Question

A textile fiber manufacturer is investigating a new drapery yarn, which the company claims has a mean thread elongation of 12 kilograms with a standard deviation of 0.5 kilograms. The company wishes to test the hypothesis Upper H Subscript 0 Baseline colon mu equals 12 against Upper H Subscript 1 Baseline colon mu less-than 12 using a random sample of n equals 4 specimens. Calculate the P-value if the observed statistic is x overBar equals 11.1.

Answer 1

Answer:

$z=\frac{11.1-12}{\frac{0.5}{\sqrt{4}}}=-3.6$    

Calculate the P-value    

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

$p_v =2*P(z<-3.6)=0.000318$

Step-by-step explanation:

Data given and notation    

$\bar X=11.1$ represent the sample mean

$\sigma=0.5$ represent the population standard deviation

$n=4$ sample size    

$\mu_o =12$ 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 determine if the mean is equal to 12 or not, the system of hypothesis would be:    

Null hypothesis:$\mu = 12$    

Alternative hypothesis:$\mu \neq 12$    

We know the population deviation, and the sample size <30, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:    

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)    

z-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:    

$z=\frac{11.1-12}{\frac{0.5}{\sqrt{4}}}=-3.6$    

Calculate the P-value    

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

$p_v =2*P(z<-3.6)=0.000318$