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 95% confidence interval for a mean of 50 is 46 to 54. What is the most accurate interpretation?

  1. A95% of the individual participants in the sample scored between 46 and 54 on the measure
  2. BThe result is not significant
  3. CThe procedure used produces intervals that contain the true mean 95% of the time
  4. DThere is a 95% chance the sample mean is 50
2.Research methods in psychologyHard

A researcher wants to show that a new anxiety scale measures something different from depression. Low correlations with a depression scale would support:

  1. AFace validity
  2. BDiscriminant validity
  3. CPredictive validity
  4. DConvergent validity
3.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 13, 47, 16, 52, 24.

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

Exam scores are normally distributed with mean 100 and standard deviation 20. Find the z-score z of a score of 125.

μ = 100σ = 20x = 125
z =
5.ProbabilityMedium
xp(x)00.310.2520.1530.3μ = 1.45

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

x: 0, 1, 2, 3p(x): 0.3, 0.25, 0.15, 0.3
Var(X) =
6.ProbabilityHard
n = 11, p = 0.2: P(X ≤ 1) = ?k01234567891011

11 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?

n = 11p = 0.2
P =
7.Normal distributionMedium
32 min35.1 minX + Y?

X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 35.1 min. Round z to two decimals.

μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 35.1 min
P =
8.Normal distributionHard
0.05 mm0D − d?

Hole diameters D are normal with mean 25.05 mm and standard deviation 0.02 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 = 25.05 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
9.Confidence intervalsMedium
n = 15, sd​ = 7.90L = ?Ud̄ = 5.3E99%

The blood pressure of 15 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 5.3 and standard deviation 7.9. Find the lower limit of the 99% confidence interval for the mean difference.

n = 15d̄ = 5.3sd = 7.999%t(14) = 2.977
lower limit = mmHg
10.Confidence intervalsHard
p ≈ 0.4, n = ?p̂ − Ep̂ + Ep̂E = 0.0190%

A pilot study suggests that about 0.4 of products are defective. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.01?

p ≈ 0.4E = 0.0190%
n =
11.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 41 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 41s = 20n = 16
t =
12.Hypothesis tests & confidence intervalsHard
σ = 12, n = ?x̄ − Ex̄ + Ex̄E = 199%

How large must a sample be so that the 99% 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 = 199%
n =
13.Linear regression & correlationMedium
xy230329438537639750xy246304050

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 6
r =
14.Linear regression & correlationHard
xy146255360468564xy24506070ŷ = 51

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 51?

n = 5ŷ = 51
x =
15.Chi-square testsMedium
busbikecarTotalstudents2414240staff26112360Total502525100

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 = 1002 × 3
χ² =
16.Chi-square testsHard

A chi-square test of independence is carried out on a contingency table with 4 rows and 4 columns (not counting the totals). How many degrees of freedom df does the test have?

4 × 4
df =

3Solutions

1.Research methods in psychology

Answer: C · The procedure used produces intervals that contain the true mean 95% of the time

Why this is the answer

A 95% CI gives a range of plausible values for the population parameter. Narrower intervals, from larger samples, mean more precise estimates.

Answer

C · The procedure used produces intervals that contain the true mean 95% of the time
2.Research methods in psychology

Answer: B · Discriminant validity

Why this is the answer

Discriminant (divergent) validity is shown when a measure correlates weakly with measures of different constructs. Convergent validity is high correlation with measures of the same construct.

Answer

B · Discriminant validity
3.Descriptive statistics

Answer: s = 17.98

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 = 152/5 = 30.4

Squared deviations

Σ(x − x̄)² = (−17.4)² + 16.6² + (−14.4)² + 21.6² + (−6.4)² = 1293

Standard deviation

s = √(1293/4) = 17.98

Answer

s = 17.98
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 = (125 − 100)/20 = 1.25

Answer

z = 1.25
5.Probability

Answer: Var(X) = 1.447

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 + 1·0.25 + 2·0.15 + 3·0.3 = 1.45

Mean of the squares

E(X²) = 0·0.3 + 1·0.25 + 4·0.15 + 9·0.3 = 3.55

Variance

Var(X) = 3.55 − 1.45² = 1.447

Answer

Var(X) = 1.447
6.Probability

Answer: P = 0.3221

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) = 0.811 = 0.0859
P(X = 1) = 11·0.2·0.810 = 0.2362

Add them

P = 0.0859 + 0.2362 = 0.3221

Answer

P = 0.3221
7.Normal distribution

Answer: P = 0.1075

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 sum

μ = 32 min, σ = √(1.5² + 2²) = 2.5 min

Standardise

z = (35.1 − 32)/2.5 = 1.24

Upper tail

P = 1 − Φ(1.24) = 1 − 0.8925 = 0.1075

Answer

P = 0.1075
8.Normal distribution

Answer: P = 0.9875

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 = 25.05 − 25 = 0.05 mm, σ = 0.02236 mm

Standardise

z = (0 − 0.05)/0.02236 = −2.24

Probability

P = 1 − Φ(−2.24) = 0.9875

Answer

P = 0.9875
9.Confidence intervals

Answer: lower limit = −0.7724 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 = 14, t = 2.977

Margin of error

E = 2.977·7.9/√15 = 6.072

Interval

5.3 ± 6.072 → [−0.7724, 11.37]

Answer

lower limit = −0.7724 mmHg
10.Confidence intervals

Answer: n = 6495

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 1.645²·0.4·0.6/0.01² = 6494

Round up

n = 6495

Answer

n = 6495
11.Hypothesis tests & confidence intervals

Answer: t = −1.8

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 = 20/√16 = 20/4 = 5

Test statistic

t = (41 − 50)/5 = −1.8

Answer

t = −1.8
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 = 956

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 = (2.576·12/1)² = 955.6

Round up

n = 956

Answer

n = 956
13.Linear regression & correlation

Answer: r = 0.9104

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 = 27/6 = 4.5
ȳ = 223/6 = 37.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 64.5
Syy = Σ(y − ȳ)² = 286.8

Correlation coefficient

r = 64.5/√(17.5·286.8) = 0.9104

Answer

r = 0.9104
14.Linear regression & correlation

Answer: x = 1.449

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 = 293/5 = 58.6

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 = 58.6 − 4.9·3 = 43.9
ŷ = 43.9 + 4.9·x

Solve for x

x = (51 − 43.9)/4.9 = 1.449

Answer

x = 1.449
15.Chi-square tests

Answer: χ² = 14.67

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: 40·50/100 = 20, 40·25/100 = 10, 40·25/100 = 10
staff: 60·50/100 = 30, 60·25/100 = 15, 60·25/100 = 15

Chi-square statistic

χ² = (24 − 20)²/20 + (14 − 10)²/10 + (2 − 10)²/10 + (26 − 30)²/30 + (11 − 15)²/15 + (23 − 15)²/15
χ² = 14.67

Answer

χ² = 14.67

df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is rejected.

16.Chi-square tests

Answer: df = 9

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 = (4 − 1)·(4 − 1) = 9

Answer

df = 9

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