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

In a content analysis, inter-rater reliability is checked by:

  1. AAsking participants afterwards whether the researcher's coding of their material is correct
  2. BHaving two researchers code the same material and comparing results
  3. CCoding the material twice in one day
  4. DUsing a larger sample of materials
2.Research methods in psychologyHard

A participant takes a drug or a sugar pill but does not know which, while the researcher does. What design is this and what bias remains possible?

  1. ASingle-blind; demand characteristics are impossible
  2. BDouble-blind; no bias is possible
  3. COpen trial; placebo effects are fully controlled
  4. DSingle-blind; experimenter bias is still possible
3.Descriptive statisticsMedium
1015202530354045505560

Find the sample standard deviation s of the data: 33, 41, 23, 60, 19, 10, 24.

n = 7
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 95?

μ = 70σ = 20x = 95
P =
5.ProbabilityMedium
xp(x)00.3520.2540.260.2μ = 2.5

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

x: 0, 2, 4, 6p(x): 0.35, 0.25, 0.2, 0.2
σ =
6.ProbabilityHard
n = 6, p = 0.4: P(X = 5) = ?k0123456

A test has 6 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 = 6p = 0.4k = 5
P =
7.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.8198

Drag z so that the shaded area between −z and z under the standard normal curve is 0.8198. Find z to two decimals.

A = 0.8198
z =
8.Normal distributionHard
10 g0X − Y?

The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
9.Confidence intervalsMedium
n = 12, s = 13.5L = ?Ux̄ = 65E95%

A sample of 12 items has mean 65 and standard deviation s = 13.5. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 12x̄ = 65s = 13.595%t(11) = 2.201
lower limit =
10.Confidence intervalsHard
n = 21Lσ = ?s = 0.695%

The diameters of 21 machined shafts have a sample standard deviation of 0.6 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 20, 2.5% of the χ² distribution lies below 9.591 and 2.5% lies above 34.17.

n = 21s = 0.6 µmχ² = 9.591, 34.17
σ = µm
11.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 127.9 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 127.9s = 15n = 16
t =
12.Hypothesis tests & confidence intervalsHard
σ = 10, n = ?x̄ − Ex̄ + Ex̄E = 1.599%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 1.5? The standard deviation is 10. Find the smallest sample size n.

σ = 10E = 1.599%
n =
13.Linear regression & correlationMedium
xy123224319420514xy24152025

The table gives 5 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 15Σy = 100Σx² = 55Σy² = 2062Σxy = 278
r =
14.Linear regression & correlationHard
xy125222324421516xy242025ŷ = 19

The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 19?

n = 5ŷ = 19
x =
15.Chi-square testsMedium
busbikecarTotalstudents30232780staff303753120Total606080200

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 = 2002 × 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 test of independence is carried out on a contingency table with 2 rows and 2 columns (not counting the totals). Using the table, find the critical value χ²crit at significance level α = 0.01.

α = 0.01
χ²crit =

3Solutions

1.Research methods in psychology

Answer: B · Having two researchers code the same material and comparing results

Why this is the answer

If independent coders produce similar codes for the same data (high agreement or correlation), the coding scheme is reliable.

Answer

B · Having two researchers code the same material and comparing results
2.Research methods in psychology

Answer: D · Single-blind; experimenter bias is still possible

Why this is the answer

In a single-blind design participants do not know their condition, which controls placebo expectations. The researcher's knowledge can still influence results, so double-blind designs are preferred.

Answer

D · Single-blind; experimenter bias is still possible
3.Descriptive statistics

Answer: s = 16.51

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 = 210/7 = 30

Squared deviations

Σ(x − x̄)² = 3² + 11² + (−7)² + 30² + (−11)² + (−20)² + (−6)² = 1636

Standard deviation

s = √(1636/6) = 16.51

Answer

s = 16.51
4.Descriptive statistics

Answer: P = 0.8944

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 = 1.25

Standard normal table

Φ(1.25) = 0.8944
P = 0.8944

Answer

P = 0.8944
5.Probability

Answer: σ = 2.269

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 + 2·0.25 + 4·0.2 + 6·0.2 = 2.5

Mean of the squares

E(X²) = 0·0.35 + 4·0.25 + 16·0.2 + 36·0.2 = 11.4

Variance

Var(X) = 11.4 − 2.5² = 5.15
σ = √5.15 = 2.269

Answer

σ = 2.269
6.Probability

Answer: P = 0.03686

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) = 6·0.45·0.61 = 6·0.01024·0.6 = 0.03686

Answer

P = 0.03686
7.Normal distribution

Answer: z = 1.34

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 + 0.8198)/2 = 0.9099

Read the z-table backwards

Find 0.9099 inside the table and read off z.

z = 1.34

Answer

z = 1.34
8.Normal distribution

Answer: P = 0.8413

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 = 500 − 490 = 10 g, σ = 10 g

Standardise

z = (0 − 10)/10 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413
9.Confidence intervals

Answer: lower limit = 56.42

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 = 11, t = 2.201

Margin of error

E = 2.201·13.5/√12 = 8.578

Interval

65 ± 8.578 → [56.42, 73.58]

Answer

lower limit = 56.42
10.Confidence intervals

Answer: σ = 0.8664 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 20·0.6² = 7.2

Limits for σ²

7.2/34.17 = 0.2107
7.2/9.591 = 0.7507

Limit for σ

σ = √0.7507 = 0.8664 µm

Answer

σ = 0.8664 µm

95% interval for σ: [0.459, 0.8664] µm.

11.Hypothesis tests & confidence intervals

Answer: t = 2.107

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 = 15/√16 = 15/4 = 3.75

Test statistic

t = (127.9 − 120)/3.75 = 2.107

Answer

t = 2.107
t0.025; df = 15 = 2.131

|t| < 2.131, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: n = 295

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).

Solve for n

n = (2.576·10/1.5)² = 294.9

Round up

n = 295

Answer

n = 295
13.Linear regression & correlation

Answer: r = −0.8835

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 = 278 − 15·100/5 = −22
Sxx = 55 − 15²/5 = 10
Syy = 2062 − 100²/5 = 62

Correlation coefficient

r = −22/√(10·62) = −0.8835

Answer

r = −0.8835
14.Linear regression & correlation

Answer: x = 4.368

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 = 108/5 = 21.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −19

Regression line

b = −19/10 = −1.9
a = 21.6 − (−1.9)·3 = 27.3
ŷ = 27.3 − 1.9·x

Solve for x

x = (19 − 27.3)/(−1.9) = 4.368

Answer

x = 4.368
15.Chi-square tests

Answer: χ² = 3.872

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

students: 80·60/200 = 24, 80·60/200 = 24, 80·80/200 = 32
staff: 120·60/200 = 36, 120·60/200 = 36, 120·80/200 = 48

Chi-square statistic

χ² = (30 − 24)²/24 + (23 − 24)²/24 + (27 − 32)²/32 + (30 − 36)²/36 + (37 − 36)²/36 + (53 − 48)²/48
χ² = 3.872

Answer

χ² = 3.872

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

16.Chi-square tests

Answer: χ²crit = 6.635

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)·(2 − 1) = 1

Table

χ²crit(df = 1, α = 0.01) = 6.635

Answer

χ²crit = 6.635

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

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