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Pregnancy In \(1998,\) a San Diego reproductive clinic reported 42 live births to 157 women under the age of 38 but only 7 live births for 89 clients aged 38 and older. Is this strong evidence of a difference in the effectiveness of the clinic's methods for older women? a) Was this an experiment? Explain. b) Test an appropriate hypothesis and state your conclusion in context. c) If you concluded there was a difference, estimate that difference with a confidence interval and interpret your interval in context.

Short Answer

Expert verified
a) This was not an experiment, but an observational study. b) There is significant evidence of a difference in effectiveness. c) Younger women's success rate is 8.4%-29.2% higher.

Step by step solution

01

Determine if it was an experiment

An experiment involves manipulating one or more variables to observe the effect on a response variable. In this case, the reproductive clinic did not manipulate any variables; instead, they reported observational data based on their clients' age groups. This means it is not an experiment but rather an observational study.
02

Define Hypotheses

We want to test if there's a difference in the effectiveness of the clinic's methods for women under 38 and women 38 and older. Let: \[ H_0: p_1 = p_2 \] (the proportion of live births to women under 38 is equal to those 38 or older) and \[ H_a: p_1 eq p_2 \] (the proportion of live births for women under 38 differs from those 38 or older).
03

Calculate the Test Statistic

Compute the proportions for each group: \[ \hat{p}_1 = \frac{42}{157} \approx 0.267 \] and \[ \hat{p}_2 = \frac{7}{89} \approx 0.079. \] Use the pooled proportion \[ \hat{p} = \frac{42 + 7}{157 + 89} \approx 0.196. \] Calculate the test statistic \( z \) using: \[ z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{157} + \frac{1}{89}\right)}}. \] Substituting the values in gives \[ z \approx \frac{0.267 - 0.079}{\sqrt{0.196 \cdot 0.804 \cdot \left(\frac{1}{157} + \frac{1}{89}\right)}} \approx 3.28. \]
04

Find P-value and Make a Conclusion

Using the standard normal distribution, a \( z \) of 3.28 corresponds to a p-value of about 0.001, which is less than typical significance levels like 0.05. Thus, we reject the null hypothesis in favor of the alternative, meaning there is statistically significant evidence to suggest a difference in effectiveness based on age.
05

Construct Confidence Interval

To estimate the difference in proportions, we use the formula for a confidence interval for two proportions: \[ (\hat{p}_1 - \hat{p}_2) \pm z^* \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}, \] where \( z^* \) is the critical value for a 95% confidence interval, approximately 1.96. Plug in the values to get: \[ (0.267 - 0.079) \pm 1.96 \sqrt{\frac{0.267(1-0.267)}{157} + \frac{0.079(1-0.079)}{89}} \approx 0.188 \pm 0.104. \] Thus, the 95% confidence interval for the difference in proportions is approximately \( (0.084, 0.292) \).
06

Interpret Confidence Interval

The 95% confidence interval \((0.084, 0.292)\) suggests that women under 38 have a higher success rate, with the rate of live births being between 8.4% and 29.2% more than for women aged 38 and older. This indicates a meaningful difference in the clinic's effectiveness based on client age.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Observational Study
An observational study is distinct from an experiment due to how data is collected. In an observational study, researchers observe subjects and measure variables of interest without assigning treatments to the subjects. In the given reproductive clinic scenario, the clinic merely observed the outcomes of pregnancies for women in two different age groups. There was no manipulation or control over any variables by the researchers, making it an observational study.
Observational studies are valuable because they allow researchers to gather insights from naturally occurring situations. However, unlike experiments, they often cannot definitively establish cause and effect because of potential confounding factors. Confounding factors are external variables that might influence the outcome, making it harder to draw firm conclusions about causality. Thus, while observational studies can reveal associations or differences, such as those in the pregnancy success rates across age groups, they must be interpreted with caution when determining causation.
Confidence Interval
A confidence interval is a range of values, derived from the sample data, that is believed to cover the true population parameter with a certain level of confidence. It provides a measure of uncertainty around the estimated effect. In the context of the reproductive clinic data, the confidence interval was used to estimate the difference in the proportion of live births between women under 38 and those 38 and older.
The calculated 95% confidence interval is \(0.084, 0.292\). This interval suggests that we are 95% confident that the true difference in success rates (proportion of live births) between the two age groups lies between 8.4% and 29.2%.
When interpreting confidence intervals, it's crucial to remember that a wide interval suggests more uncertainty about the estimate of the population parameter, whereas a narrow interval indicates more precision. The 95% level is commonly used, indicating high confidence, but not absolute certainty, that the interval contains the true population difference.
Test Statistic
A test statistic is a standardized value that is calculated from sample data during a hypothesis test. It helps determine whether to reject the null hypothesis. In statistics, the test statistic follows a known distribution—the standard normal distribution (z-distribution) in our case.
For the reproductive clinic study, the test statistic was used to compare the proportions of successful pregnancies between two age groups. It was calculated using the difference in sample proportions and the pooled proportion. The formula used was:
\[ z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \]
Substituting the known values resulted in a test statistic of approximately 3.28. This standardized value reflects how many standard deviations the observed difference is from the null hypothesis of no difference. Hence, the more extreme the test statistic, the more evidence there is against the null hypothesis.
P-Value
The p-value is a critical component of hypothesis testing. It represents the probability of obtaining test results at least as extreme as the observed data, assuming that the null hypothesis is true. A smaller p-value indicates stronger evidence against the null hypothesis.
In this example, the p-value associated with the test statistic of 3.28 was found to be approximately 0.001. This p-value is much smaller than common significance levels, such as 0.05, which are often used to determine statistical significance.
With a p-value of 0.001, we have strong evidence to reject the null hypothesis, indicating that there is a statistically significant difference in the effectiveness of the clinic's methods between the two age groups. In decision-making, a low p-value suggests that the observed differences are unlikely due to random chance, pointing toward a genuine disparity in outcomes related to the age of the women.

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