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

Demand characteristics are:

  1. ACharacteristics of the sample
  2. BStatistical requirements for a test
  3. CCues that make participants guess the aim and change their behaviour
  4. DThe formal demands an ethics committee makes before a study can begin
2.Research methods in psychologyHard

A researcher conducts five t-tests and wants to keep the overall Type I error rate at 0.05 using the Bonferroni correction. What significance level should each test use?

  1. A0.25
  2. B0.025
  3. C0.01
  4. D0.05
3.Descriptive statisticsMedium
283032343638404244

Find the sample standard deviation s of the data: 29, 28, 44, 43, 39, 44.

n = 6
s =
4.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 500 and standard deviation 25. What proportion P of values is below 515?

μ = 500σ = 25x = 515
P =
5.ProbabilityMedium
xp(x)00.3520.2540.0560.35μ = 2.8

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

x: 0, 2, 4, 6p(x): 0.35, 0.25, 0.05, 0.35
Var(X) =
6.ProbabilityHard
xp(x)00.310.3520.130.25μ

X has the distribution shown, and Y = 3X + 1. Find the standard deviation σY of Y.

x: 0, 1, 2, 3p(x): 0.3, 0.35, 0.1, 0.25Y = 3X + 1
σY =
7.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.9774

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9774. Find z to two decimals.

A = 0.9774
z =
8.Normal distributionHard
μ = 65x = ?25%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 25% of them lie (the 25th percentile).

μ = 65σ = 10P = 25%
x = points
9.Confidence intervalsMedium
n = 9, s = 19LUx̄ = 32.5E = ?95%

A sample of 9 items has mean 32.5 and standard deviation s = 19. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.

n = 9x̄ = 32.5s = 1995%t(8) = 2.306
E =
10.Confidence intervalsHard
n = 21Lσ = ?s = 2.395%

The diameters of 21 machined shafts have a sample standard deviation of 2.3 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 20, 2.5% of the χ² distribution lies below 9.591 and 2.5% lies above 34.17.

n = 21s = 2.3 µmχ² = 9.591, 34.17
σ = µm
11.Hypothesis tests & confidence intervalsMedium
−1.541.54p = ?

The mean of a population is claimed to be 80; the population standard deviation is known to be 36. A random sample of 144 observations has a mean of 84.62. Test H₀: μ = 80 against H₁: μ ≠ 80. Find the p-value p.

μ₀ = 80σ = 36n = 144x̄ = 84.62H₁: μ ≠ μ₀
p =
12.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 100 against H₁: μ < 100 is done at α = 0.05 with a sample of 25; the standard deviation is known to be 2.5. In reality the mean is 98.35. Find the power 1 − β of the test in percent.

μ₀ = 100μ = 98.35σ = 2.5n = 25α = 0.05
1 − β = %
13.Linear regression & correlationMedium
xy2434516598711075xy510506070

The table gives 5 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
14.Linear regression & correlationHard
xy17661973216723752589xy2025708090x = 18

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 18.

n = 5x = 18
ŷ =
15.Chi-square testsMedium
yesnoTotalmen172340women35540Total522880

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 = 802 × 2
χ² =
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 goodness-of-fit test compares observed counts in 7 categories with fully specified expected proportions (no parameters are estimated from the data). Using the table, find the critical value χ²crit at significance level α = 0.01.

α = 0.01
χ²crit =

3Solutions

1.Research methods in psychology

Answer: C · Cues that make participants guess the aim and change their behaviour

Why this is the answer

When participants work out the hypothesis they may act to confirm or sabotage it, threatening validity. Deception and blind procedures reduce this.

Answer

C · Cues that make participants guess the aim and change their behaviour
2.Research methods in psychology

Answer: C · 0.01

Why this is the answer

The Bonferroni correction divides α by the number of comparisons: 0.05 / 5 = 0.01. It is simple but conservative, which increases the risk of Type II errors.

