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 observational research, event sampling means:

  1. AChoosing participants at random
  2. BRecording only events the participants report
  3. CRecording every time a specified behaviour occurs
  4. DRecording behaviour only at fixed time intervals
2.Research methods in psychologyHard

A one-way ANOVA comparing four groups is significant. What should the researcher do next to find which groups differ?

  1. AConclude that all four groups differ from each other
  2. BRun post hoc tests such as Tukey's HSD
  3. CRun a chi-square test on the means
  4. DReport the result as a correlation
3.Descriptive statisticsMedium
253035404550

Find the sample standard deviation s of the data: 52, 23, 51, 37, 32.

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

A quantity is normally distributed with mean 50 and standard deviation 15. What proportion P of values is below 27.5?

μ = 50σ = 15x = 27.5
P =
5.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5P(six) = 1/6 on each roll

A fair die is rolled 5 times. What is the probability P of getting at least one six?

n = 5
P =
6.ProbabilityHard
n = 6, p = 0.7: P(X ≤ 1) = ?k0123456

6 components are tested, and each one fails with probability 0.7, independently. What is the probability P that at most one fails?

n = 6p = 0.7
P =
7.Normal distributionMedium
360 g354.6 gT = X₁ + … + X₉?

The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 9 of them exceeds 354.6 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 9> 354.6 g
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 = 21, s = 12.5LUx̄ = 15.5E = ?95%

A sample of 21 items has mean 15.5 and standard deviation s = 12.5. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.

n = 21x̄ = 15.5s = 12.595%t(20) = 2.086
E =
10.Confidence intervalsHard
n = 25σ = ?Us = 3.495%

The diameters of 25 machined shafts have a sample standard deviation of 3.4 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 24, 2.5% of the χ² distribution lies below 12.401 and 2.5% lies above 39.364.

n = 25s = 3.4 µmχ² = 12.401, 39.364
σ = µm
11.Hypothesis tests & confidence intervalsMedium
−2.1452.145t = ?df = 14

Two teaching methods are compared. Group 1: 6 students, mean score 34, standard deviation 5.5. Group 2: 10 students, mean score 17.8, standard deviation 11.5. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 6x̄₁ = 34s₁ = 5.5n₂ = 10x̄₂ = 17.8s₂ = 11.5
t =
12.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 81 measurements has mean 202.3. The population standard deviation is 10. Find the lower limit L of the 99% confidence interval for the mean.

x̄ = 202.3σ = 10n = 8199%
L =
13.Linear regression & correlationMedium
xy2841261882010251233xy510102030

The table gives 6 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 42Σy = 116Σx² = 364Σy² = 2646Σxy = 978
r =
14.Linear regression & correlationHard
xy3385467459501150xy510404550ŷ = 50

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. For which value x does the line predict ŷ = 50?

n = 5ŷ = 50
x =
15.Chi-square testsMedium
coughno coughTotalsmokers493180non-smokers4159100Total9090180

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 = 1802 × 2
χ² =
16.Chi-square testsHard

A chi-square test of independence is carried out on a contingency table with 2 rows and 11 columns (not counting the totals). How many degrees of freedom df does the test have?

2 × 11
df =

3Solutions

1.Research methods in psychology

Answer: C · Recording every time a specified behaviour occurs

Why this is the answer

Event sampling counts each occurrence of target behaviours. Time sampling records what happens at set intervals, such as every 30 seconds.

Answer

C · Recording every time a specified behaviour occurs
2.Research methods in psychology

Answer: B · Run post hoc tests such as Tukey's HSD

Why this is the answer

A significant F shows that not all means are equal. Post hoc tests or planned contrasts identify which pairs differ while controlling the error rate.

Answer

B · Run post hoc tests such as Tukey's HSD
3.Descriptive statistics

Answer: s = 12.47

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 = 195/5 = 39

Squared deviations

Σ(x − x̄)² = 13² + (−16)² + 12² + (−2)² + (−7)² = 622

Standard deviation

s = √(622/4) = 12.47

Answer

s = 12.47
4.Descriptive statistics

Answer: P = 0.06681

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 = 0.06681

Answer

P = 0.06681
5.Probability

Answer: P = 0.5981

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.

Complement

P(no six) = (5/6)5 = 0.4019
P = 1 − 0.4019 = 0.5981

Answer

P = 0.5981
6.Probability

Answer: P = 0.01094

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) = 0.36 = 0.000729
P(X = 1) = 6·0.7·0.35 = 0.01021

Add them

P = 0.000729 + 0.01021 = 0.01094

Answer

P = 0.01094
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 total

μT = 360 g, σT = 1.2·√9 = 3.6 g

Standardise

z = (354.6 − 360)/3.6 = −1.5

Probability

P = 1 − Φ(−1.5) = 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 = 5.69

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 = 20, t = 2.086

Margin of error

E = 2.086·12.5/√21 = 5.69

Answer

E = 5.69
10.Confidence intervals

Answer: σ = 2.655 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 = 24·3.4² = 277.4

Limits for σ²

277.4/39.364 = 7.048
277.4/12.401 = 22.37

Limit for σ

σ = √7.048 = 2.655 µm

Answer

σ = 2.655 µm

95% interval for σ: [2.655, 4.73] µm.

11.Hypothesis tests & confidence intervals

Answer: t = 3.205

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.

Pooled variance

sp² = (5·5.5² + 9·11.5²)/14 = 95.82
sp = 9.789

Standard error

SE = 9.789·√(1/6 + 1/10) = 5.055

Test statistic

t = (34 − 17.8)/5.055 = 3.205

Answer

t = 3.205

With df = 14, the two-sided 5% critical value is 2.145; |t| is larger, so H₀ is rejected.

12.Hypothesis tests & confidence intervals

Answer: L = 199.4

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 = 2.576·10/√81 = 2.862

Interval limit

L = 202.3 − 2.862 = 199.4

Answer

L = 199.4
13.Linear regression & correlation

Answer: r = 0.9879

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 = 978 − 42·116/6 = 166
Sxx = 364 − 42²/6 = 70
Syy = 2646 − 116²/6 = 403.3

Correlation coefficient

r = 166/√(70·403.3) = 0.9879

Answer

r = 0.9879
14.Linear regression & correlation

Answer: x = 10

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 = 35/5 = 7
ȳ = 229/5 = 45.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 56

Regression line

b = 56/40 = 1.4
a = 45.8 − 1.4·7 = 36
ŷ = 36 + 1.4·x

Solve for x

x = (50 − 36)/1.4 = 10

Answer

x = 10
15.Chi-square tests

Answer: χ² = 7.29

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

smokers: 80·90/180 = 40, 80·90/180 = 40
non-smokers: 100·90/180 = 50, 100·90/180 = 50

Chi-square statistic

χ² = (49 − 40)²/40 + (31 − 40)²/40 + (41 − 50)²/50 + (59 − 50)²/50
χ² = 7.29

Answer

χ² = 7.29

df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is rejected.

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 = (2 − 1)·(11 − 1) = 10

Answer

df = 10

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