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irga5000 [103]
3 years ago
14

Suppose a production line operates with a mean filling weight of 16 ounces per container. Since over- or under-filling can be da

ngerous, a quality control inspector samples 30 items to determine whether or not the filling weight has to be adjusted. The sample revealed a mean of 16.32 ounces. From past data, the population standard deviation is known to be 0.8 ounces. Using a 0.10 level of significance, can it be concluded that the process is out of control (not equal to 16 ounces).
Step 1: State hypotheses:
Step 2: State the test statistic. Since we know the population standard deviation and the sample is large our test statistics is
Step 3: State the critical region(s):
Step 4: Conduct the experiment/study:
Step 5: Reach conclusions and state in English:
Step 6: Calculate the p-value associated with this test. How does this the p-value support your conclusions in Step 5?
Mathematics
1 answer:
miss Akunina [59]3 years ago
4 0

Answer:

From the question we are told that

   The  population mean is  \mu  =  16

    The sample size is  n  =  30  

     The  sample mean is  \= x =  16.32

     The  population standard deviation is  \sigma  =  0.8

      The  level of significance is  \alpha  = 0.10

Step 1: State hypotheses:

The  null hypothesis is  H_o :  \mu = 16

The alternative hypothesis is  H_a :  \mu \ne  16

Step 2: State the test statistic. Since we know the population standard deviation and the sample is large our test statistics is

          t = \frac{ \= x  -\mu }{ \frac{\sigma}{ \sqrt{n} } }

=>       t = \frac{ 16.32  -16  }{ \frac{0.8 }{ \sqrt{30} } }

=>       t =2.191

Generally the degree of freedom is mathematically represented as

          df  =  n - 1

=>      df  =  30  - 1

=>      df  =29

Step 3: State the critical region(s):

From the student t-distribution table the critical value corresponding to  \alpha  = 0.10  is

         t = 1.311

Generally the critical regions is mathematically represented as  

         - 1.311 < T <  1.311

Step 4: Conduct the experiment/study:

Generally the from the value obtained we see that the t value is outside the critical region so  the decision is [Reject the null hypothesis ]  

Step 5: Reach conclusions and state in English:

  There is  sufficient evidence to show that the filling weight has to be adjusted

Step 6: Calculate the p-value associated with this test. How does this the p-value support your conclusions in Step 5?

From the student t-distribution table the probability value to the right corresponding to t =2.191 at  a degree of freedom of  df  =29  is

        P( t > 2.191) =  0.0183

Generally the p-value is mathematically represented as

       p-value  = 2 * P( t >  2.191 )

=>    p-value  = 2 * 0.0183

=>    p-value  =  0.0366

Generally  looking at the value obtained we see that p- value <  \alpha hence

The decision rule is

Reject the null hypothesis

Step-by-step explanation:

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