statistical approaches

Topics: Statistical hypothesis testing, Normal distribution, Arithmetic mean Pages: 28 (2319 words) Published: October 26, 2014
❶ A quality control engineer is in charge of testing whether or not 95% of the Blu-ray disc players produced by his company conform to specifications. To do this, the engineer randomly selects a batch of 13 Blu-ray players from each day’s production. The day’s production is acceptable provided no more than 1 Blu-ray player fails to meet specifications. Otherwise, the entire day’s production has to be tested.

(a) What is the probability that the engineer incorrectly passes a day’s production as acceptable if only 85% of the day’s Blu-ray players actually conform to specification? From the given information, we can know both problems are n Bernoulli Trials, where n  13 . Suppose p is probability that the engineer passes a day’s production as acceptable; and x is the quantity of defectives, then:

p = P( x  1) = P ( x  0)  P ( x  1) =

13!
13!
(1  0.85)0 (0.85)130 
(1  0.85)1 (0.85)131
0! (13  0)!
1! (13  1)!

=0 . 1 2+
0 09 . 2 7 7
=4
0.39=
839.83%

Thus, the probability that the engineer incorrectly passes a day’s production as acceptable if only 85% of the day’s Blu-ray players actually conform to specification is 39.83%. (b) What is the probability that the engineer unnecessarily requires the entire day’s production to be tested if in fact 95% of the Blu-ray players conform to specifications? Suppose p is the probability that the engineer requires the entire day’s production to be tested, and x is the quantity of defectives, then:

p  P( x  2)  1  P( x  1)  1  P( x  0)  P( x  1)  1 

13!
13!
(1  0.95) 0 (0.95)130 
(1  0.95)1 (0.95)131
0!(13  0)!
1!(13  1)!

 1 - 0.5133 - 0.3512  0.1355  13.55%

Thus, the probability that the engineer unnecessarily requires the entire day’s production to be tested if in fact 95% of the Blu-ray players conform to specifications is 13.55%. ❷ The reliability of an electrical fuse is the probability that a fuse, chosen at random from production, will function under its designed conditions. A random sample of 1000 fuses was tested and x = 27 defectives were observed. Calculate the approximate probability of observing 27 or more defectives, assuming that the fuse reliability is 0.98. According to the given information, we can use Normal Approximation to the Binomial Distributions to solve this problem. And, we have   np and   np(1  p) , where n=1000, p=1-0.98=0.02

We suppose the approximate probability of observing 27 or more defectives is p0 , then : p0  P( x  27)  1  P( x  27)  1  (

27  0.5  1000  0.02
1000  0.02(1  0.02)

)  1  (1.468)  1  0.929  0.071

❸A product-fill operation produces net weights that are normally distributed with mean μ = 7.9 ounces and standard deviation σ= 0.5 ounces.
(a) Estimate the percent of the containers that have a net weight less than 8.5 ounces.

1/7

P( x  8.5)  (

8.5  



)  (

8.5  7.9
)  (1.2)  0.88493  88.493%
0.5

(b) What is the probability that a sample of fifteen randomly selected containers will have an average net weight less than 8.5 ounces?
P( x  8.5)  (

8.5  



n

)  (

8.5  7.9
0.5 15

)  (4.65)  1  100%

❹An engineer wants to measure the bias in a pH meter. She uses the meter to measure the pH in thirteen neutral substances (pH = 6.97) and obtains the following data: 6.90 7.00 7.03 7.01 6.97 7.00 6.95 7.00 6.99 7.04 6.97 7.07 7.04 (a) Check the assumption of normality for the pH meter data. The normality for pH meter data is displayed as following figure:

The plot is linear, indicating that we can be sure that we can model the sample by a normal distribution.
(b) Is there sufficient evidence to support the claim that the pH meter is not correctly calibrated at the 5% level of Significance.
The appropriate hypotheses are:
H0

:   6.97

H1

:   6.97

Since  is unknown, the appropriate test statistic is the t-test and its value is given by t0 

x  0
s

n



6.9977...
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