Question

In article presents measures of penetration resistance for a certain fine-grained soil. fifteen measurements, expressed as a multiple of a standard quantity, had a mean of 2.64 and a standard deviation of 1.02. can you conclude that the mean penetration resistance is greater than 2.5? Find the p-value and state a conclusion.

Answer 1

Answer:

$t=\frac{2.64-2.5}{\frac{1.02}{\sqrt{15}}}=0.532$    

$p_v =P(t_{(14)}>0.532)=0.302$  

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis at 5% of significance

Step-by-step explanation:

Data given and notation  

$\bar X=2.64$ represent the sample mean

$s=1.02$ represent the sample standard deviation for the sample  

$n=15$ sample size  

$\mu_o =2.5$ represent the value that we want to test

$\alpha$ 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.  

We need to conduct a hypothesis in order to check if the mean is greater than 2.5, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 2.5$  

Alternative hypothesis:$\mu > 2.5$  

If we analyze the size for the sample is < 30 and we don't know the population deviation so 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".  

Calculate the statistic

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

$t=\frac{2.64-2.5}{\frac{1.02}{\sqrt{15}}}=0.532$    

P-value

The first step is calculate the degrees of freedom, on this case:  

$df=n-1=15-1=14$  

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

$p_v =P(t_{(14)}>0.532)=0.302$  

Conclusion  

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis at 5% of significance