/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 Many consumers are turning to ge... [FREE SOLUTION] | 91Ó°ÊÓ

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Many consumers are turning to generics as a way of reducing the cost of prescription medications. The article "Commercial Information on Drugs: Confusing to the Physician?" (J. of Drug Issues, 1988: 245-257) gives the results of a survey of 102 doctors. Only 47 of those surveyed knew the generic name for the drug methadone. Does this provide strong evidence for concluding that fewer than half of all physicians know the generic name for methadone? Carry out a test of hypotheses using a significance level of \(.01\) using the \(P\)-value method.

Short Answer

Expert verified
There is not enough evidence to conclude that fewer than half of all physicians know the generic name for methadone at the 0.01 significance level.

Step by step solution

01

State Hypotheses

We start by defining the null and alternative hypotheses. The null hypothesis, \( H_0 \), is that half or more of all physicians know the generic name for methadone, which is mathematically expressed as \( p \geq 0.5 \). The alternative hypothesis, \( H_a \), is that fewer than half of all physicians know the generic name, expressed as \( p < 0.5 \).
02

Determine Test Statistic

The test statistic for a proportion can be calculated using the formula: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} \] where \( \hat{p} \) is the sample proportion, \( p_0 \) is the hypothesized population proportion, and \( n \) is the sample size. Here, \( \hat{p} = \frac{47}{102} \approx 0.4608 \), \( p_0 = 0.5 \), and \( n = 102 \).
03

Calculate Test Statistic

Substitute the known values into the test statistic formula: \( z = \frac{0.4608 - 0.5}{\sqrt{\frac{0.5 \times 0.5}{102}}} \). Calculating the denominator: \( \sqrt{\frac{0.25}{102}} \approx 0.0495 \). Then, \( z = \frac{-0.0392}{0.0495} \approx -0.79 \).
04

Find P-Value

Using standard normal distribution tables or a calculator, we find the \( P \)-value corresponding to the calculated \( z \)-value of \(-0.79\). Since this is a lower-tailed test, we look for \( P(Z < -0.79) \). The \( P \)-value is approximately \( 0.2148 \).
05

Compare P-Value with Significance Level

The \( P \)-value \( 0.2148 \) is greater than the significance level \( \alpha = 0.01 \). Since \( P > \alpha \), we fail to reject the null hypothesis.
06

Conclusion

At the \( 0.01 \) significance level, there is not strong enough evidence to conclude that fewer than half of all physicians know the generic name for methadone.

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Key Concepts

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

Significance Level
The significance level, often represented by \( \alpha \), is a critical threshold used in hypothesis testing. It helps you determine whether the evidence from your sample is strong enough to reject the null hypothesis. In simpler terms, it's the probability of making a Type I error—incorrectly rejecting the null hypothesis when it is true.
In our specific example, we are working with a significance level of \( 0.01 \). This means we are setting a stringent requirement for evidence. Only if the evidence is strong enough to bring the \( P \)-value below \( 0.01 \), will we reject the null hypothesis.
Why choose \( 0.01 \)? A lower significance level, like \( 0.01 \), indicates a high level of confidence in the results. It reduces the chance of a false positive, though it may also increase the chance of a false negative (Type II error). Balancing these risks often depends on the context of your study and how much error you are willing to accept.
Null and Alternative Hypotheses
The null and alternative hypotheses are the foundational concepts of hypothesis testing. They form the basis for any test you carry out to infer about a population from a sample.
- **Null Hypothesis \( (H_0) \):** This hypothesis represents the status quo or a statement of no effect. In our example, it states that half or more of all physicians know the generic name for methadone. Mathematically, this is represented as \( p \geq 0.5 \). The null hypothesis is what you initially assume to be true.- **Alternative Hypothesis \( (H_a) \):** This is what you hope to support by your test. The alternative hypothesis suggests a new effect or situation. For the doctors’ knowledge survey, the alternative hypothesis claims that fewer than half know the generic name, represented as \( p < 0.5 \).
Setting these hypotheses clearly is crucial because they guide the selection of the statistical test and the interpretation of data. The test result will help you determine which hypothesis to support based on your evidence.
P-value Method
The \( P \)-value method is a popular approach used in hypothesis testing to help decide whether to reject the null hypothesis. A \( P \)-value tells you how likely you would observe the sampled data, or something more extreme, assuming the null hypothesis were true.
Here's a step-by-step on how this works:1. **Calculate the Test Statistic:** This involves computing a value from your sample data that can be compared to a standard distribution. In our case, we used a proportion \( z \)-test, calculating a \( z \)-value.2. **Find the \( P \)-value:** Using statistical tables or software, find the \( P \)-value corresponding to the test statistic. This \( P \)-value gives you the probability of observing your data, assuming the null hypothesis is true. For example, a \( z \)-value of \(-0.79\) led us to a \( P \)-value of approximately \( 0.2148 \).3. **Compare the \( P \)-value to \( \alpha \):** Decide whether to reject the null hypothesis based on whether the \( P \)-value is less than \( \alpha \). In our scenario, because \( 0.2148 > 0.01 \), we fail to reject the null hypothesis.
Using the \( P \)-value method is intuitive because it provides a specific probability, allowing one to easily interpret the strength of evidence provided by the data.

