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Given that \( H_0: \mu = 40 \) against \( H_1: \mu < 40 \), if mice have an average life of 38 months with a standard deviation of 5.8 months, and the distribution of life spans is approximately normal:

How large a sample is required in order for the probability of committing a Type II error to be 0.1 when the true mean is 35.9 months? Assume a level of significance of 0.05.

Answer :

Answer: 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

Final answer:

To determine the sample size required to have a probability of committing a type II error of 0.1, you use the power of the test. The formula is Sample Size = (Zα + Zβ)² * σ² / (μ0 - μ1)². Plugging in the values, a sample size of at least 16 is required.

Explanation:

To determine the sample size required to have a probability of committing a type II error of 0.1, we need to use the power of the test. The power of a test is the probability of correctly rejecting the null hypothesis when the alternative hypothesis is true.

In this case, the null hypothesis is that the true mean is 40 months, while the alternative hypothesis is that the true mean is less than 40 months. The true mean provided is 35.9 months. To calculate the sample size, we need to use the formula:

Sample Size = (Zα + Zβ)² * σ² / (μ0 - μ1)²

Where Zα is the critical z-value for the given level of significance, Zβ is the critical z-value for the desired power of the test, σ is the population standard deviation, and μ0 and μ1 are the mean under the null and alternative hypotheses, respectively.

Since the level of significance is 0.05, the critical z-value for a one-tailed test is -1.645. The critical z-value for a power of 0.9 is approximately 1.28.

Plugging in the values, we get:

Sample Size = (1.645 + 1.28)² * 5.8² / (40 - 35.9)²

Sample Size ≈ 15.39 ≈ 16

Therefore, a sample size of at least 16 is required to have a probability of committing a type II error of 0.1 when the true mean is 35.9 months.

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