Topic 5: Hypothesis Testing

Question 1

What is the simple null vs composite null, and the composite one sided vs 2 sided alternative (2,2)

Answer 1

-H₀: θ = θ₀ is the simple null
-H₀: θ ≤ or ≥ θ₀ is the composite null

-H₁: θ ≠ θ₀ is the composite 2 sided alternative
-H₁: θ < or > θ₀ is the composite 1 sided alternative

Question 2

What is a type one error (2)

Answer 2

-A type 1 error is when you reject a true H₀ (finding innocent as guilty)
-P(T₁ error) = a = significance level (how often you are willing to incorrectly reject the null hypothesis

Question 3

What is a type 2 error (2)

Answer 3

-A type 2 error is the probability of accepting a false H₀ (finding guilty as innocent)
-P(T₂ error) = β

Question 4

What is the power of a test (2)

Answer 4

-The power of a test is the probability of correctly rejecting a false H₀
-This is 1 - β

Question 5

What is the p value (1)

Answer 5

-The probability of getting the result you do with the null hypothesis

Question 6

What is the decision rule (2)

Answer 6

-If the p value < a, you reject H₀, but you don’t reject if the p value > a
-The higher a is, the more liberal you are

Question 7

What are the 5 steps in a hypothesis test (5)

Answer 7

-Specify the null
-Specify the alternative
-Choose the significance level and corresponding critical region
-Calculate the test order null hypothesis
-Compare the test statistic with the critical value

Question 8

What is the normal distribution of a sample n with known variance, and how do we standardise this (2)

Answer 8

-x̄ ~ N(µ, σ²/n)
-Z = (x̄ - µ)/√σ²/n

Question 9

How would you set up the single mean hypothesis test if the distribution is not normal, n > 25 and with a known variance (3)

Answer 9

-Suppose there’s n observations from a non-normal distribution, with unknown mean µ and known variance σ²
-By the CLT (if n > 25) we can say x̄ ~^a N(µ, σ²/n)
-We then do the 5 step procedure

Question 10

How would you set up a single mean hypothesis test with an unknown distribution and unknown variance, and n > 25 (3)

Answer 10

-Suppose there’s n observations from a non-normal distribution, with unknown mean µ and unknown variance σ², with sample variance estimator S²
-By the CLT (if n > 25) we can say x̄ ~^a N(µ, S²/n)
-We then do the 5 step procedure

Question 11

How do we set up a single mean hypothesis test with a bernoulli distribution and CLT (4)

Answer 11

-Take a bernoulli distribution, where P(X = 1) = π, P(X = 0) = 1 - π
-E(X) = π, V(X) = π(1-π), so E(x̄) = π, V(x̄) = π(1-π)/n (unknown mean and population variance)
-By the CLT with H₀: µ = π⁰, x̄ ~^a N(π⁰, π⁰(1 - π⁰)/n)
-Now, the 5 step procedure is the same as before

Question 12

How do we set up a single mean hypothesis test with a normal distribution and unknown variance (5)

Answer 12

-Suppose n observations from a normal distribution, with unknown σ² and sample variance estimator S²
-E(x̄) = µ, V(x̄) = σ²/n
-If we then divide the normal standardisation formula with the Chi-squared formula ((n-1)S²/σ²(n-1), this is equal to (X^_ - µ)/(S²/√n)
-This approximates to a standard normal distribution/(chi squared with n-1 DoF/n-1) which is the t-ratio
-Hence, we run the five step procedure but with a t_n-1 distribution

Question 13

What is the standardisation process for a t distribution (1)

Answer 13

-(x̄ - µ)/(√S²/n) ~ t_n-1

Question 14

How can we apply the CLT to a t distribution (1)

Answer 14

-For n > 25, the t distribution is approximated by a standard normal distribution

Question 15

What are the expected value’s and variance when testing for the difference in means (2)

Answer 15

-E(X̄₁ - X̄₂) = µ₁ - µ₂
-V(X̄₁ - X̄₂) = σ²₁/n₁ + σ²₂/n₂

Question 16

What is the 5-step procedure for testing the difference of means with a normal underlying distribution and known population variances (1,5)

