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Illusion [34]
3 years ago
6

A manufacturer of paper coffee cups would like to estimate the proportion of cups that are defective (tears, broken seems, etc.)

from a large batch of cups. They take a random sample of 200 cups from the batch of a few thousand cups and found 18 to be defective. The goal is to perform a hypothesis test to determine if the proportion of defective cups made by this machine is more than 8%.
Required:
a. Calculate a 95% confidence interval for the true proportion of defective cups made by this machine.
b. What is the sample proportion?
c. What is the critical value for this problem?
d. What is the standard error for this problem?
Mathematics
1 answer:
Maksim231197 [3]3 years ago
7 0

Answer:

a

  The 95% confidence interval is  0.0503  <   p < 0.1297

b

The sample proportion is  \r p =  0.09

c

The critical value is  Z_{\frac{\alpha }{2} } =  1.96

d

 The standard error is  SE   =0.020

Step-by-step explanation:

From the question we are told that

   The  sample size is  n =  200

     The number of defective is  k =  18

The null hypothesis is  H_o  :  p  =  0.08

The  alternative hypothesis is  H_a  :  p > 0.08

Generally the sample proportion is mathematically evaluated as

            \r p =  \frac{18}{200}

            \r p =  0.09

Given that the confidence level is  95% then the level  of significance is mathematically evaluated as

        \alpha  =  100 -  95

        \alpha  =  5\%

        \alpha  =  0.05

Next we obtain the critical value of  \frac{ \alpha }{2} from the normal distribution table, the value is  

        Z_{\frac{\alpha }{2} } =  1.96

Generally the standard of error is mathematically represented as

          SE   =  \sqrt{\frac{\r p (1 -  \r p)}{n} }

substituting values

         SE   =  \sqrt{\frac{0.09  (1 -  0.09)}{200} }

        SE   =0.020

The  margin of error is  

       E =  Z_{\frac{ \alpha }{2} }  * SE

=>    E =  1.96  *  0.020

=>   E =  0.0397

The  95% confidence interval is mathematically represented as

     \r p  -  E  <  \mu <  p <  \r p  + E

=>   0.09 - 0.0397  <  \mu <  p < 0.09 + 0.0397

=>  0.0503  <   p < 0.1297

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