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 Cronbach's alpha of 0.88 for a 15-item anxiety scale indicates:

  1. AThe scale predicts future anxiety well
  2. BThe scale has a significant effect on anxiety
  3. CScores are stable over six months
  4. DGood internal consistency of the items
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. AMeasure grades more accurately
  2. BRandomly assign students to different sleep schedules
  3. CSurvey more students
  4. DAsk students to report their sleep twice and average the two reports
3.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 35, 42, 12, 49, 58.

n = 5
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 above 40?

μ = 70σ = 20x = 40
P =
5.ProbabilityMedium
xp(x)00.310.320.0530.2540.1μ = 1.55

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

x: 0, 1, 2, 3, 4p(x): 0.3, 0.3, 0.05, 0.25, 0.1
σ =
6.ProbabilityHard
n = 9, p = 0.25: P(X ≤ 1) = ?k0123456789

9 components are tested, and each one fails with probability 0.25, independently. What is the probability P that at most one fails?

n = 9p = 0.25
P =
7.Normal distributionMedium
1020 kg999.2 kgT = X₁ + … + X₁₂?

The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 12 of them exceeds 999.2 kg. Round z to two decimals.

μ = 85 kgσ = 12 kgn = 12> 999.2 kg
P =
8.Normal distributionHard
μ = 500x = ?90%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 90% of them lie (the 90th percentile).

μ = 500σ = 40P = 90%
x = h
9.Confidence intervalsMedium
n = 9, sd​ = 4.50L = ?Ud̄ = 2.8E95%

The blood pressure of 9 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 2.8 and standard deviation 4.5. Find the lower limit of the 95% confidence interval for the mean difference.

n = 9d̄ = 2.8sd = 4.595%t(8) = 2.306
lower limit = mmHg
10.Confidence intervalsHard
σ = 19, n = ?x̄ − Ex̄ + Ex̄E = 0.595%

The population standard deviation is 19. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 0.5?

σ = 19E = 0.595%
n =
11.Hypothesis tests & confidence intervalsMedium
−2.0932.093t = ?df = 19

20 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 −4.67 and the standard deviation of the differences is 11. Find the paired test statistic t for H₀: μd = 0.

n = 20d̄ = −4.67sd = 11
t =
12.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 100 measurements has mean 201.7. The population standard deviation is 8. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 201.7σ = 8n = 10095%
L =
13.Linear regression & correlationMedium
xy128221319416518xy242025

The table gives 5 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
14.Linear regression & correlationHard
xy122223319418515616xy2461520x = 2.5

The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 2.5.

n = 6x = 2.5
ŷ =
15.Chi-square testsMedium
no defectminormajorTotalday shift37141566night shift21402586Total585440152

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

row total = 86column total = 54N = 152
E =
16.Chi-square testsHard

A chi-square goodness-of-fit test compares observed counts in 11 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?

k = 11
df =

3Solutions

1.Research methods in psychology

Answer: D · Good internal consistency of the items

Why this is the answer

Cronbach's alpha estimates internal consistency; values around 0.7 or above are usually considered acceptable. It does not by itself show validity.

Answer

D · Good internal consistency of the items
2.Research methods in psychology

Answer: B · 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

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

Answer: s = 17.43

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 = 196/5 = 39.2

Squared deviations

Σ(x − x̄)² = (−4.2)² + 2.8² + (−27.2)² + 9.8² + 18.8² = 1215

Standard deviation

s = √(1215/4) = 17.43

Answer

s = 17.43
4.Descriptive statistics

Answer: P = 0.9332

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 = −1.5

Standard normal table

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

Answer

P = 0.9332
5.Probability

Answer: σ = 1.396

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 = 0·0.3 + 1·0.3 + 2·0.05 + 3·0.25 + 4·0.1 = 1.55

Mean of the squares

E(X²) = 0·0.3 + 1·0.3 + 4·0.05 + 9·0.25 + 16·0.1 = 4.35

Variance

Var(X) = 4.35 − 1.55² = 1.948
σ = √1.948 = 1.396

Answer

σ = 1.396
6.Probability

Answer: P = 0.3003

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) = 0.759 = 0.07508
P(X = 1) = 9·0.25·0.758 = 0.2253

Add them

P = 0.07508 + 0.2253 = 0.3003

Answer

P = 0.3003
7.Normal distribution

Answer: P = 0.6915

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 total

μT = 1020 kg, σT = 12·√12 = 41.57 kg

Standardise

z = (999.2 − 1020)/41.57 = −0.5

Probability

P = 1 − Φ(−0.5) = 0.6915

Answer

P = 0.6915
8.Normal distribution

Answer: x = 551.3 h

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) = 0.9 → z = 1.282

Back to x

x = 500 + 1.282·40 = 551.26 h

Answer

x = 551.26 h
9.Confidence intervals

Answer: lower limit = −0.659 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 = 8, t = 2.306

Margin of error

E = 2.306·4.5/√9 = 3.459

Interval

2.8 ± 3.459 → [−0.659, 6.259]

Answer

lower limit = −0.659 mmHg
10.Confidence intervals

Answer: n = 5548

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 = (1.96·19/0.5)² = 5547

Round up

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

n = 5548

Answer

n = 5548
11.Hypothesis tests & confidence intervals

Answer: t = −1.899

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 = 11/√20 = 2.46

Test statistic

t = −4.67/2.46 = −1.899

Answer

t = −1.899

With df = 19, the two-sided 5% critical value is 2.093; |t| is smaller, so H₀ is not rejected.

12.Hypothesis tests & confidence intervals

Answer: L = 200.1

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 = 1.96·8/√100 = 1.568

Interval limit

L = 201.7 − 1.568 = 200.1

Answer

L = 200.1
13.Linear regression & correlation

Answer: r² = 0.7336

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 = 15/5 = 3
ȳ = 102/5 = 20.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −25
Syy = Σ(y − ȳ)² = 85.2

Coefficient of determination

r² = −25²/(10·85.2) = 0.7336

Answer

r² = 0.7336

About 73.36% of the variation in y is explained by the line.

14.Linear regression & correlation

Answer: ŷ = 20.4

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 = 21/6 = 3.5
ȳ = 113/6 = 18.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −27.5

Regression line

b = −27.5/17.5 = −1.571
a = 18.83 − (−1.571)·3.5 = 24.33
ŷ = 24.33 − 1.571·x

Prediction

ŷ = 24.33 + (−1.571)·2.5 = 20.4

Answer

ŷ = 20.4
15.Chi-square tests

Answer: E = 30.55

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 = 86·54/152 = 30.55

Answer

E = 30.55
16.Chi-square tests

Answer: df = 10

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 = 11 − 1 = 10

Answer

df = 10

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