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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

Expert verified
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

01

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)
02

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.
03

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.
04

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.

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Key Concepts

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

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.

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Most popular questions from this chapter

Taste Testing In a head-to-head taste test of storebrand foods versus national brands, Consumer Reports found that it was hard to tell the difference. \({ }^{4}\) If the national brand is indeed better than the store brand, it should be judged as better more than \(50 \%\) of the time. a. State the null and alternative hypotheses to be tested. Is this a one- or a two-tailed test? b. Suppose that, of the 25 food categories used for the taste test, the national brand was found to be better than the store brand in 7 of the taste comparisons, while in 10 pairs, the tasters could taste no difference between the two. Use the sign test to test the hypothesis in part a with \(\alpha \approx .05 .\) What practical conclusions can you draw?

A political scientist is studying the relationship between the voter image of a conservative political candidate and the distance (in kilometers) between the residences of the voter and the candidate. Each of 12 voters rated the candidate on a scale of \(1-20\). a. Calculate Spearman's rank correlation coefficient \(r_{s}\) b. Do these data provide sufficient evidence to indicate a negative rank correlation between rating and distance?

Use the information given in Exercises \(8-9\) to calculate Spearman's rank correlation coefficient \(r_{s} .\) Do the data present sufficient evidence to indicate an association between variables \(A\) and \(B\) ? Use \(\alpha=.05 .\) $$\begin{array}{l|rrrrrr}\text { A } & 1.2 & .8 & 2.1 & 3.5 & 2.7 & 1.5 \\\\\hline \text { B } & 1.0 & 1.3 & .1 & -.8 & -.2 & .6\end{array}$$

What three statistical tests are available for testing for a difference in location for two populations when the data are paired? What assumptions are required for each of these tests?

A psychology class performed an experiment to determine whether a recall score in which instructions to form images of 25 words were given differs from an initial recall score for which no imagery instructions were given. Twenty students participated in the experiment with the results listed in the table. $$ \begin{array}{ccc|ccc} \hline & \text { With } & \text { Without } & & \text { With } & \text { Without } \\ \text { Student } & \text { Imagery } & \text { Imagery } & \text { Student } & \text { Imagery } & \text { Imagery } \\ \hline 1 & 20 & 5 & 11 & 17 & 8 \\ 2 & 24 & 9 & 12 & 20 & 16 \\ 3 & 20 & 5 & 13 & 20 & 10 \\ 4 & 18 & 9 & 14 & 16 & 12 \\ 5 & 22 & 6 & 15 & 24 & 7 \\ 6 & 19 & 11 & 16 & 22 & 9 \\ 7 & 20 & 8 & 17 & 25 & 21 \\ 8 & 19 & 11 & 18 & 21 & 14 \\ 9 & 17 & 7 & 19 & 19 & 12 \\ 10 & 21 & 9 & 20 & 23 & 13 \\ \hline \end{array} $$ a. What three testing procedures can be used to test for differences in the distribution of recall scores with and without imagery? What assumptions are required for the parametric procedure? Do these data satisfy these assumptions? b. Use both the sign test and the Wilcoxon signed-rank test to test for differences in the distributions of recall scores under these two conditions. c. Compare the results of the tests in part b. Are the conclusions the same? If not, why not?

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