given that H0: μ=40 against H1: μ < 40 if mice have an average life of 38 months with a standard deviation of 5.8 months. If the distribution of life spans is approximately normal, how large a sample is required in order that the probability of committing a type II error be 0.1 when the true mean is 35.9 months? Assume that level of significance 0.05.

sample required n = 18

Step-by-step explanation:

Given that the value under under null hypothesis is 40 while the value under the alternative is less than 40, specifically 35.9

∴ H₀ : u = 40

  H₁  : u = 35.9

therefore β = ( 35.9 - 40 ) = -4.1

The level of significance ∝ = 0.05

Probability of committing type 11 error P = 0.1

standard deviation α = 5.8

Therefore our z-vales (z table)

Z₀.₅ = 1.645

Z₀.₁ = 1.282

NOW let n be sample size

n = {( Z₀.₅ + Z₀.₁ )² × α²} / β²

n = {( 1.645 + 1.282 )² × 5.8²} / (- 4.1)²

n = 17.14485

Since we are talking about sample size; it has to be a whole number

therefore

sample required n = 18

-10i^2-4i

step-by-step explanation:

evaluate for p=2,q=5i

(2−4)(2+5i)i

(2−4)(2+5i)i

=−10i2−4i

answer: r there any options?

step-by-step explanation: