Question

Reserve Problems Chapter 9 Section 1 Problem 5 The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 120, and the standard deviation is 5. Calculate the probability of a type II error if the true mean heat evolved is 123 and (Round your answers to two decimal places (e.g. 98.76).) (a) a = 0.01 and n= 4B=L (b) a = 0.01 and n=6 B = LINK TO TEXT (c) Compare the values of ß calculated in previous parts. What conclusions can you draw? As n increases, B increases. As n increases, ß decreases. LINK TO TEXT

Answer 1

type ii error = probability of fail to reject the null hypotheis when null is false

we fail to reject the null if
120 - z *5/sqrt(n) < Xbar < 120 + z * 5/sqrt(n)

z = 2.576 for 0.01 alpha

for mu = 123

type ii error =
P(120 - z *5/sqrt(n) < Xbar < 120 + z * 5/sqrt(n) | mu = 123)
= P(-3/(5/sqrt(n)) - 2.576 <Z < -3/(5/sqrt(n)) + 2.576)

when n = 4
beta = P(-3.776 < Z < 1.376)
= 0.9155
=0.92

when n= 6
beta = P(-4.0456 < Z < 1.1063)
= 0.8657
=0.87

c)
as n increases, beta decreases