Question

In order to comply with the Environmental protection Agency (EPA) regulations of the Clean Water Act, a large agricultural company wants to know the average nitrogen concentration in the soil of an agricultural region it plans to purchase. The seller claims that the average nitrogen level does not exceed 0.49 units. To test this claim at 0.05 level of significance, nitrogen concentration of soil samples were recorded at 51 sites in that agricultural region. The sample mean was found to be 0.505 and the sample standard deviation 0.12.

Answer 1

Answer:

$t=\frac{0.505-0.49}{\frac{0.12}{\sqrt{51}}}=0.893$  

$p_v =P(t_{50}>0.893)=0.1881$  

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean is not significantly higher than 0.49 units.  

Step-by-step explanation:

Data given and notation

$\bar X=0.505$ represent the sample mean  

$s=0.12$ represent the standard deviation for the sample

$n=51$ sample size  

$\mu_o =0.49$ represent the value that we want to test  

$\alpha=0.05$ represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses to be tested  

We need to conduct a hypothesis in order to determine if the average nitrogen level dos not exced 0.49 units, the system of hypothesis would be:

Null hypothesis:$\mu \leq 0.49$  

Alternative hypothesis:$\mu > 0.49$  

Compute the test statistic  

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

We can replace in formula (1) the info given like this:  

$t=\frac{0.505-0.49}{\frac{0.12}{\sqrt{51}}}=0.893$  

Now we need to find the degrees of freedom for the t distirbution given by:

$df=n-1=51-1=50$

What do you conclude?  

Compute the p-value  

Since is a right tailed test the p value would be:  

$p_v =P(t_{50}>0.893)=0.1881$  

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean is not significantly higher than 0.49 units.