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 repeated measures design, order effects can be controlled by:

  1. ACounterbalancing
  2. BUsing a larger sample only
  3. CRandom sampling
  4. DSingle-blind procedures
2.Research methods in psychologyHard

A researcher manipulates mood by showing a sad film, then asks participants to rate how sad they feel before the main task. Why is this rating included?

  1. AAs the main dependent variable of the study
  2. BTo debrief participants about the aims
  3. CAs a manipulation check to confirm the IV worked as intended
  4. DTo counterbalance the order of conditions
3.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 53, 24, 14, 24, 23.

n = 5
s =
4.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 60 and standard deviation 8. What proportion P of values is above 64.8?

μ = 60σ = 8x = 64.8
P =
5.ProbabilityMedium
xp(x)00.310.320.3530.05μ = 1.15

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

x: 0, 1, 2, 3p(x): 0.3, 0.3, 0.35, 0.05
Var(X) =
6.ProbabilityHard
xp(x)00.3510.320.130.25μ

X has the distribution shown, and Y = 2X + 10. Find the standard deviation σY of Y.

x: 0, 1, 2, 3p(x): 0.35, 0.3, 0.1, 0.25Y = 2X + 10
σY =
7.Normal distributionMedium
144 kg146.8 kgT = X₁ + … + X₁₂?

The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 12 of them exceeds 146.8 kg. Round z to two decimals.

μ = 12 kgσ = 0.8 kgn = 12> 146.8 kg
P =
8.Normal distributionHard
557085100115130145pointsbottom 70%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 70% is to the left of it. Find the cut-off x.

μ = 100 pointsσ = 15 points70%
x = points
9.Confidence intervalsMedium
n = 7, s = 6L = ?Ux̄ = 27.5E95%

A sample of 7 items has mean 27.5 and standard deviation s = 6. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 7x̄ = 27.5s = 695%t(6) = 2.447
lower limit =
10.Confidence intervalsHard
n = 30Lσ = ?s = 2.295%

The diameters of 30 machined shafts have a sample standard deviation of 2.2 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 29, 2.5% of the χ² distribution lies below 16.047 and 2.5% lies above 45.722.

n = 30s = 2.2 µmχ² = 16.047, 45.722
σ = µm
11.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 121 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 121s = 8n = 100
t =
12.Hypothesis tests & confidence intervalsHard
σ = 12, n = ?x̄ − Ex̄ + Ex̄E = 195%

How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 12. Find the smallest sample size n.

σ = 12E = 195%
n =
13.Linear regression & correlationMedium
xy14701678188720952211224119xy15202580100120

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 114Σy = 561Σx² = 2236Σy² = 54283Σxy = 11014
r =
14.Linear regression & correlationHard
xy129235339440551xy24304050x = 2.5

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. Predict ŷ for x = 2.5.

n = 5x = 2.5
ŷ =
15.Chi-square testsMedium
lowmediumhighTotalurban10372471rural4122871Total515932142

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

row total = 71column total = 51N = 142
E =
16.Chi-square testsHard

A chi-square goodness-of-fit test checks whether counts in 7 categories follow a Poisson distribution whose mean is estimated from the same data. How many degrees of freedom df does the test have?

k = 7
df =

3Solutions

1.Research methods in psychology

Answer: A · Counterbalancing

Why this is the answer

Counterbalancing distributes practice and fatigue effects evenly across conditions.

Answer

A · Counterbalancing
2.Research methods in psychology

Answer: C · As a manipulation check to confirm the IV worked as intended

Why this is the answer

A manipulation check verifies that the IV produced the intended psychological state. Without it, a null result could mean the manipulation failed.

Answer

C · As a manipulation check to confirm the IV worked as intended
3.Descriptive statistics

Answer: s = 14.81

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 = 138/5 = 27.6

Squared deviations

Σ(x − x̄)² = 25.4² + (−3.6)² + (−13.6)² + (−3.6)² + (−4.6)² = 877.2

Standard deviation

s = √(877.2/4) = 14.81

Answer

s = 14.81
4.Descriptive statistics

Answer: P = 0.2743

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 = 1 − 0.7257 = 0.2743

Answer

P = 0.2743
5.Probability

Answer: Var(X) = 0.8275

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 + 2·0.35 + 3·0.05 = 1.15

Mean of the squares

E(X²) = 0·0.3 + 1·0.3 + 4·0.35 + 9·0.05 = 2.15

Variance

Var(X) = 2.15 − 1.15² = 0.8275

Answer

Var(X) = 0.8275
6.Probability

Answer: σY = 2.356

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.

Spread of X

μ = 1.25, Var(X) = 1.388, σX = 1.178

Linear change

σY = |2|·1.178 = 2.356

Answer

σY = 2.356

The mean does move: E(Y) = 2·1.25 + 10 = 12.5.

7.Normal distribution

Answer: P = 0.1562

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 = 144 kg, σT = 0.8·√12 = 2.771 kg

Standardise

z = (146.8 − 144)/2.771 = 1.01

Probability

P = 1 − Φ(1.01) = 0.1562

Answer

P = 0.1562
8.Normal distribution

Answer: x = 107.8 points

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) = 0.7

z from the table

z ≈ 0.52

Back to the original units

x = 100 + 0.52·15 = 107.8 points

Answer

x = 107.8 points
9.Confidence intervals

Answer: lower limit = 21.95

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 = 6, t = 2.447

Margin of error

E = 2.447·6/√7 = 5.549

Interval

27.5 ± 5.549 → [21.95, 33.05]

Answer

lower limit = 21.95
10.Confidence intervals

Answer: σ = 2.958 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 = 29·2.2² = 140.4

Limits for σ²

140.4/45.722 = 3.07
140.4/16.047 = 8.747

Limit for σ

σ = √8.747 = 2.958 µm

Answer

σ = 2.958 µm

95% interval for σ: [1.752, 2.958] µm.

11.Hypothesis tests & confidence intervals

Answer: t = 1.25

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 = 8/√100 = 8/10 = 0.8

Test statistic

t = (121 − 120)/0.8 = 1.25

Answer

t = 1.25
t0.025; df = 99 = 1.984

|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: n = 554

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 = (1.96·12/1)² = 553.2

Round up

n = 554

Answer

n = 554
13.Linear regression & correlation

Answer: r = 0.992

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 = 11014 − 114·561/6 = 355
Sxx = 2236 − 114²/6 = 70
Syy = 54283 − 561²/6 = 1830

Correlation coefficient

r = 355/√(70·1830) = 0.992

Answer

r = 0.992
14.Linear regression & correlation

Answer: ŷ = 36.35

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 = 194/5 = 38.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 49

Regression line

b = 49/10 = 4.9
a = 38.8 − 4.9·3 = 24.1
ŷ = 24.1 + 4.9·x

Prediction

ŷ = 24.1 + 4.9·2.5 = 36.35

Answer

ŷ = 36.35
15.Chi-square tests

Answer: E = 25.5

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 = 71·51/142 = 25.5

Answer

E = 25.5
16.Chi-square tests

Answer: df = 5

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 = 7 − 1 − 1 = 5

Answer

df = 5

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