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

Which test is suitable for comparing the means of two independent groups with interval data that are roughly normal?

  1. AWilcoxon signed-rank test
  2. BChi-square test
  3. CPearson correlation
  4. DIndependent samples t-test
2.Research methods in psychologyHard

A meta-analysis combines 40 studies of a therapy. Why might its overall effect size still be too large?

  1. APublication bias means unpublished null studies may be missing
  2. BCombining studies always removes all bias
  3. CMeta-analyses cannot include randomised trials
  4. DMeta-analyses always give smaller effects than the single studies they include
3.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 53, 54, 14, 54, 17.

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

Exam scores are normally distributed with mean 60 and standard deviation 8. Find the z-score z of a score of 70.

μ = 60σ = 8x = 70
z =
5.ProbabilityMedium
xp(x)00.320.340.1560.25μ = 2.7

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

x: 0, 2, 4, 6p(x): 0.3, 0.3, 0.15, 0.25
σ =
6.ProbabilityHard
n = 7, p = 0.6: P(X = 1) = ?k01234567

A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.6, independently. What is the probability P of exactly 1 correct answers?

n = 7p = 0.6k = 1
P =
7.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.7939

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.7939. Find z to two decimals.

A = 0.7939
z =
8.Normal distributionHard
μ = 100x = ?25%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 25% of them lie (the 25th percentile).

μ = 100σ = 15P = 25%
x = points
9.Confidence intervalsMedium
x̄E₁ = 10.790%, n = 100x̄E₂ = ?95%, n = 400

A 90% confidence interval for a mean, based on 100 observations and a known σ, has a margin of error of 10.7. What is the margin of error E₂ of a 95% interval based on 400 observations?

90%n₁ = 100E₁ = 10.795%n₂ = 400
E₂ =
10.Confidence intervalsHard
n = 20σ = ?Us = 11.895%

The diameters of 20 machined shafts have a sample standard deviation of 11.8 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 19, 2.5% of the χ² distribution lies below 8.907 and 2.5% lies above 32.852.

n = 20s = 11.8 µmχ² = 8.907, 32.852
σ = µm
11.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 103.4 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 103.4s = 8n = 25
t =
12.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 81 measurements has mean 48.1. The population standard deviation is 10. Find the lower limit L of the 99% confidence interval for the mean.

x̄ = 48.1σ = 10n = 8199%
L =
13.Linear regression & correlationMedium
xy332440538649752858xy46830405060

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

n = 6Σx = 33Σy = 269Σx² = 199Σy² = 12537Σxy = 1568
r =
14.Linear regression & correlationHard
xy147716861898201112211924127xy15202580100120x = 19

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (19, 106) is made. Find its residual e = y − ŷ.

n = 6(19, 106)
e =
15.Chi-square testsMedium
no defectminormajorTotalday shift9272864night shift27302077Total365748141

The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “night shift, no defect”.

row total = 77column total = 36N = 141
E =
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 3 columns (not counting the totals). Using the table, find the critical value χ²crit at significance level α = 0.05.

α = 0.05
χ²crit =

3Solutions

1.Research methods in psychology

Answer: D · Independent samples t-test

Why this is the answer

The independent t-test compares two group means. Chi-square is for frequencies; Wilcoxon is for related samples with non-parametric data.

Answer

D · Independent samples t-test
2.Research methods in psychology

Answer: A · Publication bias means unpublished null studies may be missing

Why this is the answer

If null results are less often published, the pooled effect is inflated. Funnel plots and searches for unpublished data help detect and reduce this.

Answer

A · Publication bias means unpublished null studies may be missing
3.Descriptive statistics

Answer: s = 20.94

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 = 192/5 = 38.4

Squared deviations

Σ(x − x̄)² = 14.6² + 15.6² + (−24.4)² + 15.6² + (−21.4)² = 1753

Standard deviation

s = √(1753/4) = 20.94

Answer

s = 20.94
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 = (70 − 60)/8 = 1.25

Answer

z = 1.25
5.Probability

Answer: σ = 2.304

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 + 2·0.3 + 4·0.15 + 6·0.25 = 2.7

Mean of the squares

E(X²) = 0·0.3 + 4·0.3 + 16·0.15 + 36·0.25 = 12.6

Variance

Var(X) = 12.6 − 2.7² = 5.31
σ = √5.31 = 2.304

Answer

σ = 2.304
6.Probability

Answer: P = 0.0172

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 = 1) = 7·0.61·0.46 = 7·0.6·0.004096 = 0.0172

Answer

P = 0.0172
7.Normal distribution

Answer: z = −0.82

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 − 0.7939 = 0.2061

Read the z-table backwards

Find 0.2061 inside the table and read off z.

z = −0.82

Answer

z = −0.82
8.Normal distribution

Answer: x = 89.88 points

Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.

z for the percentile

Φ(z) = 0.25 → z = −0.6745

Back to x

x = 100 + −0.6745·15 = 89.88 points

Answer

x = 89.88 points
9.Confidence intervals

Answer: E₂ = 6.374

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 = (1.96/1.645)·√(100/400) = 0.5957

New margin

E₂ = 10.7·0.5957 = 6.374

Answer

E₂ = 6.374
10.Confidence intervals

Answer: σ = 8.974 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 = 19·11.8² = 2646

Limits for σ²

2646/32.852 = 80.53
2646/8.907 = 297

Limit for σ

σ = √80.53 = 8.974 µm

Answer

σ = 8.974 µm

95% interval for σ: [8.974, 17.23] µm.

11.Hypothesis tests & confidence intervals

Answer: t = 2.125

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/√25 = 8/5 = 1.6

Test statistic

t = (103.4 − 100)/1.6 = 2.125

Answer

t = 2.125
t0.025; df = 24 = 2.064

|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: L = 45.24

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 = 48.1 − 2.862 = 45.24

Answer

L = 45.24
13.Linear regression & correlation

Answer: r = 0.9688

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 = 1568 − 33·269/6 = 88.5
Sxx = 199 − 33²/6 = 17.5
Syy = 12537 − 269²/6 = 476.8

Correlation coefficient

r = 88.5/√(17.5·476.8) = 0.9688

Answer

r = 0.9688
14.Linear regression & correlation

Answer: e = 3

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 = 114/6 = 19
ȳ = 618/6 = 103

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 362

Regression line

b = 362/70 = 5.171
a = 103 − 5.171·19 = 4.743
ŷ = 4.743 + 5.171·x

Predicted value

ŷ = 4.743 + 5.171·19 = 103

Residual

e = 106 − 103 = 3

Answer

e = 3

The point lies above the line.

15.Chi-square tests

Answer: E = 19.66

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 = 77·36/141 = 19.66

Answer

E = 19.66
16.Chi-square tests

Answer: χ²crit = 5.991

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)·(3 − 1) = 2

Table

χ²crit(df = 2, α = 0.05) = 5.991

Answer

χ²crit = 5.991

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

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