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. AGood internal consistency of the items
  2. BThe scale has a significant effect on anxiety
  3. CScores are stable over six months
  4. DThe scale predicts future anxiety well
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. CReport the result as a correlation
  4. DRun a chi-square test on the means
3.Descriptive statisticsMedium
36384042444648

Find the sample standard deviation s of the data: 48, 46, 38, 36, 35.

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

Exam scores are normally distributed with mean 170 and standard deviation 8. Find the z-score z of a score of 180.

μ = 170σ = 8x = 180
z =
5.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6?7?8P(six) = 1/6 on each roll

A fair die is rolled 8 times. What is the probability P of getting at least one six?

n = 8
P =
6.ProbabilityHard
€p(x)−50.69200.32450.01

A game costs €5 to play. You win €25 with probability 0.3 and €250 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €5€25: p = 0.3€250: p = 0.01
E(G) = €
7.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.6808

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.6808. Find z to two decimals.

A = 0.6808
z =
8.Normal distributionHard
0.02 mm0D − d?

Hole diameters D are normal with mean 25.02 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.02 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
9.Confidence intervalsMedium
x̄E₁ = 899%, n = 64x̄E₂ = ?99%, n = 256

A 99% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 8. What is the margin of error E₂ if the same study uses 256 observations instead?

99%n₁ = 64E₁ = 8n₂ = 256
E₂ =
10.Confidence intervalsHard
μ₁ − μ₂, t(10) = 3.1690LU = ?x̄₁ − x̄₂E99%

Two fertilisers are tested. Plot group 1: 7 plots, mean yield 31 kg, standard deviation 9 kg. Plot group 2: 5 plots, mean 29.5 kg, standard deviation 3.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 7x̄₁ = 31s₁ = 9n₂ = 5x̄₂ = 29.5s₂ = 3.599%t(10) = 3.169
upper limit = kg
11.Hypothesis tests & confidence intervalsMedium
t −2.0302.03

A sample of 36 students has a mean score of 496.5 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 496.5s = 15n = 36
t =
12.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 80 against H₁: μ < 80 is done at α = 0.01 with a sample of 64; the standard deviation is known to be 16. In reality the mean is 72.2. Find the probability of a Type II error β in percent.

μ₀ = 80μ = 72.2σ = 16n = 64α = 0.01
β = %
13.Linear regression & correlationMedium
xy3685777829991110813114xy51080100

The table gives 6 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 48Σy = 548Σx² = 454Σy² = 51738Σxy = 4724
r =
14.Linear regression & correlationHard
xy21241762483510401250xy5102040x = 9

The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (9, 34) is made. Find its residual e = y − ŷ.

n = 6(9, 34)
e =
15.Chi-square testsMedium
no defectminormajorTotalday shift342845107night shift20182765Total544672172

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 = 65column total = 46N = 172
E =
16.Chi-square testsHard

A chi-square goodness-of-fit test compares observed counts in 7 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?

k = 7
df =

3Solutions

1.Research methods in psychology

Answer: A · 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

A · Good internal consistency of the items
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 = 5.983

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 = 203/5 = 40.6

Squared deviations

Σ(x − x̄)² = 7.4² + 5.4² + (−2.6)² + (−4.6)² + (−5.6)² = 143.2

Standard deviation

s = √(143.2/4) = 5.983

Answer

s = 5.983
4.Descriptive statistics

Answer: z = 1.25

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 = (180 − 170)/8 = 1.25

Answer

z = 1.25
5.Probability

Answer: P = 0.7674

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)8 = 0.2326
P = 1 − 0.2326 = 0.7674

Answer

P = 0.7674
6.Probability

Answer: E(G) = 5 €

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.

Net outcomes

G: 20 (p = 0.3), 245 (p = 0.01), −5 (p = 0.69)

Expected value

E(G) = 20·0.3 + 245·0.01 − 5·0.69 = 5 €

Answer

E(G) = 5 €

A positive value means you win money on average.

7.Normal distribution

Answer: z = −0.47

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 − 0.6808 = 0.3192

Read the z-table backwards

Find 0.3192 inside the table and read off z.

z = −0.47

Answer

z = −0.47
8.Normal distribution

Answer: P = 0.8665

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 = 25.02 − 25 = 0.02 mm, σ = 0.01803 mm

Standardise

z = (0 − 0.02)/0.01803 = −1.11

Probability

P = 1 − Φ(−1.11) = 0.8665

Answer

P = 0.8665
9.Confidence intervals

Answer: E₂ = 4

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.

Ratio of margins

E₂/E₁ = (2.576/2.576)·√(64/256) = 0.5

New margin

E₂ = 8·0.5 = 4

Answer

E₂ = 4
10.Confidence intervals

Answer: upper limit = 15.07 kg

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.

Pooled standard deviation

sp² = (6·9² + 4·3.5²)/10 = 53.5
sp = 7.314

Margin of error

E = 3.169·7.314·√(1/7 + 1/5) = 13.57

Interval

1.5 ± 13.57 → [−12.07, 15.07]

Answer

upper limit = 15.07 kg
11.Hypothesis tests & confidence intervals

Answer: t = −1.4

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).

Standard error

SE = 15/√36 = 15/6 = 2.5

Test statistic

t = (496.5 − 500)/2.5 = −1.4

Answer

t = −1.4
t0.025; df = 35 = 2.03

|t| < 2.03, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: β = 5.774 %

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 = 16/√64 = 2
x̄c = 80 − 2.326·2 = 75.35

z under the true mean

z = (75.35 − 72.2)/2 = 1.574

Type II error

power = P(Z < 1.574) = 94.23%
β = 100% − 94.23% = 5.774%

Answer

β = 5.774%
13.Linear regression & correlation

Answer: r = 0.9893

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 = 4724 − 48·548/6 = 340
Sxx = 454 − 48²/6 = 70
Syy = 51738 − 548²/6 = 1687

Correlation coefficient

r = 340/√(70·1687) = 0.9893

Answer

r = 0.9893
14.Linear regression & correlation

Answer: e = −3.381

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 = 42/6 = 7
ȳ = 178/6 = 29.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 270

Regression line

b = 270/70 = 3.857
a = 29.67 − 3.857·7 = 2.667
ŷ = 2.667 + 3.857·x

Predicted value

ŷ = 2.667 + 3.857·9 = 37.38

Residual

e = 34 − 37.38 = −3.381

Answer

e = −3.381

The point lies below the line.

15.Chi-square tests

Answer: E = 17.38

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 = 65·46/172 = 17.38

Answer

E = 17.38
16.Chi-square tests

Answer: df = 6

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 = 6

Answer

df = 6

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