### Statistical Analysis of Graduation Rates Among 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 from 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 5% level of significance.**Objective:**- To determine if the graduation rate of women athletes has significantly changed from the established rate of 67%.**Given Data:**- Long-term graduation rate (population proportion), $ p_0 $ = 67% or 0.67.- Sample size, $ n $ = 38.- Number of graduates in the sample, $ x $ = 21.- Sample proportion, $ \hat{p} $ = $ \frac{x}{n} $ = $ \frac{21}{38} $.**Hypotheses:**- Null Hypothesis, $ H_0 $: $ p = 0.67 $ (Graduation rate has not changed).- Alternative Hypothesis, $ H_1 $: $ p \neq 0.67 $ (Graduation rate has changed).**Level of Significance:**- $ \alpha $ = 0.05 (5%).**Test Statistic (z-score):**Use the formula for the z-score in a proportion test:\[ Z = \frac{\hat{p} - p_0}{ \sqrt{\frac{p_0 (1 - p_0)}{n}} } \]- Calculate $ \hat{p} $:\[ \hat{p} = \frac{21}{38} \approx 0.5526 \]- Calculate the standard error (SE):\[ SE = \sqrt{ \frac{0.67 \times (1 - 0.67)}{38} } \approx 0.0772 \]- Calculate the z-score:\[ Z = \frac{0.5526 - 0.67}{0.0772} \approx -1.521 \]**Decision Rule:**- Compare the calculated z-score with the critical z-value from the standard normal distribution table. For a two-tailed test at the 0.05 significance level, the critical z-values are approximately ±1.96.

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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 5% 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 5% level of significance.

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Concept explainers

2-Proportion Test

In a 2-proportion test, two separate proportions are measured at the same time, and their similarity is evaluated. It is also known as a 2-proportion z-test, and a comparison of two different proportions is done in this method.

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### Statistical Analysis of Graduation Rates Among 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 from 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 5% level of significance.

**Objective:**
- To determine if the graduation rate of women athletes has significantly changed from the established rate of 67%.

**Given Data:**
- Long-term graduation rate (population proportion), $ p_0 $ = 67% or 0.67.
- Sample size, $ n $ = 38.
- Number of graduates in the sample, $ x $ = 21.
- Sample proportion, $ \hat{p} $ = $ \frac{x}{n} $ = $ \frac{21}{38} $.

**Hypotheses:**
- Null Hypothesis, $ H_0 $: $ p = 0.67 $ (Graduation rate has not changed).
- Alternative Hypothesis, $ H_1 $: $ p \neq 0.67 $ (Graduation rate has changed).

**Level of Significance:**
- $ \alpha $ = 0.05 (5%).

**Test Statistic (z-score):**
Use the formula for the z-score in a proportion test:
\[ Z = \frac{\hat{p} - p_0}{ \sqrt{\frac{p_0 (1 - p_0)}{n}} } \]

- Calculate $ \hat{p} $:
\[ \hat{p} = \frac{21}{38} \approx 0.5526 \]

- Calculate the standard error (SE):
\[ SE = \sqrt{ \frac{0.67 \times (1 - 0.67)}{38} } \approx 0.0772 \]

- Calculate the z-score:
\[ Z = \frac{0.5526 - 0.67}{0.0772} \approx -1.521 \]

**Decision Rule:**
- Compare the calculated z-score with the critical z-value from the standard normal distribution table. For a two-tailed test at the 0.05 significance level, the critical z-values are approximately ±1.96.

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