Question

AIDS Research Scientists have shown that a newly developed vaccine can shield rhesus monkeys from infection by the SIV virus, a virus closely related to the HIV virus which affects humans. In their work, researchers gave each of \(n=6\) rhesus monkeys five inoculations with the SIV vaccine and one week after the last vaccination, each monkey received an injection of live SIV. Two of the six vaccinated monkeys showed no evidence of SIV infection for as long as a year and a half after the SIV injection. \({ }^{5}\) Scientists were able to isolate the SIV virus from the other four vaccinated monkeys, although these animals showed no sign of the disease. Does this information contain sufficient evidence to indicate that the vaccine is effective in protecting monkeys from SIV? Use \(\alpha=.10 .\)

Short answer

Answer: No, there is not enough evidence to conclude that the vaccine is effective in protecting monkeys from SIV at a significance level of 0.10, as the probability of observing at least 2 successes (protected monkeys) in the study is 0, which is less than the given significance level. More studies with larger sample sizes should be conducted to obtain more conclusive evidence on the vaccine's effectiveness.

Step-by-step solution

Set up the hypothesis test

The Null hypothesis (H0) and the Alternative hypothesis (Ha) for this problem are given as: H0: \(p = 0\) (The vaccine has no effect) Ha: \(p > 0\) (The vaccine effectively protects monkeys from SIV)

Calculate the test statistic

The test statistic for a hypothesis test for proportions can be calculated using the formula: \(z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\) Here, \(\hat{p}\) is the sample proportion (2/6) \(n\) is the number of monkeys (6) \(p_0\) is the proportion under the null hypothesis (0) Now, let's plug in the values and calculate the test statistic: \(z = \frac{\frac{2}{6} - 0}{\sqrt{\frac{0(1-0)}{6}}} = \frac{\frac{1}{3}}{\sqrt{\frac{0}{6}}}\) Since the denominator is 0, we cannot calculate the z-value. This indicates that we need a different approach.

Perform a one-sample proportion Binomial test

Since the sample size is small, we can perform a one-sample proportion Binomial test instead. Using the binomial probability formula, we calculate the probability of observing at least 2 successes (protected monkeys) out of 6 trials, given the null hypothesis (\(p=0\)) is true. The binomial probability formula is: \(P(X \geq x) = \sum_{k=x}^{n} {n \choose k}p^k(1-p)^{n-k}\) Here, \(n\) is the number of monkeys (6) \(k\) is the number of successes (at least 2 protected) \(p\) is the success probability under the null hypothesis (0) Since the success probability is 0, the vaccine is considered not effective for all monkeys in the calculation. Therefore, the probability of observing at least 2 successes is also equal to 0 in this case.

Compare the probability to the significance level

According to the alternative hypothesis (Ha), the vaccine is effective when the proportion of protected monkeys is greater than 0. We found that the probability of observing at least 2 successes (protected monkeys) in our study is also 0. The probability we calculated (0) is less than the given significance level (\(\alpha = 0.10\)), which means that the data is not enough to reject the null hypothesis and conclude that the vaccine is effective in protecting monkeys from SIV. More studies with larger sample sizes should be performed to gain more conclusive evidence on the effectiveness of the vaccine.

Key concepts

Binomial Probability Formula

The binomial probability formula is a fundamental tool when dealing with discrete probability distributions in statistics. It is particularly useful in determining the likelihood of a specific number of successes in a fixed number of independent trials.

For a given number of trials, denoted by \(n\), the probability of observing exactly \(k\) successes, where the success probability is \(p\), the binomial probability is given by:
\[ P(X = k) = {n \choose k}p^k(1-p)^{n-k} \]
This equation uses the combination function \({n \choose k}\), which computes the number of ways to choose \(k\) successes out of \(n\) trials. The term \(p^k\) represents the probability of having \(k\) successes, while \((1-p)^{n-k}\) is the probability of the remaining trials resulting in failures.

When conducting a hypothesis test for proportions, this formula allows us to calculate the probability of outcomes under the null hypothesis. In cases where the sample size is small, or the expected number of successes is low, using the binomial formula over a normal approximation can be more accurate.

One-Sample Proportion Test

An essential aspect of statistical analysis is the one-sample proportion test, which is used to infer if a sample proportion of successes from one group differs significantly from a known or hypothesized population proportion.

The test statistic for a one-sample proportion test is typically calculated by:
\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \]
where \(\hat{p}\) is the observed sample proportion, \(p_0\) is the hypothesized population proportion under the null hypothesis, and \(n\) is the sample size.

This z-score tells us how many standard deviations our sample proportion is from the hypothesized population proportion. However, if the sample size or the success probability under the null hypothesis is very small, such as in the given problem, the z-score cannot be used. Indeed, we might instead use a binomial test as an alternative. This test employs the aforementioned binomial probability formula to consider all possible outcomes that could support the alternative hypothesis.

Null and Alternative Hypothesis

When engaging in hypothesis testing, researchers formulate two competing hypotheses: the null hypothesis \((H_0)\) and the alternative hypothesis \((H_a)\).

The null hypothesis is a statement of no effect or no difference, which in the case of proportion tests, often posits that a certain population proportion equals a specific value. For instance, stating \(H_0: p = 0\) suggests that a treatment or intervention has no effect.

On the other hand, the alternative hypothesis suggests that there is an effect or difference. It can be one-sided, indicating that the parameter is either greater than or less than the null hypothesis value, like \(H_a: p > 0\), or two-sided, implying that the parameter is simply different from the value stated in the null hypothesis.

Determining whether to reject the null hypothesis involves calculating a test statistic and comparing it with a critical value or p-value. If evidence against the null hypothesis is strong enough (typically if the p-value is less than the chosen significance level \(\alpha\)), the null is rejected in favor of the alternative. Otherwise, there is insufficient evidence to do so, and the null hypothesis is not rejected.