Significance testing
06 Theory - Significance testing
Theory 1 - Significance testing
Significance test
Ingredients of a significance test (unary hypothesis test):
— Null hypothesis event
- Identify a Claim
- Then:
is background assumption (supposing Claim isn’t known) - Goal is to invalidate
in favor of Claim — Rejection Region event (decision rule)
is written in terms of decision statistic and significance level is unlikely assuming . is more likely if Claim — Able to compute this
- Usually: inferred from
or - Adjust
to achieve Significance level
Suppose we are given a null hypothesis
and a rejection region . The significance level of
is: Sometimes the condition is dropped and we write
, e.g. when a background model without assuming is not known. Null hypothesis implies a distribution
Usually
is unspecified, yet determines a known distribution. In this case
will not take the form of an event in a sample space, . At a minimum,
must determine . We do NOT need these details:
- Background sample space
- Non-conditional distribution (full model):
or - Complement conditionals:
or
In basic statistical inference theory, there are two kinds of error.
- Type I error concludes with rejecting
when is true. - Type II error concludes with maintaining
when is false. Type I error is usually a bigger problem. We want to consider
as “innocent until proven guilty.”
is true is false Maintain null hypothesis Made right call Wrong acceptance Reject null hypothesis Wrong rejection Made right call
To design a significance test at
, we must identify , and specify with the property that . When
is written using a variable , we must choose between: Link to original
- One-tail rejection region:
with or with - Two-tail rejection region:
with
07 Illustration
Example - One-tail test: Weighted die
One-tail test: Weighted die
Your friend gives you a single regular die, and say she is worried that it has been weighted to prefer the outcome of 2. She wants you to test it.
Design a significance test for the data of 20 rolls of the die to determine whether the die is weighted. Use significance level
. Solution
Let
count the number of 2s that come up. The Claim: “the die is weighted to prefer 2” The null hypothesis
: “the die is normal” Assuming
is true, then , and therefore:
⚠️ Notice that “prefer 2” implies the claim is for more 2s than normal.
Therefore: Choose a one-tail rejection region.
Need
such that:
Solve for
by computing conditional CDF values:
0 1 2 3 4 5 6 7 0.026 0.130 0.329 0.567 0.769 0.898 0.963 0.989 Therefore, choose
:
, but . Final answer: Link to original
Two-tail test: Circuit voltage
Two-tail test: Circuit voltage
A boosted AC circuit is supposed to maintain an average voltage of
with a standard deviation of . Nothing else is known about the voltage distribution. Design a two-tail test incorporating the data of 40 independent measurements to determine if the expected value of the voltage is truly
. Use . Solution
Use
as the decision statistic, i.e. the sample mean of 40 measurements of . The Claim to test:
The null hypothesis
: Rejection region:
where
is chosen so that
Assuming
, we expect that: Recall Chebyshev’s inequality:
Now solve:
Therefore the rejection region should be:
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One-tail test with a Gaussian: Weight loss drug
One-tail test with a Gaussian: Weight loss drug
Assume that in the background population in a specific demographic, the distribution of a person’s weight
satisfies . Suppose that a pharmaceutical company has developed a weight-loss drug and plans to test it on a group of 64 individuals. Design a test at the
significance level to determine whether the drug is effective. Solution
Since the drug is tested on 64 individuals, we use the sample mean
as the decision statistic. The Claim: “the drug is effective in reducing weight”
The null hypothesis
: “no effect: weights on the drug still follow ” Assuming
is true, then . ⚠️ One-tail test because the drug is expected to reduce weight (unidirectional). Rejection region:
Calculate:
⚠️ Standardized
is approximately normal! (The standardization of
removes the effect of . As if it’s the summation.) So, standardize and apply CLT:
Solve:
Therefore, the rejection region:
Link to original
Binary hypothesis testing
01 Theory - Binary testing, MAP and ML
Theory 1 - Binary testing, MAP and ML
Binary hypothesis test
Ingredients of a binary hypothesis test:
and — Complementary hypotheses
- Maybe also know the prior probabilities
and - Goal: determine which case we are in,
or and — Complementary events of the Decision Rule
- Directionality: given
, is likely; given , is likely - Decision Rule: outcome
, accept ; outcome , accept - Usually:
written in terms of decision statistic using a design - We cover three designs:
- MAP and ML (minimize ‘error probability’)
- MC (minimizes ‘error cost’)
- Designs use
and (or , ) to construct and MAP design
Suppose we know:
and
- Both prior probabilities
and (or and )
- Both conditional distributions
The maximum a posteriori probability (MAP) design for a decision statistic
: Discrete case:
Continuous case:
And
. The MAP design minimizes the total probability of error.
