The mean of a population of raw scores is 60 (o, = 16). Your X = 66 (with N= 40). Using the .05 criterion with the region of rejection in the lower tail, should you reject that the sample repre- sents this population and why?
The mean of a population of raw scores is 60 (o, = 16). Your X = 66 (with N= 40). Using the .05 criterion with the region of rejection in the lower tail, should you reject that the sample repre- sents this population and why?
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
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![### Statistical Analysis Question
**Problem Statement:**
The mean of a population of raw scores is 60 (with a standard deviation, σ, of 16). Your sample mean (X̄) is 66, with a sample size (N) of 40. Considering a 0.05 significance level, with the region of rejection in the lower tail, should you reject the idea that the sample represents this population? Explain your reasoning.
**Analysis Process:**
1. **Parameters and Hypotheses:**
- Population mean (μ) = 60
- Population standard deviation (σ) = 16
- Sample mean (X̄) = 66
- Sample size (N) = 40
- Significance level (α) = 0.05
2. **Identify the Null Hypothesis (H0):**
- H0: The sample mean is equal to the population mean (X̄ = μ).
3. **Identify the Alternative Hypothesis (H1):**
- H1: The sample mean is not equal to the population mean (X̄ ≠ μ), specifically, lower tail in this context implies we check if X̄ < μ.
4. **Standard Error Calculation:**
\[
\text{Standard Error (SE)} = \frac{\sigma}{\sqrt{N}} = \frac{16}{\sqrt{40}} = \frac{16}{6.32} \approx 2.53
\]
5. **Calculate the Z-score:**
\[
Z = \frac{X̄ - μ}{SE} = \frac{66 - 60}{2.53} \approx 2.37
\]
6. **Decision Rule:**
- Determine the critical value for a one-tailed test at α = 0.05. For the lower tail, the critical Z-value is approximately -1.645.
- If calculated Z > critical Z (and considering the direction of the alternative, i.e., lower tail), do not reject H0.
7. **Decision:**
- Since 2.37 is greater than -1.645 and lies in the acceptance region for a lower tail test, there is not enough evidence to reject the null hypothesis. Thus, you should not reject that the sample represents the population based on this test.
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Transcribed Image Text:### Statistical Analysis Question
**Problem Statement:**
The mean of a population of raw scores is 60 (with a standard deviation, σ, of 16). Your sample mean (X̄) is 66, with a sample size (N) of 40. Considering a 0.05 significance level, with the region of rejection in the lower tail, should you reject the idea that the sample represents this population? Explain your reasoning.
**Analysis Process:**
1. **Parameters and Hypotheses:**
- Population mean (μ) = 60
- Population standard deviation (σ) = 16
- Sample mean (X̄) = 66
- Sample size (N) = 40
- Significance level (α) = 0.05
2. **Identify the Null Hypothesis (H0):**
- H0: The sample mean is equal to the population mean (X̄ = μ).
3. **Identify the Alternative Hypothesis (H1):**
- H1: The sample mean is not equal to the population mean (X̄ ≠ μ), specifically, lower tail in this context implies we check if X̄ < μ.
4. **Standard Error Calculation:**
\[
\text{Standard Error (SE)} = \frac{\sigma}{\sqrt{N}} = \frac{16}{\sqrt{40}} = \frac{16}{6.32} \approx 2.53
\]
5. **Calculate the Z-score:**
\[
Z = \frac{X̄ - μ}{SE} = \frac{66 - 60}{2.53} \approx 2.37
\]
6. **Decision Rule:**
- Determine the critical value for a one-tailed test at α = 0.05. For the lower tail, the critical Z-value is approximately -1.645.
- If calculated Z > critical Z (and considering the direction of the alternative, i.e., lower tail), do not reject H0.
7. **Decision:**
- Since 2.37 is greater than -1.645 and lies in the acceptance region for a lower tail test, there is not enough evidence to reject the null hypothesis. Thus, you should not reject that the sample represents the population based on this test.
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