Question

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

Answer 1

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: