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4. Women athletes at a university have a long term graduation rate of 67%. Over the past several years, a random sample of 38 women athletes at the school showed that 21 eventually graduated. Does this indicate that the population proportion of women athletes who graduate from this university has changed (either way)? Use a 1% level of significance.

4. Women athletes at a university have a long term graduation rate of 67%. Over the past several years, a random sample of 38 women athletes at the school showed that 21 eventually graduated. Does this indicate that the population proportion of women athletes who graduate from this university has changed (either way)? Use a 1% level of significance.

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
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### Statistical Analysis Problem: Evaluating Graduation Rates of Women Athletes **Problem Statement:** Women athletes at a university have a long-term graduation rate of 67%. Over the past several years, a random sample of 38 women athletes at the school showed that 21 eventually graduated. Does this indicate that the population proportion of women athletes who graduate from this university has changed (either way)? Use a 1% level of significance. **Explanation for Educational Website:** This problem involves conducting a hypothesis test to determine if there is a significant change in the graduation rate of women athletes at a university. The long-term graduation rate is considered to be 67%, but a recent sample suggests a different rate. To address this question statistically: 1. **Null Hypothesis (H₀)**: The population proportion of women athletes who graduate is 67% (p = 0.67). 2. **Alternative Hypothesis (H₁)**: The population proportion of women athletes who graduate is not 67% (p ≠ 0.67). We use the sample data (38 women athletes, with 21 graduates) to test these hypotheses. **Significance Level:** The problem specifies a 1% level of significance (α = 0.01), meaning we will reject the null hypothesis if the probability of observing our sample data under the null hypothesis is less than 1%. **Steps:** 1. **Calculate the sample proportion (p̂):** \[ \hat{p} = \frac{21}{38} \approx 0.553 \] 2. **Standard Error (SE) of the sample proportion:** \[ SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.67 \times (1-0.67)}{38}} \approx 0.077 \] 3. **Z-score calculation:** \[ Z = \frac{\hat{p} - p}{SE} = \frac{0.553 - 0.67}{0.077} \approx -1.52 \] 4. **Determine the critical Z-value at a 1% significance level:** For a two-tailed test at α = 0.01, the critical Z-value is approximately ±2.58. 5. **Decision:** Since -1.52 does not
Transcribed Image Text:### Statistical Analysis Problem: Evaluating Graduation Rates of Women Athletes **Problem Statement:** Women athletes at a university have a long-term graduation rate of 67%. Over the past several years, a random sample of 38 women athletes at the school showed that 21 eventually graduated. Does this indicate that the population proportion of women athletes who graduate from this university has changed (either way)? Use a 1% level of significance. **Explanation for Educational Website:** This problem involves conducting a hypothesis test to determine if there is a significant change in the graduation rate of women athletes at a university. The long-term graduation rate is considered to be 67%, but a recent sample suggests a different rate. To address this question statistically: 1. **Null Hypothesis (H₀)**: The population proportion of women athletes who graduate is 67% (p = 0.67). 2. **Alternative Hypothesis (H₁)**: The population proportion of women athletes who graduate is not 67% (p ≠ 0.67). We use the sample data (38 women athletes, with 21 graduates) to test these hypotheses. **Significance Level:** The problem specifies a 1% level of significance (α = 0.01), meaning we will reject the null hypothesis if the probability of observing our sample data under the null hypothesis is less than 1%. **Steps:** 1. **Calculate the sample proportion (p̂):** \[ \hat{p} = \frac{21}{38} \approx 0.553 \] 2. **Standard Error (SE) of the sample proportion:** \[ SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.67 \times (1-0.67)}{38}} \approx 0.077 \] 3. **Z-score calculation:** \[ Z = \frac{\hat{p} - p}{SE} = \frac{0.553 - 0.67}{0.077} \approx -1.52 \] 4. **Determine the critical Z-value at a 1% significance level:** For a two-tailed test at α = 0.01, the critical Z-value is approximately ±2.58. 5. **Decision:** Since -1.52 does not
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