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Coffee cans are to be filled with 16 oz. of coffee. The mean content of cans filled on a production line is monitored. It is known from past experience that the contents have a Normal distribution with a known standard deviation of 0.1 oz. A sample of 9 cans is taken every hour and their mean content is measured
a) Set up the hypotheses to test whether the mean content is 16 oz. Should the alternative be one-sided or two-sided? Why?
b) Give a decision rule in terms of the sample mean for a 0.05-level test.
c) If the true mean content during that the test derived in (b) will correctly detect this deviation from the target value of 16 oz? particular period is 16.1 oz., what is the probability

Sagot :

akiran007

Answer:

a) H0: μ= 16  against the claim Ha: μ≠ 16

b) H0 will be rejected if it does not lie in the interval  [15.934; 16.065 ]

c) 0.8531

Step-by-step explanation:

a) The null and alternate hypothesis are

H0: μ= 16  against the claim Ha: μ≠ 16

The test chosen is two sided because the cans have more or less content than 16 oz.

b) Here n = 9 , σ= 0.1 and α 0.05 and z∝/2 = ± 1.96

z∝/2 = x − µ/ σ/√n

z ∝/2= x-16/0.0333

± 1.96 *0.33= x-16

± 0.0653= x- 16

or x= 16± 0.0653

x=  15.934; 16.065

H0 will be rejected if it does not lie in the interval  [15.934; 16.065 ]

c) The probability can be computed by  transforming x into z

z= x-µ/ σ/√n

z= 15.934-16.1  /0.1/√9

z= [−4.95]

z2= x-µ/ σ/√n

z= 16.065-16.1 /0.1/√9

z2=[ −1.05]

P (15.934<x< 16.065 )= P (−4.95, −1.05)

                              =P ( z < −4.95 ) + P ( z > −1.05)

                               =  0+ 0.3531+0.5

                             = 0.8531

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