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 student scores 130 on a test with mean 100 and standard deviation 15. What is the z-score?

  1. A1.5
  2. B2.0
  3. C0.5
  4. D3.0
2.Research methods in psychologyHard

Which non-parametric test compares two related samples, such as before-and-after scores on an ordinal scale?

  1. AChi-square test of independence
  2. BMann–Whitney U test
  3. CWilcoxon signed-rank test
  4. DIndependent t-test
3.Descriptive statisticsMedium
101520253035404550

Find the sample standard deviation s of the data: 36, 13, 11, 27, 10, 54.

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

Exam scores are normally distributed with mean 170 and standard deviation 5. Find the z-score z of a score of 162.5.

μ = 170σ = 5x = 162.5
z =
5.ProbabilityMedium
xp(x)00.3510.0520.2530.35μ = 1.6

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

x: 0, 1, 2, 3p(x): 0.35, 0.05, 0.25, 0.35
σ =
6.ProbabilityHard
n = 9, p = 0.4: P(X = 5) = ?k0123456789

A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 5 correct answers?

n = 9p = 0.4k = 5
P =
7.Normal distributionMedium
105 s95.4 sX + Y?

X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 95.4 s. Round z to two decimals.

μX = 60, σX = 5μY = 45, σY = 4X + Y > 95.4 s
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
n = 12, sd​ = 1.20LUd̄ = 2.2E = ?95%

The blood pressure of 12 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 2.2 and standard deviation 1.2. Find the margin of error E of the 95% confidence interval for the mean difference.

n = 12d̄ = 2.2sd = 1.295%t(11) = 2.201
E = mmHg
10.Confidence intervalsHard
σ = 31, n = ?x̄ − Ex̄ + Ex̄E = 1.599%

The population standard deviation is 31. 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 = 31E = 1.599%
n =
11.Hypothesis tests & confidence intervalsMedium
t −2.0302.03

A sample of 36 students has a mean score of 201 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 201s = 5n = 36
t =
12.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 90 of 300 patients on treatment A recover, and 103 of 250 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 90n₁ = 300x₂ = 103n₂ = 250
z =
13.Linear regression & correlationMedium
xy2114176248351042xy51010203040

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
14.Linear regression & correlationHard
xy1783209823106261222914232160xy202530100150x = 24.5

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. A new observation (24.5, 122) is made. Find its residual e = y − ŷ.

n = 6(24.5, 122)
e =
15.Chi-square testsMedium
lowmediumhighTotalurban10242660rural144046100Total246472160

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 × 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 checks whether counts in 6 categories follow a Poisson distribution whose mean is estimated from the same data. Using the table, find the critical value χ²crit at significance level α = 0.05.

α = 0.05
χ²crit =

3Solutions

1.Research methods in psychology

Answer: B · 2.0

Why this is the answer

(130 − 100) / 15 = 2. The score is two standard deviations above the mean, higher than about 97–98% of people.

Answer

B · 2.0
2.Research methods in psychology

Answer: C · Wilcoxon signed-rank test

Why this is the answer

The Wilcoxon signed-rank test is for paired ordinal data; Mann–Whitney U is its counterpart for independent groups.

Answer

C · Wilcoxon signed-rank test
3.Descriptive statistics

Answer: s = 17.5

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.

Mean

x̄ = 151/6 = 25.17

Squared deviations

Σ(x − x̄)² = 10.83² + (−12.17)² + (−14.17)² + 1.833² + (−15.17)² + 28.83² = 1531

Standard deviation

s = √(1531/5) = 17.5

Answer

s = 17.5
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 = (162.5 − 170)/5 = −1.5

Answer

z = −1.5
5.Probability

Answer: σ = 1.281

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 + 1·0.05 + 2·0.25 + 3·0.35 = 1.6

Mean of the squares

E(X²) = 0·0.35 + 1·0.05 + 4·0.25 + 9·0.35 = 4.2

Variance

Var(X) = 4.2 − 1.6² = 1.64
σ = √1.64 = 1.281

Answer

σ = 1.281
6.Probability

Answer: P = 0.1672

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 = 5) = 126·0.45·0.64 = 126·0.01024·0.1296 = 0.1672

Answer

P = 0.1672
7.Normal distribution

Answer: P = 0.9332

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 = 105 s, σ = √(5² + 4²) = 6.403 s

Standardise

z = (95.4 − 105)/6.403 = −1.5

Upper tail

P = 1 − Φ(−1.5) = 1 − 0.06681 = 0.9332

Answer

P = 0.9332
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: E = 0.7624 mmHg

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.

Critical value

df = 11, t = 2.201

Margin of error

E = 2.201·1.2/√12 = 0.7624

Answer

E = 0.7624 mmHg
10.Confidence intervals

Answer: n = 2835

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 = (2.576·31/1.5)² = 2834

Round up

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

n = 2835

Answer

n = 2835
11.Hypothesis tests & confidence intervals

Answer: t = 1.2

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).

Standard error

SE = 5/√36 = 5/6 = 0.8333

Test statistic

t = (201 − 200)/0.8333 = 1.2

Answer

t = 1.2
t0.025; df = 35 = 2.03

|t| < 2.03, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: z = −2.74

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.

Sample proportions

p̂₁ = 90/300 = 0.3
p̂₂ = 103/250 = 0.412
p̂ = 193/550 = 0.3509

Standard error under H₀

SE = √(0.3509·0.6491·(1/300 + 1/250)) = 0.04087

Test statistic

z = (0.3 − 0.412)/0.04087 = −2.74

Answer

z = −2.74

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

13.Linear regression & correlation

Answer: r² = 0.9895

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 = 129/5 = 25.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 160
Syy = Σ(y − ȳ)² = 646.8

Coefficient of determination

r² = 160²/(40·646.8) = 0.9895

Answer

r² = 0.9895

About 98.95% of the variation in y is explained by the line.

14.Linear regression & correlation

Answer: e = 3.5

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 = 147/6 = 24.5
ȳ = 711/6 = 118.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 799.5

Regression line

b = 799.5/157.5 = 5.076
a = 118.5 − 5.076·24.5 = −5.867
ŷ = −5.867 + 5.076·x

Predicted value

ŷ = −5.867 + 5.076·24.5 = 118.5

Residual

e = 122 − 118.5 = 3.5

Answer

e = 3.5

The point lies above the line.

15.Chi-square tests

Answer: χ² = 0.237

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

urban: 60·24/160 = 9, 60·64/160 = 24, 60·72/160 = 27
rural: 100·24/160 = 15, 100·64/160 = 40, 100·72/160 = 45

Chi-square statistic

χ² = (10 − 9)²/9 + (24 − 24)²/24 + (26 − 27)²/27 + (14 − 15)²/15 + (40 − 40)²/40 + (46 − 45)²/45
χ² = 0.237

Answer

χ² = 0.237

df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is not rejected.

16.Chi-square tests

Answer: χ²crit = 9.488

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 = 6 − 1 − 1 = 4

Table

χ²crit(df = 4, α = 0.05) = 9.488

Answer

χ²crit = 9.488

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

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