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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 compares children's aggression before and after television arrives in a remote town. The researcher did not control the arrival of television. This is a:

  1. ACorrelational survey of television viewing
  2. BNatural experiment
  3. CLaboratory experiment
  4. DField experiment
2.Research methods in psychologyHard

Increasing the sample size in a study most directly:

  1. ARemoves all bias from the sample
  2. BIncreases statistical power to detect a real effect
  3. CIncreases the effect size
  4. DGuarantees a statistically significant result if any effect exists
3.Descriptive statisticsMedium
152025303540

Find the sample standard deviation s of the data: 29, 39, 13, 30, 12, 40, 33.

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

Exam scores are normally distributed with mean 60 and standard deviation 10. Find the z-score z of a score of 45.

μ = 60σ = 10x = 45
z =
5.ProbabilityMedium
xp(x)00.3510.2520.1530.140.15μ = 1.45

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

x: 0, 1, 2, 3, 4p(x): 0.35, 0.25, 0.15, 0.1, 0.15
σ =
6.ProbabilityHard
xp(x)00.210.220.3530.25μ

X has the distribution shown, and Y = 0.5X + 10. Find the standard deviation σY of Y.

x: 0, 1, 2, 3p(x): 0.2, 0.2, 0.35, 0.25Y = 0.5X + 10
σY =
7.Normal distributionMedium
750 g764.8 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 764.8 g. Round z to two decimals.

μX = 250, σX = 4μY = 500, σY = 9X + Y > 764.8 g
P =
8.Normal distributionHard
157164171178185192199cmbottom 60%

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 60% is to the left of it. Find the cut-off x.

μ = 178 cmσ = 7 cm60%
x = cm
9.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?99%

The battery life of two phone models is compared. Model 1: 30 phones, mean 83.5 h, known standard deviation 9 h. Model 2: 30 phones, mean 71 h, known standard deviation 18 h. Find the margin of error E of the 99% confidence interval for μ₁ − μ₂.

n₁ = 30x̄₁ = 83.5 hσ₁ = 9 hn₂ = 30x̄₂ = 71 hσ₂ = 18 h99%
E = h
10.Confidence intervalsHard
p ≈ 0.6, n = ?p̂ − Ep̂ + Ep̂E = 0.0599%

A pilot study suggests that about 0.6 of products are defective. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.05?

p ≈ 0.6E = 0.0599%
n =
11.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 40 parts has mean length 68 mm, and the line's known standard deviation is 15 mm. Line 2: a sample of 80 parts has mean 72.2 mm, known standard deviation 16 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 40x̄₁ = 68 mmσ₁ = 15 mmn₂ = 80x̄₂ = 72.2 mmσ₂ = 16 mm
z =
12.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 64 measurements has mean 48.1. The population standard deviation is 15. Find the lower limit L of the 99% confidence interval for the mean.

x̄ = 48.1σ = 15n = 6499%
L =
13.Linear regression & correlationMedium
xy358461560671776880xy468607080

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 33Σy = 406Σx² = 199Σy² = 27902Σxy = 2316
r =
14.Linear regression & correlationHard
xy126228326429535636xy2463035ŷ = 31

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 31?

n = 6ŷ = 31
x =
15.Chi-square testsMedium
busbikecarTotalstudents34231067staff35242988Total694739155

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

row total = 67column total = 39N = 155
E =
16.Chi-square testsHard

A chi-square goodness-of-fit test compares observed counts in 10 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?

k = 10
df =

3Solutions

1.Research methods in psychology

Answer: B · Natural experiment

Why this is the answer

In a natural experiment the IV varies because of events outside the researcher's control. Ecological validity is high, but random allocation is not possible.

Answer

B · Natural experiment
2.Research methods in psychology

Answer: B · Increases statistical power to detect a real effect

Why this is the answer

Larger samples reduce standard error, increasing power. They do not change the true effect size or remove systematic bias.

Answer

B · Increases statistical power to detect a real effect
3.Descriptive statistics

Answer: s = 11.37

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 = 196/7 = 28

Squared deviations

Σ(x − x̄)² = 1² + 11² + (−15)² + 2² + (−16)² + 12² + 5² = 776

Standard deviation

s = √(776/6) = 11.37

Answer

s = 11.37
4.Descriptive statistics

Answer: z = −1.5

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 = (45 − 60)/10 = −1.5

Answer

z = −1.5
5.Probability

Answer: σ = 1.431

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 + 1·0.25 + 2·0.15 + 3·0.1 + 4·0.15 = 1.45

Mean of the squares

E(X²) = 0·0.35 + 1·0.25 + 4·0.15 + 9·0.1 + 16·0.15 = 4.15

Variance

Var(X) = 4.15 − 1.45² = 2.047
σ = √2.047 = 1.431

Answer

σ = 1.431
6.Probability

Answer: σY = 0.5309

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 = 1.65, Var(X) = 1.127, σX = 1.062

Linear change

σY = |0.5|·1.062 = 0.5309

Answer

σY = 0.5309

The mean does move: E(Y) = 0.5·1.65 + 10 = 10.82.

7.Normal distribution

Answer: P = 0.06681

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 = (764.8 − 750)/9.849 = 1.5

Upper tail

P = 1 − Φ(1.5) = 1 − 0.9332 = 0.06681

Answer

P = 0.06681
8.Normal distribution

Answer: x = 179.8 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) = 0.6

z from the table

z ≈ 0.25

Back to the original units

x = 178 + 0.25·7 = 179.8 cm

Answer

x = 179.8 cm
9.Confidence intervals

Answer: E = 9.465 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 = 83.5 − 71 = 12.5

Standard error

SE = √(9²/30 + 18²/30) = 3.674

Margin of error

E = 2.576·3.674 = 9.465

Answer

E = 9.465 h

The interval does not contain 0, so the means differ significantly.

10.Confidence intervals

Answer: n = 638

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 2.576²·0.6·0.4/0.05² = 637

Round up

n = 638

Answer

n = 638
11.Hypothesis tests & confidence intervals

Answer: z = −1.414

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.

Standard error of the difference

SE = √(15²/40 + 16²/80) = √8.825 = 2.971

Test statistic

z = (68 − 72.2)/2.971 = −1.414

Answer

z = −1.414

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: L = 43.27

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 = 2.576·15/√64 = 4.83

Interval limit

L = 48.1 − 4.83 = 43.27

Answer

L = 43.27
13.Linear regression & correlation

Answer: r = 0.9576

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.

Centred sums

Sxy = 2316 − 33·406/6 = 83
Sxx = 199 − 33²/6 = 17.5
Syy = 27902 − 406²/6 = 429.3

Correlation coefficient

r = 83/√(17.5·429.3) = 0.9576

Answer

r = 0.9576
14.Linear regression & correlation

Answer: x = 3.973

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 = 180/6 = 30

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 37

Regression line

b = 37/17.5 = 2.114
a = 30 − 2.114·3.5 = 22.6
ŷ = 22.6 + 2.114·x

Solve for x

x = (31 − 22.6)/2.114 = 3.973

Answer

x = 3.973
15.Chi-square tests

Answer: E = 16.86

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 = 67·39/155 = 16.86

Answer

E = 16.86
16.Chi-square tests

Answer: df = 9

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 = 10 − 1 = 9

Answer

df = 9

ExamProblems · examproblems.com/courses/research-methods-psychology