Answer

C · 0.01
3.Descriptive statistics

Answer: s = 7.468

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 = 227/6 = 37.83

Squared deviations

Σ(x − x̄)² = (−8.833)² + (−9.833)² + 6.167² + 5.167² + 1.167² + 6.167² = 278.8

Standard deviation

s = √(278.8/5) = 7.468

Answer

s = 7.468
4.Descriptive statistics

Answer: P = 0.7257

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 = 0.7257

Answer

P = 0.7257
5.Probability

Answer: Var(X) = 6.56

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 = 0·0.35 + 2·0.25 + 4·0.05 + 6·0.35 = 2.8

Mean of the squares

E(X²) = 0·0.35 + 4·0.25 + 16·0.05 + 36·0.35 = 14.4

Variance

Var(X) = 14.4 − 2.8² = 6.56

Answer

Var(X) = 6.56
6.Probability

Answer: σY = 3.434

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 = 1.3, Var(X) = 1.31, σX = 1.145

Linear change

σY = |3|·1.145 = 3.434

Answer

σY = 3.434

The mean does move: E(Y) = 3·1.3 + 1 = 4.9.

7.Normal distribution

Answer: z = 2.28

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 + 0.9774)/2 = 0.9887

Read the z-table backwards

Find 0.9887 inside the table and read off z.

z = 2.28

Answer

z = 2.28
8.Normal distribution

Answer: x = 58.26 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 = 65 + −0.6745·10 = 58.26 points

Answer

x = 58.26 points
9.Confidence intervals

Answer: E = 14.6

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 = 8, t = 2.306

Margin of error

E = 2.306·19/√9 = 14.6

Answer

E = 14.6
10.Confidence intervals

Answer: σ = 3.321 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 = 20·2.3² = 105.8

Limits for σ²

105.8/34.17 = 3.096
105.8/9.591 = 11.03

Limit for σ

σ = √11.03 = 3.321 µm

Answer

σ = 3.321 µm

95% interval for σ: [1.76, 3.321] µm.

11.Hypothesis tests & confidence intervals

Answer: p = 0.1236

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.

Test statistic

SE = σ/√n = 36/√144 = 3
z = (84.62 − 80)/3 = 1.54

Tail area

Φ(1.54) = 0.9382
1 − Φ(1.54) = 0.0618

p-value

p = 2·0.0618 = 0.1236

Answer

p = 0.1236

p > 0.05, so H₀ is not rejected at the 5% level.

12.Hypothesis tests & confidence intervals

Answer: 1 − β = 95.1 %

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 = 2.5/√25 = 0.5
x̄c = 100 − 1.645·0.5 = 99.18

z under the true mean

z = (99.18 − 98.35)/0.5 = 1.655

Power

power = P(Z < 1.655) = 95.1%
1 − β = 95.1%

Answer

1 − β = 95.1%
13.Linear regression & correlation

Answer: r² = 0.9844

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 = 30/5 = 6
ȳ = 299/5 = 59.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 168
Syy = Σ(y − ȳ)² = 716.8

Coefficient of determination

r² = 168²/(40·716.8) = 0.9844

Answer

r² = 0.9844

About 98.44% of the variation in y is explained by the line.

14.Linear regression & correlation

Answer: ŷ = 66.8

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 = 105/5 = 21
ȳ = 370/5 = 74

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 96

Regression line

b = 96/40 = 2.4
a = 74 − 2.4·21 = 23.6
ŷ = 23.6 + 2.4·x

Prediction

ŷ = 23.6 + 2.4·18 = 66.8

Answer

ŷ = 66.8
15.Chi-square tests

Answer: χ² = 17.8

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

men: 40·52/80 = 26, 40·28/80 = 14
women: 40·52/80 = 26, 40·28/80 = 14

Chi-square statistic

χ² = (17 − 26)²/26 + (23 − 14)²/14 + (35 − 26)²/26 + (5 − 14)²/14
χ² = 17.8

Answer

χ² = 17.8

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

16.Chi-square tests

Answer: χ²crit = 16.81

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

Table

χ²crit(df = 6, α = 0.01) = 16.812

Answer

χ²crit = 16.812

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

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