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

In a positively skewed distribution, such as income:

  1. AThe mean and median are always equal
  2. BThe mean is greater than the median
  3. CThere are no extreme scores
  4. DThe mean is less than the median
2.Research methods in psychologyHard

A memory study uses a repeated measures design where everyone learns list A then list B. Recall is better for list B. What is the most likely flaw?

  1. AThe use of an ordinal level of measurement
  2. BToo large a sample size
  3. CIndividual differences between the groups
  4. DOrder effects such as practice, because conditions were not counterbalanced
3.Descriptive statisticsMedium
20253035404550

Find the sample standard deviation s of the data: 41, 19, 42, 30, 43, 50, 43.

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

Exam scores are normally distributed with mean 70 and standard deviation 5. Find the z-score z of a score of 64.

μ = 70σ = 5x = 64
z =
5.ProbabilityMedium
xp(x)00.3510.120.3530.2μ = 1.4

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

x: 0, 1, 2, 3p(x): 0.35, 0.1, 0.35, 0.2
σ =
6.ProbabilityHard
n = 12, p = 0.7: P(X = 2) = ?k0123456789101112

A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.7, independently. What is the probability P of exactly 2 correct answers?

n = 12p = 0.7k = 2
P =
7.Normal distributionMedium
164181.5P = ?

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 164 and 181.5 cm? Give P as a decimal.

μ = 178σ = 7x1 = 164x2 = 181.5
P =
8.Normal distributionHard
μ = 65x = ?10%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 10% of them lie (the 10th percentile).

μ = 65σ = 10P = 10%
x = points
9.Confidence intervalsMedium
n = 25, s = 6.5L = ?Ux̄ = 92E95%

A sample of 25 items has mean 92 and standard deviation s = 6.5. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 25x̄ = 92s = 6.595%t(24) = 2.064
lower limit =
10.Confidence intervalsHard
σ = 21, n = ?x̄ − Ex̄ + Ex̄E = 390%

The population standard deviation is 21. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 3?

σ = 21E = 390%
n =
11.Hypothesis tests & confidence intervalsMedium
−2.3652.365t = ?df = 7

8 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 9.38 and the standard deviation of the differences is 7.8. Find the paired test statistic t for H₀: μd = 0.

n = 8d̄ = 9.38sd = 7.8
t =
12.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 200 against H₁: μ > 200 is done at α = 0.05 with a sample of 64; the standard deviation is known to be 24. In reality the mean is 205.1. Find the power 1 − β of the test in percent.

μ₀ = 200μ = 205.1σ = 24n = 64α = 0.05
1 − β = %
13.Linear regression & correlationMedium
xy21142162683310411251xy5102040

The table gives 6 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
14.Linear regression & correlationHard
xy274961481510191223xy510101520x = 5

The table gives 6 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 = 6x = 5
ŷ =
15.Chi-square testsMedium
improvednot improvedTotalnew drug4476120placebo43640Total48112160

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 = 1602 × 2
χ² =
16.Chi-square testsHard
yesnoTotalline 132840line 2352560line 3172340Total8456140dfα = 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: B · The mean is greater than the median

Why this is the answer

A few very high values in the right tail raise the mean above the median, which is why the median is often reported for skewed data.

Answer

B · The mean is greater than the median
2.Research methods in psychology

Answer: D · Order effects such as practice, because conditions were not counterbalanced

Why this is the answer

When all participants experience conditions in the same order, improvement may reflect practice rather than the list. Counterbalancing (ABBA or half AB, half BA) controls this.

