10.43

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 = 2.75, Var(X) = 1.087, σX = 1.043

Linear change

σY = |10|·1.043 = 10.43

Answer

σY = 10.43

The mean does move: E(Y) = 10·2.75 + 4 = 31.5.

7.Normal distribution

Answer: P = 0.02275

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 = 32 min, σ = √(1.5² + 2²) = 2.5 min

Standardise

z = (37 − 32)/2.5 = 2

Upper tail

P = 1 − Φ(2) = 1 − 0.9772 = 0.02275

Answer

P = 0.02275
8.Normal distribution

Answer: P = 0.7389

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 = 180 − 170 = 10 cm, σ = 15.62 cm

Standardise

z = (0 − 10)/15.62 = −0.64

Probability

P = 1 − Φ(−0.64) = 0.7389

Answer

P = 0.7389
9.Confidence intervals

Answer: E₂ = 2.938

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 = (1.645/1.96)·√(25/25) = 0.8393

New margin

E₂ = 3.5·0.8393 = 2.938

Answer

E₂ = 2.938
10.Confidence intervals

Answer: E = 7.005 kg

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.

Pooled standard deviation

sp² = (7·8.5² + 13·9.5²)/20 = 83.95
sp = 9.162

Margin of error

E = 1.725·9.162·√(1/8 + 1/14) = 7.005

Answer

E = 7.005 kg
11.Hypothesis tests & confidence intervals

Answer: t = −2

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.

Pooled variance

sp² = (7·5.5² + 7·5.5²)/14 = 30.25
sp = 5.5

Standard error

SE = 5.5·√(1/8 + 1/8) = 2.75

Test statistic

t = (96.5 − 102)/2.75 = −2

Answer

t = −2

With df = 14, the two-sided 5% critical value is 2.145; |t| is smaller, so H₀ is not rejected.

12.Hypothesis tests & confidence intervals

Answer: L = 118

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 = 1.645·10/√100 = 1.645

Interval limit

L = 119.6 − 1.645 = 118

Answer

L = 118
13.Linear regression & correlation

Answer: r = −0.9853

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 = 115/6 = 19.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −54.5
Syy = Σ(y − ȳ)² = 174.8

Correlation coefficient

r = −54.5/√(17.5·174.8) = −0.9853

Answer

r = −0.9853
14.Linear regression & correlation

Answer: ŷ = 61.75

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 = 25/5 = 5
ȳ = 324/5 = 64.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 122

Regression line

b = 122/40 = 3.05
a = 64.8 − 3.05·5 = 49.55
ŷ = 49.55 + 3.05·x

Prediction

ŷ = 49.55 + 3.05·4 = 61.75

Answer

ŷ = 61.75
15.Chi-square tests

Answer: E = 34.71

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 = 84·81/196 = 34.71

Answer

E = 34.71
16.Chi-square tests

Answer: χ² = 5.694

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

plant A: 16, 24
plant B: 16, 24
plant C: 24, 36

Chi-square statistic

χ² = (22 − 16)²/16 + (18 − 24)²/24 + (12 − 16)²/16 + (28 − 24)²/24 + (22 − 24)²/24 + (38 − 36)²/36
χ² = 5.694

Decision

df = 2, χ²crit = 5.991
5.694 < 5.991

Answer

χ² = 5.694

χ² is below the critical value, so H₀ is not rejected: there is no significant evidence of a relationship.

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