The U.S. Department of Agriculture claims that the mean consumption of coffee by a person in the United States is 24.2 gallons per year. A random sample of 101 people in the United States shows that the mean coffee consumption is 23.5 gallons per year with a standard deviation of 3.2 gallons. At α=0.05, can you reject the claim? Using a P-value, evaluate the claim. Using Rejection Regions for a z-test, evaluate the claim.

Answer:

The decision rule is  

Reject the null hypothesis  

The conclusion is  

There is sufficient evidence to reject the claim  

Step-by-step explanation:

From the question we are told that

   The population mean is  [tex]\mu = 24.2 \ gallons \ per \ year[/tex]

    The sample size is  n  =  101

     The sample mean is  [tex]\= x = 23.5 \ gallons\ per\ year[/tex]

     The standard deviation is  [tex]\sigma = 3.2 \ gallons[/tex]

The null hypothesis is  [tex]H_o : \mu = 24.2[/tex]

 The alternative hypothesis is [tex]H_a : \mu \ne 24.2[/tex]

Generally the test statistics is mathematically represented as    

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

=>             [tex]z = \frac{ 23.5 - 24.2 }{ \frac{ 3.2}{ \sqrt{ 101} } }[/tex]  

=>             [tex]z = -2.198[/tex]

From the z table  the area under the normal curve to the left corresponding to    -2.198  is  

         [tex]P(Z < -2.198 ) = 0.013975[/tex]

Gnerally the  p-value is mathematically represented as

         [tex]p-value = 0.013975 * 2[/tex]

=>      [tex]p-value = 0.02795[/tex]

From the value obtained we have that  [tex]p-value \ < \ \alpha[/tex] hence  

The decision rule is  

Reject the null hypothesis  

The conclusion is  

There is sufficient evidence to reject the claim  

Using the Rejection Regions for a z-test method

Generally from the z table  the critical value  of [tex]\frac{\alpha }{2}[/tex]  is  

     [tex]z_{critical} = \pm 1.96[/tex]

Not we are using  [tex]\frac{\alpha }{2}[/tex]  because it is a two - tailed test

   Now comparing the critical value and the test statistics we see that the

region covered by  the test statistics (i.e [tex]2.198 < z < -2.198[/tex])   is greater than the region covered by  the critical value  (i.e [tex]1.96 < z < -1.96[/tex])

Hence

The decision rule is  

Reject the null hypothesis  

The conclusion is  

There is sufficient evidence to reject the claim  

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