ML design
Suppose we don’t know the priors, we know only:
and (or and )
- Both conditional distributions
The maximum likelihood (ML) design for
: ML is a simplified version of MAP.
(Set and to .)
The probability of a false alarm, a Type I error, is called
. The probability of a miss, a Type II error, is called . Total probability of error:
Wrong meanings of
Suppose
sets off a smoke alarm, and is ‘no fire’ and is ‘yes fire’. Then
is the odds that we get an alarm assuming there is no fire. This is not the odds of experiencing a false alarm (no context). That would be
. This is not the odds of a given alarm being a false one. That would be
.
02 Illustration
Example - ML test: Smoke detector
ML test: Smoke detector
Suppose that a smoke detector sensor is configured to produce
when there is smoke, and otherwise. But there is background noise with distribution . Design an ML test for the detector electronics to decide whether to activate the alarm.
What are the three error probabilities? (Type I, Type II, Total.)
Solution
First, establish the conditional distributions:
Density functions:
The ML condition becomes:
Therefore,
is , while is . The decision rule is: activate alarm when
.
Type I error:
Type II error:
Total error:
Link to original
Example - MAP test: Smoke detector
MAP test: Smoke detector
Suppose that a smoke detector sensor is configured to produce
when there is smoke, and otherwise. But there is background noise with distribution . Suppose that the background chance of smoke is
. Design a MAP test for the alarm. What are the three error probabilities? (Type I, Type II, Total.)
Solution
First, establish priors:
The MAP condition becomes:
Therefore,
is , while is . The decision rule is: activate alarm when
.
Type I error:
Type II error:
Total error:
Link to original
03 Theory - MAP criterion proof
Theory 2 - MAP criterion proof
Link to originalExplanation of MAP criterion - discrete case
First, we show that the MAP design selects for
all those which render more likely than . This will be used in the next step to show that MAP minimizes probability of error. Observe this calculation:
Recall the MAP criterion:
Divide both sides by
and apply the above Calculation in reverse: This is what we sought to prove.
Next, we verify that the MAP design minimizes the total probability of error.
The total probability of error is:
Expand this with summation notation (assuming the discrete case):
Now, how do we choose the set
(and thus ) in such a way that this sum is minimized? Since all terms are positive, and any
may be placed in or in freely and independently of all other choices, the total sum is minimized when we minimize the impact of placing each . So, for each
, we place it in if: That is equivalent to the MAP criterion.
04 Theory - MC design
Theory 3 - MC design
- Write
for cost of false alarm, i.e. cost when is true but decided .
- Probability of incurring cost
is . - Write
for cost of miss, i.e. cost when is true but decided .
- Probability of incurring cost
is . Expected value of cost incurred
MC design
Suppose we know:
- Both prior probabilities
and - Both conditional distributions
and (or and ) The minimum cost (MC) design for a decision statistic
: Discrete case:
Continuous case:
Then
. The MC design minimizes the expected value of the cost of error.
Link to originalMC minimizes expected cost
Inside the argument that MAP minimizes total probability of error, we have this summation:
The expected value of the cost has a similar summation:
Following the same reasoning, we see that the cost is minimized if each
is placed into precisely when the MC design condition is satisfied, and otherwise it is placed into .
05 Illustration
Example - MC Test: Smoke detector
MC Test: Smoke detector
Suppose that a smoke detector sensor is configured to produce
when there is smoke, and otherwise. But there is background noise with distribution . Suppose that the background chance of smoke is
. Suppose the cost of a miss is the cost of a false alarm. Design an MC test for the alarm. Compute the expected cost.
Solution
We have priors:
And we have costs:
(The ratio of these numbers is all that matters in the inequalities of the condition.)
The MC condition becomes:
Therefore,
is , while is . The decision rule is: activate alarm when
.
Type I error:
Type II error:
Total error:
PMF of total cost:
Therefore
Link to original.