Question

A Belgian company (Saluc’s Aramith pool ball factory) is the world’s leading producer of billiard balls with a 80% market share worldwide. High-technology machinery and computerized equipment allow the company to produce billiard balls that require tight dimensional tolerances. Perhaps the most important dimension is the diameter of the balls. They must have a diameter that is 2.25 inches for the U.S. market. If the diameter is too large or too small, the playing characteristics of the billiard balls are affected.

Each hour the quality control engineers select a random sample of 25 billiard balls from the production line and measure the diameter of each ball. The standard deviation is tightly controlled as well. Assume σ = 0.05 inches.

Suppose the quality control engineers are interested in testing that the production line is producing billiard balls with an average diameter of 2.25 inches versus the mean diameter of billiard balls has changed. They choose a 5% significance level for the test.

A) The assistant manager of the quality control team however believes the production machinery has aged and is producing “larger” billiard balls with a population mean diameter of 2.26 inches. Everything else being same, what is the probability of type II error and the power of the test when true population mean is 2.26? For questions (B) and (C) we will assume the assistant manager’s belief about the population mean is correct. True population mean = 2.26 inches. However, the original hypothesis test is still the one being tested. All the original information applies for questions (B) and (C). Any (and only) changes will be specified for each the question.

(B) Everything else being same, what is the power of the test if the population standard deviation of diameter changes to 0.01. How does this compare to your answer in part (A)? Intuitively explain why the answers are different.

(C) What is the probability of TYPE I error for the original test? Everything else being the same, if I wanted the probability of Type I error to be 2%, what will be the probability of type II error? How does it compare to the probability of type II error in part (A)? Intuitively explain why the answers are different.

Answer 1

a)

true mean ,    µ =    2.26              
                      
hypothesis mean,   µo =    2.25              
significance level,   α =    0.05              
sample size,   n =   25              
std dev,   σ =    0.0500              
                      
δ=   µ - µo =    0.01              
                      
std error of mean,   σx = σ/√n =    0.0500   / √    25   =   0.01000
                      
Zα/2   = ±   1.960   (two tailed test)          

ß =   P(Z < Zα/2 - δ/σx) - P(Z < -Zα/2-δ/σx)                                          
= P(Z <    1.960   - (   0.01   /    0.0100   )) - P ( Z <   -1.960   - (   0.01   /   0.0100   ))
   = P ( Z <    0.9600   ) - P ( Z <   -2.9600   )                          
   =   0.8315   -    0.0015   =   0.8299   [ Excel function: =NORMSDIST(z) ]                  

so, P(Type II error) ,ß=0.8299

power =    1 - ß =   0.1701

B)

std error of mean,   σx = σ/√n =    0.0100   / √    25   =   0.00200
ß =   P(Z < Zα/2 - δ/σx) - P(Z < -Zα/2-δ/σx)                                          
= P(Z <    1.960   - (   0.01   /    0.0020   )) - P ( Z <   -1.960   - (   0.01   /   0.0020   ))
   = P ( Z <    -3.0400   ) - P ( Z <   -6.9600   )                          
   =   0.0012   -    0.0000   =   0.0012   [ Excel function: =NORMSDIST(z) ]                  

so, P(Type II error) ,ß=0.0012

power =    1 - ß =   0.9988

decreasing the std dev, decreases the std error , hence, Type II error get increases and Power get increase

c)

Type I error = 0.05

---------

α=0.05

Zα/2   = ±   2.326   (two tailed test)
ß =   P(Z < Zα/2 - δ/σx) - P(Z < -Zα/2-δ/σx)                                          
= P(Z <    2.326   - (   0.01   /    0.0020   )) - P ( Z <   -2.326   - (   0.01   /   0.0020   ))
   = P ( Z <    -2.6737   ) - P ( Z <   -7.3263   )                          
   =   0.0038   -    0.0000   =   0.0038   [ Excel function: =NORMSDIST(z) ]                  
Type II error = 0.0038

probability is smaller as compare to part A)

decreasing the Type I error , Probability of type II error get decreases