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 cross-sectional study compares 20-, 40- and 60-year-olds on technology use at the same time. Which problem may confound the age comparison?

  1. ACohort effects
  2. BOrder effects
  3. CFloor effects
  4. DPractice effects
2.Research methods in psychologyHard

A researcher wants to test whether a new teaching method causes better grades, but cannot randomise students. Which design gives the strongest causal evidence in this situation?

  1. AAsking students whether they think the method worked
  2. BMeasuring only the class using the new method after teaching
  3. CCorrelating grades with how much students like the teacher
  4. DComparing matched classes measured before and after, with a comparison group
3.Descriptive statisticsMedium
202530354045

Find the sample standard deviation s of the data: 49, 33, 45, 16, 44, 31.

n = 6
s =
4.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 70 and standard deviation 8. Find the z-score z of a score of 62.

μ = 70σ = 8x = 62
z =
5.ProbabilityMedium
xp(x)100.35200.35300.1400.2μ = 21.5

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

x: 10, 20, 30, 40p(x): 0.35, 0.35, 0.1, 0.2
Var(X) =
6.ProbabilityHard
n = 6, p = 0.25: P(X = 1) = ?k0123456

A test has 6 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 = 6p = 0.25k = 1
P =
7.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0505

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0505. Find z to two decimals.

A = 0.0505
z =
8.Normal distributionHard
157164171178185192199cmbottom 10%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded bottom 10% is to the left of it. Find the cut-off x.

μ = 178 cmσ = 7 cm10%
x = cm
9.Confidence intervalsMedium
n = 25, s = 11.5LU = ?x̄ = 113.5E99%

A sample of 25 items has mean 113.5 and standard deviation s = 11.5. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.

n = 25x̄ = 113.5s = 11.599%t(24) = 2.797
upper limit =
10.Confidence intervalsHard
σ = 11, n = ?x̄ − Ex̄ + Ex̄E = 399%

The population standard deviation is 11. 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 3?

σ = 11E = 399%
n =
11.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 118.9 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 118.9s = 4n = 25
t =
12.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 64 measurements has mean 50.7. The population standard deviation is 4. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 50.7σ = 4n = 6495%
L =
13.Linear regression & correlationMedium
xy364464569686784898xy46880100

The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 6
r =
14.Linear regression & correlationHard
xy142716341838204422652472xy1520254060x = 15

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 15.

n = 6x = 15
ŷ =
15.Chi-square testsMedium
no defectminormajorTotalday shift26342787night shift13372373Total397150160

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

row total = 87column total = 50N = 160
E =
16.Chi-square testsHard
passfailTotalage 18–30542680age 31–504555100age 51+213960Total120120240dfα = 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 = 0.01 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 = 2403 × 2α = 0.01
χ² =

3Solutions

1.Research methods in psychology

Answer: A · Cohort effects

Why this is the answer

Different age groups were raised in different historical conditions, so differences may reflect generation (cohort) rather than ageing itself.

Answer

A · Cohort effects
2.Research methods in psychology

Answer: D · Comparing matched classes measured before and after, with a comparison group

Why this is the answer

A non-equivalent groups pretest-posttest design controls for initial differences and for change over time, though it is weaker than a randomised experiment.

Answer

D · Comparing matched classes measured before and after, with a comparison group
3.Descriptive statistics

Answer: s = 12.23

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 = 218/6 = 36.33

Squared deviations

Σ(x − x̄)² = 12.67² + (−3.333)² + 8.667² + (−20.33)² + 7.667² + (−5.333)² = 747.3

Standard deviation

s = √(747.3/5) = 12.23

Answer

s = 12.23
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 = (62 − 70)/8 = −1

Answer

z = −1
5.Probability

Answer: Var(X) = 122.8

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 = 10·0.35 + 20·0.35 + 30·0.1 + 40·0.2 = 21.5

Mean of the squares

E(X²) = 100·0.35 + 400·0.35 + 900·0.1 + 1600·0.2 = 585

Variance

Var(X) = 585 − 21.5² = 122.8

Answer

Var(X) = 122.8
6.Probability

Answer: P = 0.356

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) = 6·0.251·0.755 = 6·0.25·0.2373 = 0.356

Answer

P = 0.356
7.Normal distribution

Answer: z = 1.64

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 of z

Φ(z) = 1 − 0.0505 = 0.9495

Read the z-table backwards

Find 0.9495 inside the table and read off z.

z = 1.64

Answer

z = 1.64
8.Normal distribution

Answer: x = 169 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) = 0.1

z from the table

z ≈ −1.28

Back to the original units

x = 178 − 1.28·7 = 169 cm

Answer

x = 169 cm
9.Confidence intervals

Answer: upper limit = 119.9

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 = 2.797

Margin of error

E = 2.797·11.5/√25 = 6.433

Interval

113.5 ± 6.433 → [107.1, 119.9]

Answer

upper limit = 119.9
10.Confidence intervals

Answer: n = 90

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 = (2.576·11/3)² = 89.21

Round up

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

n = 90

Answer

n = 90
11.Hypothesis tests & confidence intervals

Answer: t = −1.375

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 = 4/√25 = 4/5 = 0.8

Test statistic

t = (118.9 − 120)/0.8 = −1.375

Answer

t = −1.375
t0.025; df = 24 = 2.064

|t| < 2.064, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: L = 49.72

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 = 1.96·4/√64 = 0.98

Interval limit

L = 50.7 − 0.98 = 49.72

Answer

L = 49.72
13.Linear regression & correlation

Answer: r = 0.9472

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 = 33/6 = 5.5
ȳ = 465/6 = 77.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 123.5
Syy = Σ(y − ȳ)² = 971.5

Correlation coefficient

r = 123.5/√(17.5·971.5) = 0.9472

Answer

r = 0.9472
14.Linear regression & correlation

Answer: ŷ = 28.15

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 = 114/6 = 19
ȳ = 280/6 = 46.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 324

Regression line

b = 324/70 = 4.629
a = 46.67 − 4.629·19 = −41.28
ŷ = −41.28 + 4.629·x

Prediction

ŷ = −41.28 + 4.629·15 = 28.15

Answer

ŷ = 28.15
15.Chi-square tests

Answer: E = 27.19

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 = 87·50/160 = 27.19

Answer

E = 27.19
16.Chi-square tests

Answer: χ² = 16.2

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

age 18–30: 40, 40
age 31–50: 50, 50
age 51+: 30, 30

Chi-square statistic

χ² = (54 − 40)²/40 + (26 − 40)²/40 + (45 − 50)²/50 + (55 − 50)²/50 + (21 − 30)²/30 + (39 − 30)²/30
χ² = 16.2

Decision

df = 2, χ²crit = 9.210
16.2 > 9.210

Answer

χ² = 16.2

χ² is above the critical value, so H₀ is rejected: the variables appear to be related.

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