Answer:
First question
The correct option is A
Second question
The correct option is E
Step-by-step explanation:
From the question we are told that
The population mean is [tex]\mu = 50 \ mpg \ (miles \ per \ gallon )[/tex]
The sample size is n = 30
The population standard deviation is [tex]\sigma = 2.3 \ miles[/tex]
The sample mean is [tex]\= x = 49 \ mpg[/tex]
The level of significance is [tex]\alpha = 0.05[/tex]
The null hypothesis is [tex]H_o : \mu = 50 \ mpg[/tex]
The alternative hypothesis is [tex]H_a : \mu < 50 \ mpg[/tex]
Generally the test statistics is mathematically represented as
[tex]t = \frac{ \= x - \mu }{ \frac{\sigma }{ \sqrt{n} } }[/tex]
=> [tex]t = \frac{ 49 - 50 }{ \frac{2.3}{ \sqrt{30 } } }[/tex]
=> [tex]t = -2.38[/tex]
Generally the degree of freedom is mathematically represented as
[tex]df = n - 1[/tex]
[tex]df = 30 - 1[/tex]
[tex]df = 29[/tex]
Generally from the student t distribution table the probability corresponding to -2.38 to the left at a degree of freedom of [tex]df = 29[/tex] is
[tex]p -value = P( t < -2.38 ) = 0.0174[/tex]
From the value obtained we see that
[tex]p-value < \alpha[/tex] hence
The decision rule
Reject the null hypothesis
The conclusion is
There evidence to reject the manufacturer's claim because the p-value of 0.0174 is less than the significance level E