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Most popular questions from this chapter

Suppose the population distribution is normal with known \(\sigma\). Let \(\gamma\) be such that \(0<\gamma<\alpha\). For testing \(H_{0}: \mu=\mu_{0}\) versus \(H_{\mathrm{a}}: \mu \neq \mu_{0}\), consider the test that rejects \(H_{0}\) if either \(z \geq z_{\gamma}\) or \(z \leq-z_{\alpha-\gamma}\), where the test statistic is \(Z=\left(\bar{X}-\mu_{0}\right) /\) \((\sigma / \sqrt{n})\) a. Show that \(P\) (type I error) \(=\alpha\). b. Derive an expression for \(\beta\left(\mu^{\prime}\right)\). [Hint: Express the test in the form "reject \(H_{0}\) if either \(\bar{x} \geq c_{1}\) or \(\leq c_{2}\)."] c. Let \(\Delta>0\). For what values of \(\gamma\) (relative to \(\alpha\) ) will \(\beta\left(\mu_{0}+\Delta\right)<\beta\left(\mu_{0}-\Delta\right) ?\)

Pairs of \(P\)-values and significance levels, \(\alpha\), are given. For each pair, state whether the observed \(P\)-value would lead to rejection of \(H_{0}\) at the given significance level. a. \(P\)-value \(=.084, \alpha=.05\) b. \(P\)-value \(=.003, \alpha=.001\) c. \(P\)-value \(=.498, \alpha=.05\) d. \(P\)-value \(-.084, \alpha-.10\) e. \(P\)-value \(=.039, \alpha=.01\) f. \(P\)-value \(=.218, \alpha=.10\)

The times of first sprinkler activation for a series of tests with fire prevention sprinkler systems using an aqueous film-forming foam were (in sec) \(\begin{array}{lllllllllllll}27 & 41 & 22 & 27 & 23 & 35 & 30 & 33 & 24 & 27 & 28 & 22 & 24\end{array}\) (see "Use of AFFF in Sprinkler Systems," Fire Technology, 1976: 5). The system has been designed so that true average activation time is at most \(25 \mathrm{sec}\) under such conditions. Does the data strongly contradict the validity of this design specification? Test the relevant hypotheses at significance level \(.05\) using the \(P\)-value approach.

It is known that roughly \(2 / 3\) of all human beings have a dominant right foot or eye. Is there also right-sided dominance in kissing behavior? The article "Human Behavior: Adult Persistence of Head-Turning Asymmetry" (Nature, 2003: 771) reported that in a random sample of 124 kissing couples, both people in 80 of the couples tended to lean more to the right than to the left. a. If \(2 / 3\) of all kissing couples exhibit this right-leaning behavior, what is the probability that the number in a sample of 124 who do so differs from the expected value by at least as much as what was actually observed? b. Does the result of the experiment suggest that the \(2 / 3\) figure is implausible for kissing behavior? State and test the appropriate hypotheses.

The article "Analysis of Reserve and Regular Bottlings: Why Pay for a Difference Only the Critics Claim to Notice?" (Chance, Summer 2005, pp. 9-15) reported on an experiment to investigate whether wine tasters could distinguish between more expensive reserve wines and their regular counterparts. Wine was presented to tasters in four containers labeled A, B, C, and D, with two of these containing the reserve wine and the other two the regular wine. Each taster randomly selected three of the containers, tasted the selected wines, and indicated which of the three he/she believed was different from the other two. Of the \(n=855\) tasting trials, 346 resulted in correct distinctions (either the one reserve that differed from the two regular wines or the one regular wine that differed from the two reserves). Does this provide compelling evidence for concluding that tasters of this type have some ability to distinguish between reserve and regular wines? State and test the relevant hypotheses using the \(P\)-value approach. Are you particularly impressed with the ability of tasters to distinguish between the two types of wine?

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