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

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 \\e 16`

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

`t = ( \= x -\mu )/( (\sigma)/( √(n) ) )`

=>
`t = ( 16.32 -16 )/( (0.8 )/( √(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

Explanation: