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
