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. B3.0
  3. C2.0
  4. D0.5
2.Research methods in psychologyHard

A study finds that students who sleep more have higher grades. Which design change would best allow a causal conclusion?

  1. ASurvey more students
  2. BMeasure grades more accurately
  3. CRandomly assign students to different sleep schedules
  4. DAsk students to report their sleep twice and average the two reports
3.Descriptive statisticsMedium
101520253035404550

Find the sample standard deviation s of the data: 37, 52, 10, 37, 18, 39.

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

A quantity is normally distributed with mean 70 and standard deviation 20. What proportion P of values is below 82?

μ = 70σ = 20x = 82
P =
5.ProbabilityMedium
xp(x)00.3520.240.3560.1μ = 2.4

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

x: 0, 2, 4, 6p(x): 0.35, 0.2, 0.35, 0.1
Var(X) =
6.ProbabilityHard
n = 12, p = 0.4: P(X ≤ 1) = ?k0123456789101112

12 components are tested, and each one fails with probability 0.4, independently. What is the probability P that at most one fails?

n = 12p = 0.4
P =
7.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.9439

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9439. Find z to two decimals.

A = 0.9439
z =
8.Normal distributionHard
235240245250255260265gtop 25%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 25% is to the right of it. Find the cut-off x.

μ = 250 gσ = 5 g25%
x = g
9.Confidence intervalsMedium
n = 6, sd​ = 0.80LUd̄ = 2.8E = ?99%

The blood pressure of 6 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 2.8 and standard deviation 0.8. Find the margin of error E of the 99% confidence interval for the mean difference.

n = 6d̄ = 2.8sd = 0.899%t(5) = 4.032
E = mmHg
10.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.02599%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.025?

p = 0.5E = 0.02599%
n =
11.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 100 parts has mean length 84.5 mm, and the line's known standard deviation is 6 mm. Line 2: a sample of 50 parts has mean 82.6 mm, known standard deviation 8 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 100x̄₁ = 84.5 mmσ₁ = 6 mmn₂ = 50x̄₂ = 82.6 mmσ₂ = 8 mm
z =
12.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 80 against H₁: μ > 80 is done at α = 0.05 with a sample of 36; the standard deviation is known to be 15. In reality the mean is 87. Find the probability of a Type II error β in percent.

μ₀ = 80μ = 87σ = 15n = 36α = 0.05
β = %
13.Linear regression & correlationMedium
xy1573655807929105xy56080100

The table gives 5 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 5
r =
14.Linear regression & correlationHard
xy235336435538646xy246354045x = 5.5

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (5.5, 44) is made. Find its residual e = y − ŷ.

n = 5(5.5, 44)
e =
15.Chi-square testsMedium
no defectminormajorTotalday shift24222369night shift15123865Total393461134

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 = 69column total = 61N = 134
E =
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 test of independence is carried out on a contingency table with 2 rows and 4 columns (not counting the totals). 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: C · 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

C · 2.0
2.Research methods in psychology

Answer: C · Randomly assign students to different sleep schedules

Why this is the answer

Only manipulating sleep with random allocation controls third variables and allows causal inference, within ethical limits.

Answer

C · Randomly assign students to different sleep schedules
3.Descriptive statistics

Answer: s = 15.35

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 = 193/6 = 32.17

Squared deviations

Σ(x − x̄)² = 4.833² + 19.83² + (−22.17)² + 4.833² + (−14.17)² + 6.833² = 1179

Standard deviation

s = √(1179/5) = 15.35

Answer

s = 15.35
4.Descriptive statistics

Answer: P = 0.7257

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 = 0.6

Standard normal table

Φ(0.6) = 0.7257
P = 0.7257

Answer

P = 0.7257
5.Probability

Answer: Var(X) = 4.24

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 + 2·0.2 + 4·0.35 + 6·0.1 = 2.4

Mean of the squares

E(X²) = 0·0.35 + 4·0.2 + 16·0.35 + 36·0.1 = 10

Variance

Var(X) = 10 − 2.4² = 4.24

Answer

Var(X) = 4.24
6.Probability

Answer: P = 0.01959

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.

Zero and one failure

P(X = 0) = 0.612 = 0.002177
P(X = 1) = 12·0.4·0.611 = 0.01741

Add them

P = 0.002177 + 0.01741 = 0.01959

Answer

P = 0.01959
7.Normal distribution

Answer: z = 1.91

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 + 0.9439)/2 = 0.9719

Read the z-table backwards

Find 0.9719 inside the table and read off z.

z = 1.91

Answer

z = 1.91
8.Normal distribution

Answer: x = 253.4 g

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) = 0.75

z from the table

z ≈ 0.67

Back to the original units

x = 250 + 0.67·5 = 253.4 g

Answer

x = 253.4 g
9.Confidence intervals

Answer: E = 1.317 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 = 5, t = 4.032

Margin of error

E = 4.032·0.8/√6 = 1.317

Answer

E = 1.317 mmHg
10.Confidence intervals

Answer: n = 2655

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 = 2.576²·0.5·0.5/0.025² = 2654

Round up

n = 2655

Answer

n = 2655
11.Hypothesis tests & confidence intervals

Answer: z = 1.484

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 = √(6²/100 + 8²/50) = √1.64 = 1.281

Test statistic

z = (84.5 − 82.6)/1.281 = 1.484

Answer

z = 1.484

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

12.Hypothesis tests & confidence intervals

Answer: β = 12.4 %

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 = 15/√36 = 2.5
x̄c = 80 + 1.645·2.5 = 84.11

z under the true mean

z = (84.11 − 87)/2.5 = −1.155

Type II error

power = P(Z > −1.155) = 87.6%
β = 100% − 87.6% = 12.4%

Answer

β = 12.4%
13.Linear regression & correlation

Answer: r = 0.9967

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 = 399/5 = 79.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 246
Syy = Σ(y − ȳ)² = 1523

Correlation coefficient

r = 246/√(40·1523) = 0.9967

Answer

r = 0.9967
14.Linear regression & correlation

Answer: e = 2.4

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 = 20/5 = 4
ȳ = 190/5 = 38

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 24

Regression line

b = 24/10 = 2.4
a = 38 − 2.4·4 = 28.4
ŷ = 28.4 + 2.4·x

Predicted value

ŷ = 28.4 + 2.4·5.5 = 41.6

Residual

e = 44 − 41.6 = 2.4

Answer

e = 2.4

The point lies above the line.

15.Chi-square tests

Answer: E = 31.41

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 = 69·61/134 = 31.41

Answer

E = 31.41
16.Chi-square tests

Answer: χ²crit = 7.815

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 = (2 − 1)·(4 − 1) = 3

Table

χ²crit(df = 3, α = 0.05) = 7.815

Answer

χ²crit = 7.815

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

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