Question

In: Statistics and Probability

Q))A coin-operated coffee machine made by BIG Corporation was designed to discharge a mean of 7.2...

Q))A coin-operated coffee machine made by BIG Corporation was designed to discharge a mean of 7.2 ounces of coffee per cup. If it dispenses more than that on average, the corporation may lose money, and if it dispenses less, the customers may complain. Believing that the mean amount of coffeeμ dispensed by the machine is less than

7.2 ounces, BIG plans to do a statistical test of the claim that the machine is working as designed. BIG gathers a random sample of100 amounts of coffee dispensed by the machine. Suppose that the population of amounts of coffee dispensed by the machine has a standard deviation of 0.7 ounces and that BIG performs its hypothesis test using the

0.05 level of significance Based on this information, answer the questions below. Carry your intermediate computations to at least four decimal places, and round your responses as indicated.

What are the null and alternative hypotheses that BIG should use for the test?

H0:μis

?less thanless than or equal togreater thangreater than or equal tonot equal toequal to

?0.77.27.04100

H1:μis

?less thanless than or equal togreater thangreater than or equal tonot equal toequal to

?0.77.27.04100

Assuming that the actual value of µ is 7.04 ounces, what is the power of the test? Round your response to at least two decimal places.

What is the probability that BIG rejects the null hypothesis when, in fact, it is true? Round your response to at least two decimal places.

Suppose that BIG decides to perform another statistical test using the same population, the same null and alternative hypotheses, and the same level of significance, but for this second test BIG chooses a random sample of size 125 instead of a random sample of size 100. Assuming that the actual value of µ is 7.04 ounces, how does the probability that BIG commits a Type II error in this second test compare to the probability that BIG commits a Type II error in the original test?

The probability of committing a Type II error in the second test is greater

The probability of committing a Type II error in the second test is less

The probabilities of committing a Type II error are equal

Solutions

Expert Solution

Ho :   µ =   7.2
Ha :   µ ╪   7.2
======================

true mean ,    µ =    7.04                          
                                  
hypothesis mean,   µo =    7.2                          
significance level,   α =    0.05                          
sample size,   n =   100                          
std dev,   σ =    0.7000                          
                                  
δ=   µ - µo =    -0.16                          
                                  
std error of mean,   σx = σ/√n =    0.7000   / √    100   =   0.07000          
                                  
Zα/2   = ±   1.960   (two tailed test)                      
We will fail to reject the null (commit a Type II error) if we get a Z statistic between                           -1.960   and   1.960
these Z-critical value corresponds to some X critical values ( X critical), such that                                  
                                  
-1.960   ≤(x̄ - µo)/σx≤   1.960                          
7.063   ≤ x̄ ≤   7.337                          
                                  
now, type II error is ,ß =        P (   7.063   ≤ x̄ ≤   7.337   )          
       Z =    (x̄-true mean)/σx                      
       Z1 = (   7.063   -   7.04   ) /   0.07000   =   0.326
       Z2 = (   7.337   -   7.04   ) /   0.07000   =   4.246
                                  
   so, P(   0.326   ≤ Z ≤   4.246   ) = P ( Z ≤   4.246   ) - P ( Z ≤   0.326   )
                                  
       =   1.000   -   0.628   =   0.3723   [ Excel function: =NORMSDIST(z) ]  
                                  
power =    1 - ß =   0.6277                          
===========================

Standard Error , SE = σ/√n =   0.7000   / √    100   =   0.0700      
Z-test statistic= (x̅ - µ )/SE = (   7.040   -   7.2   ) /    0.0700   =   -2.286
P(type I error) = P(Z<-2.286) = 0.0223

(please try 0.05 if above gets wrong)

=================

The probability of committing a Type II error in the second test is less

because

Larger the sample size,smaller the probability of type II error.   


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