Answer

D · Order effects such as practice, because conditions were not counterbalanced
3.Descriptive statistics

Answer: s = 10.36

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 = 268/7 = 38.29

Squared deviations

Σ(x − x̄)² = 2.714² + (−19.29)² + 3.714² + (−8.286)² + 4.714² + 11.71² + 4.714² = 643.4

Standard deviation

s = √(643.4/6) = 10.36

Answer

s = 10.36
4.Descriptive statistics

Answer: z = −1.2

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 = (64 − 70)/5 = −1.2

Answer

z = −1.2
5.Probability

Answer: σ = 1.158

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 + 1·0.1 + 2·0.35 + 3·0.2 = 1.4

Mean of the squares

E(X²) = 0·0.35 + 1·0.1 + 4·0.35 + 9·0.2 = 3.3

Variance

Var(X) = 3.3 − 1.4² = 1.34
σ = √1.34 = 1.158

Answer

σ = 1.158
6.Probability

Answer: P = 0.000191

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 = 2) = 66·0.72·0.310 = 66·0.49·5.905 × 10⁻⁶ = 0.000191

Answer

P = 0.000191
7.Normal distribution

Answer: P = 0.6687

Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.

z-scores

z₁ = (164 − 178)/7 = −2
z₂ = (181.5 − 178)/7 = 0.5

Area between

P = Φ(0.5) − Φ(−2) = 0.6915 − 0.02275 = 0.6687

Answer

P = 0.6687
8.Normal distribution

Answer: x = 52.18 points

Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.

z for the percentile

Φ(z) = 0.1 → z = −1.282

Back to x

x = 65 + −1.282·10 = 52.18 points

Answer

x = 52.18 points
9.Confidence intervals

Answer: lower limit = 89.32

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 = 2.064

Margin of error

E = 2.064·6.5/√25 = 2.683

Interval

92 ± 2.683 → [89.32, 94.68]

Answer

lower limit = 89.32
10.Confidence intervals

Answer: n = 133

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 = (1.645·21/3)² = 132.6

Round up

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

n = 133

Answer

n = 133
11.Hypothesis tests & confidence intervals

Answer: t = 3.401

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 of the mean difference

SE = 7.8/√8 = 2.758

Test statistic

t = 9.38/2.758 = 3.401

Answer

t = 3.401

With df = 7, the two-sided 5% critical value is 2.365; |t| is larger, so H₀ is rejected.

12.Hypothesis tests & confidence intervals

Answer: 1 − β = 52.19 %

Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).

Critical sample mean

SE = 24/√64 = 3
x̄c = 200 + 1.645·3 = 204.9

z under the true mean

z = (204.9 − 205.1)/3 = −0.055

Power

power = P(Z > −0.055) = 52.19%
1 − β = 52.19%

Answer

1 − β = 52.19%
13.Linear regression & correlation

Answer: r² = 0.9912

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 = 42/6 = 7
ȳ = 183/6 = 30.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 267
Syy = Σ(y − ȳ)² = 1028

Coefficient of determination

r² = 267²/(70·1028) = 0.9912

Answer

r² = 0.9912

About 99.12% of the variation in y is explained by the line.

14.Linear regression & correlation

Answer: ŷ = 11.33

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 = 42/6 = 7
ȳ = 87/6 = 14.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 111

Regression line

b = 111/70 = 1.586
a = 14.5 − 1.586·7 = 3.4
ŷ = 3.4 + 1.586·x

Prediction

ŷ = 3.4 + 1.586·5 = 11.33

Answer

ŷ = 11.33
15.Chi-square tests

Answer: χ² = 10.16

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

new drug: 120·48/160 = 36, 120·112/160 = 84
placebo: 40·48/160 = 12, 40·112/160 = 28

Chi-square statistic

χ² = (44 − 36)²/36 + (76 − 84)²/84 + (4 − 12)²/12 + (36 − 28)²/28
χ² = 10.16

Answer

χ² = 10.16

df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is rejected.

16.Chi-square tests

Answer: χ² = 11.84

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

line 1: 24, 16
line 2: 36, 24
line 3: 24, 16

Chi-square statistic

χ² = (32 − 24)²/24 + (8 − 16)²/16 + (35 − 36)²/36 + (25 − 24)²/24 + (17 − 24)²/24 + (23 − 16)²/16
χ² = 11.84

Decision

df = 2, χ²crit = 5.991
11.84 > 5.991

Answer

χ² = 11.84

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

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