Question

Polymer composite materials have gained popularity because they have high strength to weight ratios and are relatively easy and inexpensive to manufacture. However, their nondegradable nature has prompted development of environmentally friendly composites using natural materials. An article reported that for a sample of 10 specimens with 2% fiber content, the sample mean tensile strength (MPa) was 51.6 and the sample standard deviation was 1.1. Suppose the true average strength for 0% fibers (pure cellulose) is known to be 48 MPa. Does the data provide compelling evidence for concluding that true average strength for the WSF/cellulose composite exceeds this value? (Use α = 0

Answer 1

Answer:

$t=\frac{51.6-48}{\frac{1.1}{\sqrt{10}}}=10.349$      

$p_v =P(t_9>10.349)=1.34x10^{-6}$  

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the the actual mean is significantly higher than 48 MPa at 5% of significance.

Step-by-step explanation:

1) Data given and notation      

$\bar X=51.6$ represent the sample mean

$s=1.1$ represent the standard deviation for the sample      

$n=10$ sample size      

$\mu_o =48$ 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.      

We need to conduct a hypothesis in order to determine if the mean score is higher than 48, the system of hypothesis would be:      

Null hypothesis:$\mu \geq 48$      

Alternative hypothesis:$\mu > 48$      

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:      

$t=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (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:      

$t=\frac{51.6-48}{\frac{1.1}{\sqrt{10}}}=10.349$      

Calculate the P-value      

First we find the degrees of freedom:

$df=n-1=10-1=9$

Since is a one-side upper test the p value would be:      

$p_v =P(t_9>10.349)=1.34x10^{-6}$  

Conclusion      

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the the actual mean is significantly higher than 48 MPa at 5% of significance.