Question

The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100, and the standard deviation is 2. Find the boundary of the critical region for the Type I error probability and sample size provided. Round your answers to two decimal places (e.g. 98.76). (a) a(alpha)=.01 and n=2 ≤ x̄≤ (b)= a=.05 and n=2 ≤ x̄≤ (c)= a=.01 and n=7 ≤ x̄≤ (d)= a=.05 and n=7 ≤ x̄≤ Please show the exact steps on how to solve this.

Answer 1

Solution:

Given: The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100, and the standard deviation is 2.

That is: Mean =

SD =

Find the boundary of the critical region for the Type I error probability and sample size provided.

Part a) and n = 2

Find z value for two tail area= 0.01

that is for 1 - 0.01/2=1 - 0.005 = 0.9950

Thus look in z table for Area = 0.9950 or its closest area and find corresponding z critical value.

From above table we can see area 0.9950 is in between 0.9949 and 0.9951 and both are at same distance from 0.9950, Hence corresponding z values are 2.57 and 2.58

Thus average of both z values is 2.575

Thus Z_c = 2.575

Thus use following z formula:

for lower limit , use z = -2.575 and upper limit use z = 2.575

and

Thus we get :

Part b) , n= 2

find z value for 1 - 0.05 / 2 = 1- 0.025 = 0.9750 area

Look in z table for Area = 0.9750 or its closest area and find z value.

Area = 0.9750 corresponds to 1.9 and 0.06 , thus z critical value = 1.96

That is : Z_c = 1.96

and

thus

Thus we get :

Part c) a=.01 and n=7

and

thus

Part d) a=.05 and n=7

and

thus