A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 420.0 gram setting. It is believed that the machine is underfilling the bags. A 33 bag sample had a mean of 417.0 grams. A level of significance of 0.01 will be used. Determine the decision rule. Assume the variance is known to be 576.00.

Respuesta :

Answer

The decision rule is

Fail to reject the null hypothesis

The conclusion is  

There is no sufficient evidence to conclude that the machine is underfilling the bags

Step-by-step-explanation:

From the question we are told that

   The  population mean is   [tex]\mu = 420.0 \ g[/tex]

    The  sample size is  n  =  33 bags

    The sample mean is  [tex]\= x = 417.0 \ g[/tex]

     The level of significance is  [tex]\alpha = 0.01[/tex]

     The  variance  is  [tex]\sigma^2 = 576.0[/tex]

The  null hypothesis is [tex]H_o : \mu = 420[/tex]

The alternative hypothesis is [tex]H_a : \mu < 420[/tex]

Generally the standard deviation is mathematically represented as

        [tex]\sigma = \sqrt{\sigma^2}[/tex]

=>     [tex]\sigma = \sqrt{576 }[/tex]

=>     [tex]\sigma = 24[/tex]

Generally the test statistics is mathematically represented as

        [tex]z = \frac{ \= x - \sigma }{ \frac{\sigma}{\sqrt{n} } }[/tex]

=>     [tex]z = \frac{417 - 420 }{ \frac{24}{ \sqrt{ 33} } }[/tex]

=>      [tex]z = -0.71807[/tex]

From the z table  the area under the normal curve to the left corresponding to    -0.71807   is  

   [tex]p-value = P(Z < -0.71807 ) = 0.23636[/tex]

Generally looking at the values we see that

              [tex]p-value > \alpha[/tex] ,  hence

The decision rule is

Fail to reject the null hypothesis

The conclusion is  

There is no sufficient evidence to conclude that the machine is underfilling the bags