Answer 16

-X̄₁ - X̄₂ ~ N(µ₁ - µ₂,σ²₁/n₁ + σ²₂/n₂)

1: H₀: µ₁ - µ₂ = D, so X̄₁ - X̄₂ ~ N(D,σ²₁/n₁ + σ²₂/n₂)
2: H₁: µ₁ - µ₂ ≠ D
3: The critical values are ± z^a/2
4: The test statistic is Z = ((x̄₁ - x̄₂) - D)/√(σ²₁/n₁ + σ²₂/n₂)
5: Decision rule, compare test statistic and critical values

Question 17

How do we set up an equality of means hypothesis test with a non-normal underlying distribution, known population variance and n>25 (2)

Answer 17

-By the CLT, we can get X̄₁ - X̄₂ ~^aN(µ₁ - µ₂, σ²₁/n₁ + σ²₂/n₂)
-Then, the 5 step procedure is as usual

Question 18

How do we set up an equality of means hypothesis test with a non-normal underlying distribution, unknown population variance and n>25 (2)

Answer 18

-By the CLT, we can get X̄₁ - X̄₂ ~^aN(µ₁ - µ₂, S²₁/n₁ + S²₂/n₂)
-Then, the 5 step procedure is as usual

Question 19

How would we do an equality of means test coming from a bernoulli distribution where n > 25 (1,5)

Answer 19

-If X₁ and X₂ come from a bernoulli distribution, then x̄₁ - x̄₂ ~^a N(µ₁ - µ₂, σ²₁/n₁ + σ²₂/n₂)

The 5 step procedure is hence:
-H₀: π₁ - π₂ = 0 and so under H₀ x̄₁ - x̄₂ ~^a N(0, π₀(1-π₀)/n₁ + π₀(1-π₀)/n₂)
-H₁: π₁ - π₂ ≠ 0
-The critical values are +-z_a/2
-The test statistic is z = (x̄₁ - x̄₂ - 0)/√(p₀(1-p₀)(1/n₁ + 1/n₂), where p₀ = (n₁x̄₁ + n₂x̄₂)/(n₁ + n₂)
-Make your decision rule

Question 20

How would we do an equality of means test if the underlying distribution is normal, and the population variances are unknown, but equal (5,1)

Answer 20

-H₀: µ₁ - µ₂ = D₀
-H₁: µ₁ - µ₂ ≠ D₀
-The critical values are from a t-distribution, denoted ±t_a/2, (n₁+n₂-2)
-The test statistic is t = ((x̄₁ - x̄₂) - D₀)/√s²₀(1/n₁ + 1/n₂), where s²₀ = ((n₁ - 1)s²₁ + (n₂ - 1)s²₂)/(n₁ + n₂ - 2)
-Make the decision rule

-Note to use this test you have to prove in a prior test that the population variances aren’t different

Question 21

How would we do an equality of means test if the underlying distribution is normal, and the population variances are unknown, but not equal (5,1)

Answer 21

-H₀: µ₁ - µ₂ = D₀
-H₁: µ₁ - µ₂ ≠ D₀
-The critical values are from a t-distribution, denoted ±t_a/2, DoF, where DoF = [(s²₁/n₁) + (s²₂/n₂)]²/(((s²₁/n₁)²/(n₁-1)) + ((s²₂/n₂)²/(n₂-1)))
-The test statistic is t = ((x̄₁ - x̄₂) - D₀)/√(s²₁/n₁ + s²₂/n₂
-Make the decision rule

-Note to use this test, you have had to have proven the population variances aren’t the same

Question 22

How do we test for a single variance (1,5)

Answer 22

-To formulate a hypothesis for testing the population variance, the distribution of the random variable X must be normally distributed

-H₀: σ² = σ²₀
-H₁: σ² > σ²₀
-The critical value is from a χ² distribution, denoted χ²_{a, n-1}
-The test statistic is (n-1)s²/σ²₀
-Make the decision rule (reject if TS > CV)

Question 23

How can we test for the equality of variance (1,5)

Answer 23

-To formulate this test, the distribution of random variables X₁ and X₂ must be normally distributed

