"A fool must now and then be right,by chance."

10.4 INFERENCE AS DECISION (Pages 567 - 577)OVERVIEW: This section introduces three conceptsthat were added to the AP Statistics syllabus for the 1998-99academic year: Type I error, Type II error, and Power of atest.

NOTE: Reading the paper *STATISTICALERRORS (TYPE I, TYPE II, POWER)* , under WRITINGSin Herkimer's Hideaway, might be useful in gaining an understandingof Type I error, Type II error, and Power.

Here is an example that illustrates the conceptsdiscussed in this section.

Example:

A Quality Control Situation:

Herkimer's company manufactures a machine-produced product weighing1500 pounds. The population of produced items has an allowablestandard deviation of 40 pounds. Samples of size 100 are periodicallyexamined to see if production standards are being maintained.

Consider the set, M, consisting of the meanweights of all samples of size 100. The Central Limit Theorem statesthat M will have a normal distribution with mean = 1500 pounds andstandard deviation = 40/sqrt(100) = 40/10 = 4 pounds.

Herkimer runs quality control tests at the 5%level of significance.

Null hypothesis H

_{0}: m = 1500.Alternate hypothesis H

_{a}: m is not equal to 1500.Type of test: 2-tail (interested in deviations,both directions)

Level of significance: 5% (2.5% in eachtail)

Critical values of z: z < -1.96 or z >1.96.

Herkimer will reject H_{0} if a sample of 100 yields amean that is not within 1.96 standard deviations of 1500 pounds.Hence, H_{0}will be rejected if a mean weight is less than 1500-1.96(4) = 1492.6pounds, or if a mean weight is more than 1500+1.96(4) = 1507.84pounds. In other words, Herkimer will fail to rejectH_{0} if a meanweight, x, is in the range 1492.16 < x < 1507.84.

Related TI-83 computations:

invNorm(.975,0,1) = 1.959963986

normalcdf(1492.16,1507.84,1500,4) = .9500043497

Now, suppose a random sample produces a mean of1509 pounds. In this case, the H_{0} would be rejected. There isa "suggestion" that production standards aren't being met. Theprobability of obtaining a mean as large as 1509 is** normalcdf(**1509,1E99,1500,4)=.012224, or about 1.2% Hence, if H

Assume now that it is extremely undesirable tohave a produced item weighing 1515 pounds or more. Herkimer realizesthat things can go wrong in a mass production process and that thiscould result in his products being too heavy. He is interested inknowing the probability that his quality control test willincorrectly accept Ho if the mean weight somehow shifts to 1515pounds. If Ho is incorrectly accepted, this is a Type IIerror.

Let's remember that H_{0} will be accepted if a samplemean weight is less than 1507.84 pounds (and greater than 1492.16pounds).

If the population mean has shifted to 1515 pounds,the z-value for 1507.84 pounds is

z_{1507.84} = (1507.84 - 1515)/4 =-1.79

Using the normal distribution table, theprobability that z < -1.79 is 0.0367. This is the probability ofmaking a **Type II error**_{0} about 3.67% of the time. (Hewill assume all is OK when it isn't.)

TI-83 computation

normalcdf(-1E99,1507.84,1515,4) = .0367269039

The ** power** of a test is theprobability that H

TI-83 computation

normalcdf(1507.84,1E99,1515,4) = .9632730961.

That is, if m = 1515 pounds, Herkimer canexpect to correctly reject H_{0} about 96.33% of thetime.

Real-life situations frequently involve Type I andType II errors. Consider the legal world and a null hypothesis "Thisaccused man is innocent." A Type I error would be determining the manis guilty when he is innocent.. A Type II error would involvedeclaring the man innocent when he is guilty. Decreasing the chanceof one error type frequently increased the chance of the other errortype. In real world situations, one must often decide which errortype is more important to minimize.

Things to remember:

(1)AType I error can only occur when a null hypothesis is true.

(You incorrectly reject a true null hypothesis.)

(2)A Type II errorcan only occur when a null hypothesis is false.

(You incorrectly fail to reject a false null hypothesis.)

(3)The Power of atest is 1 - probability (Type II error).

(This is the probability that you correctly reject a false nullhypothesis.)

(4)One needs analternative to the null hypothesis in order to calculate a Type IIerror. In the above example, the alternative hypothesis wasm = 1515.Without an alternative hypothesis, the question "what is theprobability of a Type II error?" is meaningless.