Question

Q93/Q95. A new cure has been developed for a certain type of cement that should change its mean compressive strength.
a) It is known that the standard deviation of the compressive strength is 130 kg/cm2 and that we may assume that it follows a normal distribution. 9 chunks of cement have been tested and the observed sample mean is X = 4970. Find the 95% confidence interval for the mean of the compressive strength.
b) Now, assume that we do not know the standard deviation of the normal distribution. 9 chunks of cement have been tested, and the measurements are 5001,4945,5008,5018,4991,4990,4968,5020,5003. Find the 95% confidence interval for the mean of the compressive strength.

Answer 1

a)

n=9 , =4970, c=95% , =130

formula for confidence interval is

where Zc is the Z critical value for C= 95%

Zc= 1.96

4885.068 < μ < 5054.932

Confidence interval is = ( 4885.068 , 5054.932)

b)

n=9 ,  c=95%

calculate the sample mean and sample standard deviation

=  4993.78

s= 24.1856

formula for confidence interval is

where tc is the t critical value for C= 95% with df= n-1 = 91 =8

tc= 2.306

4993.78−18.591 < μ <  4993.78+18.591

4975.189 < μ < 5012.371

Confidence interval is = ( 4975.189 , 5012.371)