A meat inspector has randomly selected 30 packs of 95% lean beef. The sample resulted in a mean of 96.2% with a sample standard deviation of 0.8%. Calculate an upper prediction bound for the leanness of a new pack using a prediction level of 99%. Assume normality. The contents of seven similar containers of sulfuric acid are 9.8, 10.2, 10.4, 9.8,10.0, 10.2, and 9.6 liters. Find a 95% confidence interval for the mean contents of all such containers, assuming an approximately normal distribution.

a

 The upper bound of the 99% prediction level is  

b

 The  95% confidence interval is

Step-by-step explanation:

Considering first question

From the question we are told that

   The sample size is  n  =  30  

   The sample mean is  

   The standard deviation is

Generally the degree of freedom is mathematically represented as

=>      

=>      

From the question we are told the confidence level is  99% , hence the level of significance is    

=>  

Generally from the t distribution table the critical value  of   at a degree of freedom of is  

Generally the  99%  prediction level is mathematically represented as

Generally the upper bound of the 99%  prediction level is mathematically represented as

=>      

=>      

Considering second question

 Generally the sample is mathematically represented as

=>            

=>              

Generally the standard deviation  is mathematically represented as

=>        

=>        

Generally the degree of freedom is mathematically represented as

=>    

=>    

From the question we are told the confidence level is  95% , hence the level of significance is    

=>  

Generally from the t distribution table the critical value  of   at a degree of freedom of  is  

Generally the margin of error is mathematically represented as  

=>    

=>    

Generally 95% confidence interval is mathematically represented as  

=>  

=>