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The chance of developing breast cancer is under 11% for women. 10 Ho: H = 11. He μ = 11 |ο Hoμ 211 Hp <11 Ho: s11, H₂ μ> 11 Ho: p=0.11, H₂p= 0.11 O Ho <11, H₂ μZ 11 Ho: p20.11, H₂; p< 0.11 Ho p≤0.11, H₂p> 0.11 Ho: p<0.11, H₂. pz 0.11 Incorrect. The given statement is considering a population proportion and not a population mean. Part (1) X Private universities' mean tuition cost is more than $20,000 per year. Ho: H=20,000, H₂ μ = 20,000 ọ Ho-l>20,000, Ο Ho: a s 20.000, O Ho: p=20,000, O Hops 20,000, O Ho:p> 20,000, Ο Ho: μ ε 20.000, O Ho: pz 20,000. Hg με 20.000 Hg a > 20.000 H₁: p= 20,000 H₂ p > 20,000 Hps 20,000 Hg με 20.000 H: p<20,000

The chance of developing breast cancer is under 11% for women. 10 Ho: H = 11. He μ = 11 |ο Hoμ 211 Hp <11 Ho: s11, H₂ μ> 11 Ho: p=0.11, H₂p= 0.11 O Ho <11, H₂ μZ 11 Ho: p20.11, H₂; p< 0.11 Ho p≤0.11, H₂p> 0.11 Ho: p<0.11, H₂. pz 0.11 Incorrect. The given statement is considering a population proportion and not a population mean. Part (1) X Private universities' mean tuition cost is more than $20,000 per year. Ho: H=20,000, H₂ μ = 20,000 ọ Ho-l>20,000, Ο Ho: a s 20.000, O Ho: p=20,000, O Hops 20,000, O Ho:p> 20,000, Ο Ho: μ ε 20.000, O Ho: pz 20,000. Hg με 20.000 Hg a > 20.000 H₁: p= 20,000 H₂ p > 20,000 Hps 20,000 Hg με 20.000 H: p<20,000

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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## Hypothesis Testing Examples ### Part (i) **Problem Statement:** The chance of developing breast cancer is under 11% for women. **Hypotheses Options:** - \( H_0: \mu = 11, \quad H_a: \mu < 11 \) - \( H_0: \mu = 11, \quad H_a: \mu \geq 11 \) - \( H_0: \mu = 0.11, \quad H_a: \mu < 0.11 \) - \( H_0: p = 0.11, \quad H_a: p < 0.11 \) - \( H_0: \mu < 11, \quad H_a: \mu \geq 11 \) - \( H_0: p \geq 0.11, \quad H_a: p < 0.11 \) - \( H_0: p < 0.11, \quad H_a: p \geq 0.11 \) - \( H_0: p < 0.11, \quad H_a: p \geq 0.11 \) (Selected and marked incorrect) **Explanation:** The statement correctly identifies that the problem is considering a population proportion (\( p \)), not a population mean (\( \mu \)). Therefore, the correct null and alternative hypotheses should involve \( p \). --- ### Part (j) **Problem Statement:** Private universities' mean tuition cost is more than $20,000 per year. **Hypotheses Options:** - \( H_0: \mu = 20,000, \quad H_a: \mu \neq 20,000 \) - \( H_0: \mu > 20,000, \quad H_a: \mu \leq 20,000 \) - \( H_0: \mu \leq 20,000, \quad H_a: \mu > 20,000 \) - \( H_0: p = 20,000, \quad H_a: p \neq 20,000 \) - \( H_0: p = 20,000, \quad H_a: p > 20,000 \) - \( H_0: \mu \leq 20,000, \quad H_a: \
Transcribed Image Text:## Hypothesis Testing Examples ### Part (i) **Problem Statement:** The chance of developing breast cancer is under 11% for women. **Hypotheses Options:** - \( H_0: \mu = 11, \quad H_a: \mu < 11 \) - \( H_0: \mu = 11, \quad H_a: \mu \geq 11 \) - \( H_0: \mu = 0.11, \quad H_a: \mu < 0.11 \) - \( H_0: p = 0.11, \quad H_a: p < 0.11 \) - \( H_0: \mu < 11, \quad H_a: \mu \geq 11 \) - \( H_0: p \geq 0.11, \quad H_a: p < 0.11 \) - \( H_0: p < 0.11, \quad H_a: p \geq 0.11 \) - \( H_0: p < 0.11, \quad H_a: p \geq 0.11 \) (Selected and marked incorrect) **Explanation:** The statement correctly identifies that the problem is considering a population proportion (\( p \)), not a population mean (\( \mu \)). Therefore, the correct null and alternative hypotheses should involve \( p \). --- ### Part (j) **Problem Statement:** Private universities' mean tuition cost is more than $20,000 per year. **Hypotheses Options:** - \( H_0: \mu = 20,000, \quad H_a: \mu \neq 20,000 \) - \( H_0: \mu > 20,000, \quad H_a: \mu \leq 20,000 \) - \( H_0: \mu \leq 20,000, \quad H_a: \mu > 20,000 \) - \( H_0: p = 20,000, \quad H_a: p \neq 20,000 \) - \( H_0: p = 20,000, \quad H_a: p > 20,000 \) - \( H_0: \mu \leq 20,000, \quad H_a: \
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