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From Algorithms to Z-Scores - matloff - University of California, Davis

From Algorithms to Z-Scores - matloff - University of California, Davis

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11.2. GENERAL TESTING BASED ON NORMALLY DISTRIBUTED ESTIMATORS 223<br />

Now recall from the derivation <strong>of</strong> (10.22) that -1.96 and 1.96 are the lower- and upper-2.5% points<br />

<strong>of</strong> the N(0,1) distribution. Thus,<br />

P (Z < −1.96 or Z > 1.96) ≈ 0.05 (11.5)<br />

Now here is the point: After we collect our data, in this case by <strong>to</strong>ssing the coin n times, we<br />

compute p from that data, and then compute Z from (11.4). If Z is smaller than -1.96 or larger<br />

than 1.96, we reason as follows:<br />

Hmmm, Z would stray that far from 0 only 5% <strong>of</strong> the time. So, either I have <strong>to</strong> believe<br />

that a rare event has occurred, or I must abandon my assumption that H0 is true.<br />

For instance, say n = 100 and we get 62 heads in our sample. That gives us Z = 2.4, in that “rare”<br />

range. We then reject H0, and announce <strong>to</strong> the world that this is an unfair coin. We say, “The<br />

value <strong>of</strong> p is significantly different from 0.5.”<br />

The 5% “suspicion criterion” used above is called the significance level, typically denoted α. One<br />

common statement is “We rejected H0 at the 5% level.”<br />

On the other hand, suppose we get 47 heads in our sample. Then Z = -0.60. Again, taking 5% as<br />

our significance level, this value <strong>of</strong> Z would not be deemed suspicious, as it occurs frequently. We<br />

would then say “We accept H0 at the 5% level,” or “We find that p is not significantly different<br />

from 0.5.”<br />

The word significant is misleading. It should NOT be confused with important. It simply is saying<br />

we don’t believe the observed value <strong>of</strong> Z is a rare event, which it would be under H0; we have<br />

instead decided <strong>to</strong> abandon our believe that H0 is true.<br />

Note by the way that Z values <strong>of</strong> -1.96 and 1.96 correspond getting 50 − 1.96 · 0.5 · √ 100 or<br />

50 + 1.96 · 0.5 · √ 100 heads, i.e. roughly 40 or 60. In other words, we can describe our rejection<br />

rule <strong>to</strong> be “Reject if we get fewer than 40 or more than 60 heads, out <strong>of</strong> our 100 <strong>to</strong>sses.”<br />

11.2 General Testing Based on Normally Distributed Estima<strong>to</strong>rs<br />

In Section 10.5, we developed a method <strong>of</strong> constructing confidence intervals for general approximately<br />

normally distributed estima<strong>to</strong>rs. Now we do the same for significance testing.<br />

Suppose θ is an approximately normally distributed estima<strong>to</strong>r <strong>of</strong> some population value θ. Then

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