Answer:
Null hypothesis: 
  
Alternative hypothesis: 
  
 
  
 
  
Step-by-step explanation:
1) Data given and notation  
Data: 105.6, 90.9, 91.2, 96.9, 96.5, 91.3, 100.1, 105.0, 99.6, 107.7, 103.3, 92.4
We can calculate the sample mean and deviation for this data with the following formulas:


The results obtained are:
 represent the sample mean
 represent the sample mean  
 represent the sample standard deviation
 represent the sample standard deviation  
 sample size
 sample size  
 represent the value that we want to test
 represent the value that we want to test  
 represent the significance level for the hypothesis test.
 represent the significance level for the hypothesis test.  
z would represent the statistic (variable of interest)  
 represent the p value for the test (variable of interest)
 represent the p value for the test (variable of interest)  
2) State the null and alternative hypotheses.  
We need to conduct a hypothesis in order to check if the mean is equal to 100pCL/L, the system of hypothesis are :  
Null hypothesis: 
  
Alternative hypothesis: 
  
Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  
 (1)
 (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".  
3) Calculate the statistic  
We can replace in formula (1) the info given like this:  
 
  
4) P-value  
First we need to find the degrees of freedom for the statistic given by:

Since is a two sided test the p value would given by:  
 
  
5) Conclusion  
If we compare the p value and the significance level assumed  we see that
 we see that  so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significant different from 100 at 5% of significance.
 so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significant different from 100 at 5% of significance.