Maths help probability of type 2 errors

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TAEuler
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Report Thread starter 2 years ago
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A simple question, but couldn't understand why I got the answer wrong.

A normal distribution with standard deviation 6.25 is being tested at the 5% significance level. The null hypothesis is H0: mean = 15.7 and the alternative is H1: mean < 15.7
a) State the probability of a type I error:
P(type I error) = 0.05 = 5%
b) Find the probability of a type II error if actually mean = 5.7:
To find the critical value:
X - N(15.7, 6.25^2)
using calculator: Inverse Normal: Left tail, 0.05 area, s.d = 6.25, mean = 15.7.
c.v. = 5.42

X - N(5.7, 6.25^2)
P(X> (or equal to) 5.42) = 1 - 0.518
= 0.482
Hence the probability of a type II error is 48.2%

c) Find the probability of a type II error if actually mean = 25.7
X - N (25.7, 6.25^2)
P(X> (or equal to) 5.42) = 1 - 0.9994123

= 5.8 x 10^-4

P(type II error) = 0.0588%

However, textbook says 0.588%, have I missed something or is it a simple incorrect answer in the textbook?
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