Complete Question
The complete question is shown on the first uploaded image
Answer:
Yes the test suggest that the true average percentage of organic matter in such soil is something other than 3%
Step-by-step explanation:
From the question we are told that
The sample mean is [tex]\= x = 2.482\%[/tex]
The standard deviation is [tex]\sigma = 1.614[/tex]
The standard error is [tex]SE = 0.295[/tex]
The sample size is [tex]n = 30[/tex]
The level of significance is [tex]\alpha = 0.05[/tex]
The null hypothesis is [tex]H_o : \mu = 3\%[/tex]
The alternative hypothesis is [tex]H_a : \mu \ne 3\%[/tex]
Now the degree of freedom is evaluated as
[tex]df = n - 1[/tex]
[tex]df = 30 - 1[/tex]
[tex]df = 29[/tex]
The test statistics is mathematically evaluated as
[tex]t = \frac{ 2.482 - 3}{ 0.295}[/tex]
[tex]t = -1.756[/tex]
The p-value is obtained from the the student t -distribution table , the value is
[tex]p-value = P( T \le t)= 2 * t_{ t, df } = t_{ -1.756 , 29 } = 2 *0.0448= 0.0896[/tex]
The reason for the 2 in the equation is because the test is a two -tailed test i.e -1.756 and 1.756
Given that the [tex]p-value > \alpha[/tex] then we fail to reject the null hypothesis
Hence the test the suggest that the true average percentage of organic matter in such soil is something other than 3%
