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It has been estimated that between 1945 and 1971 , as many as 2 million children were born to mothers treated with diethylstilbestrol (DES), a nonsteroidal estrogen recommended for pregnancy maintenance. The FDA banned this drug in 1971 because research indicated a link with the incidence of cervical cancer. The article "Effects of Prenatal Exposure to Diethylstilbestrol (DES) on Hemispheric Laterality and Spatial Ability in Human Males" (Hormones Behav., 1992: 62-75) discussed a study in which 10 males exposed to DES and their unexposed brothers underwent various tests. This is the summary data on the results of a spatial ability test: \(\bar{x}=12.6\) (exposed), \(\bar{y}=13.7\), and standard error of mean difference \(=.5\). Test at level \(.05\) to see whether exposure is associated with reduced spatial ability by obtaining the \(P\)-value.

Step by step solution

01

Define the Hypotheses

We start by defining the null hypothesis (H_0) and the alternative hypothesis (H_a). The null hypothesis states that there is no difference in spatial ability between males exposed to DES and those not exposed, or equivalently a difference of zero:\[H_0: \mu_D = 0\]The alternative hypothesis suggests that exposure to DES is associated with reduced spatial ability, implying a difference less than zero:\[H_a: \mu_D < 0\]where \(\mu_D\) represents the mean difference in test scores between exposed and unexposed males.

02

Calculate the Test Statistic

Next, calculate the test statistic using the provided sample means and standard error of the mean difference. The formula for the test statistic \(t\) is:\[ t = \frac{\bar{x} - \bar{y} - \mu_0}{SE} \]where \(SE\) is the standard error, \(\bar{x}\) and \(\bar{y}\) are the sample means for exposed and unexposed groups, and \(\mu_0 = 0\) is the hypothesized mean difference. Here:\( \bar{x} = 12.6, \bar{y} = 13.7, \text{ and } SE = 0.5. \)Substitute these values:\[ t = \frac{12.6 - 13.7 - 0}{0.5} = \frac{-1.1}{0.5} = -2.2 \]

03

Determine the Degrees of Freedom

The degrees of freedom \(df\) for this test can be determined by the number of paired observations minus one. Since there are 10 males in the study, \[ df = 10 - 1 = 9. \]

04

Find the P-value

Using the t-distribution table or a calculator, find the P-value corresponding to a t-score of -2.2 with 9 degrees of freedom. This is a one-tailed test because we are checking for reduced spatial ability (\(H_a: \mu_D < 0\)). Typically, a t-score of -2.2 with 9 degrees of freedom yields a P-value less than 0.05.

05

Make a Decision

Since the P-value obtained in Step 4 is less than the significance level of 0.05, we reject the null hypothesis \(H_0\). This suggests that exposure to DES is indeed associated with reduced spatial ability.

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Key Concepts

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

Null Hypothesis

The null hypothesis is a fundamental element in hypothesis testing. It's basically a starting assumption that nothing has changed or that there is no effect or difference. In the context of our exercise about the impact of diethylstilbestrol (DES) on spatial ability, the null hypothesis (\(H_0\)) is that there is no difference in spatial abilities between males exposed and those not exposed to DES. Mathematically, this is expressed as:
\[H_0: \mu_D = 0\]
where \(\mu_D\) is the mean difference in spatial ability test scores. A null hypothesis always assumes equality, reflecting the expectation of no effect. It's our "default position."

Alternative Hypothesis

The alternative hypothesis (\(H_a\)) poses a contrast to the null hypothesis. It suggests that there is actually an effect or a difference, which we aim to prove. In our scenario, the alternative hypothesis claims exposure to DES is linked to reduced spatial ability, implying less spatial ability in exposed males than their unexposed siblings. This can be expressed mathematically as:
\[H_a: \mu_D < 0\]
Here, \(\mu_D < 0\) indicates that the mean score for exposed males is less than that for unexposed males. The alternative hypothesis is central to hypothesis testing, as we seek sufficient evidence to support it over the null hypothesis.

T-Distribution

In hypothesis testing, the t-distribution is crucial when working with small sample sizes. The t-distribution is similar to the normal distribution but has heavier tails, meaning there's a greater chance of obtaining values far from the mean. This accounts for the increased variability assumed in smaller samples.
To evaluate our study on DES exposure, we use the t-distribution due to only 10 pairs of siblings being tested. We computed a t-statistic value using the formula:\[t = \frac{\bar{x} - \bar{y} - \mu_0}{SE}\]where \(\bar{x}\) and \(\bar{y}\) are the means of exposed and unexposed groups, \(\mu_0 = 0\), and \(SE\) is the standard error. Given our data:\[ t = \frac{12.6 - 13.7 - 0}{0.5} = -2.2 \]This t-value helps us understand how extreme our observed data is, compared to the null hypothesis.

P-Value

The P-value is a vital tool in hypothesis testing, indicating the probability of observing our data (or something more extreme) if the null hypothesis is true. A smaller P-value suggests stronger evidence against the null hypothesis.
For our study, we determined the P-value using a t-distribution with 9 degrees of freedom (10 paired observations minus one). With a calculated t-score of -2.2, we found a P-value less than 0.05 in a one-tailed test. This signifies that the likelihood of observing such a difference under the null hypothesis is low enough to reject the null hypothesis.

• **Evidence Against Null Hypothesis**: Small P-values (< 0.05) provide evidence to reject \(H_0\).
• **Decision Making**: Since our P-value was less than 0.05, it indicates that exposure to DES likely affects spatial ability negatively.

With a P-value threshold set at 0.05, our analysis concludes DES exposure is significantly associated with reduced spatial ability in males.