-H₀: σ²₁ = σ²₂
-H₁: σ²₁ ≠ σ²₂
-The critical value is from the F distribution, denoted F^a/2_{n1-1, n2-1}
-The test statistic is F = s²₁/s²₂ (always put the higher sample standard deviation on the top)
-Make the decision rule (reject if TS > CV)

Question 24

When do you use matched pairs (3)

Answer 24

-This is to test 2 events which aren’t independent
-Formulating about µ₁ - µ₂, when X₁ and X₂ are with the same sample
-Apart from artificial examples, if you can’t get matched pairs, you go as close as possible

Question 25

What is the expected value + variance of matched pairs (2)

Answer 25

-E(D) = E(X₁ - X₂) = µ₁ - µ₂ = µ_d -V(X) = V(X₁ - X₂) = σ²₁ + σ²₂ - 2σ₁₂ = σ²_d

Question 26

How do we normalise a matched pairs distribution (1,1,3,1)

Answer 26

-Assuming the underlying distribution of X₁~N(µ₁, σ²₁) and X₂~N(µ₂, σ²₂) -Đ = X̄₁ - X̄₂~N(µ₁ - µ₂, σ²_d) -Define the difference as d₁ = x¹₁ - x¹₂, ..., d_n = xⁿ₁ - xⁿ₂ -Calculate đ = ∑d_i/n -Calculate s²_d = (∑d_i - đ)²/(n-1) -Normalising the expression gives us ((X̄₁ - X̄₂) - (µ₁ - µ₂))/(√S²_d/n) ~ t_n-1

Question 27

What is the five step procedure for an equality of means test (5)

Answer 27

-H₀:µ₁ - µ₂ = µ_d = D₀ -H₁:µ₁ - µ₂ = µ_d ≠ D₀ -The critical values are +-t_{a/2, n-1} -The test statistic is ((X̄₁ - X̄₂) - (µ₁ - µ₂))/(√S²_d/n) ~ t_n-1 -Make the decision rule (Reject if TS>CV)

Question 28

What is the power of a test (1)

Answer 28

-Power = P(Reject H₀|H₀ false)

Question 29

What is the 3 step procedure for calculating the power of a test (1,2,2)

Answer 29

-Define the critical value as the point at which you reject H₀ -Find the sample mean x̄^c which would give you the exact critical value -x̄^c = +-z_a/2(√S²/n) + µ_d -Calculate the probability of rejecting that sample mean with your true mean -Calculate [P(X̄c|µ = µ₁]

Question 30

What do we use an ANOVA test for (1)

Answer 30

-ANOVA (ANalysis Of VAriance) to test for the equality of means across two or more groups

Question 31

What do we assume for an ANOVA test (3)

Answer 31

-Random sample -Normal distribution -Variance for each group is the same

Question 32

What does TSS, BSS and WSS and the relationship between them (3,1)

Answer 32

-TSS = the Total Sum of Squares (the overall amount of variation in the variable X) -WSS = the Within Sum of Squares (the amount of variation within each of the k groups) -BSS = the Between Sum of Squares (the amount of variation across the k groups) -TSS = BSS + WSS

Question 33

How do you work out the TSS (2)

Answer 33

-TSS = ∑^k_j=1∑^nj_i=1 (X^j_i - X̄)² -TSS = variance x degrees of freedom

Question 34

How do you work out the BSS (2)

Answer 34

-BSS = ∑^k_j=1 n_j(X̄^j - X̄)² -BSS = n(mean of each category - overall mean)² for each category

Question 35

How do you work out the WSS (2,3)

Answer 35

-WSS =∑kj=1∑njI=1(Xij - X̄j)2 -WSS = (each data value - mean of that category)2 for each value -However, this is inefficient to work out -One method of working out WSS is (n-1)(variance of a group) then adding this for all groups -Another method is using the fact that TSS = WSS + BSS, and working out TSS and BSS

Question 36

What is the five step procedure for an ANOVA test (5)

Answer 36

-H₀: µ¹ = µ² = ... = µ^k -H₁: µ^j ≠ µ^l for j ≠ l -The critical values are F^a_{k-1, n-k} (n is the number of values, k is the number of means) -The test statistic is F = ((BSS)/(k-1))/((WSS)/(n-k)) -Make the decision rule (Reject if TS > CV)