/*! 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 15 A recent study \(^{4}\) consider... [FREE SOLUTION] | 91Ó°ÊÓ

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A recent study \(^{4}\) considered whether dogs could be trained to detect whether a person has lung cancer or breast cancer by smelling the subject's breath. The researchers trained five ordinary household dogs to distinguish, by scent alone, exhaled breath samples of 55 lung and 31 breast cancer patients from those of 83 healthy controls. A dog gave a correct indication of a cancer sample by sitting in front of that sample when it was randomly placed among four control samples. Once trained, the dogs' ability to distinguish cancer patients from controls was tested using breath samples from subjects not previously encountered by the dogs. (The researchers blinded both dog handlers and experimental observers to the identity of breath samples.) Let \(p\) denote the probability a dog correctly detects a cancer sample placed among five samples when the other four are controls. a. Set up the null hypothesis that the dog's predictions correspond to random guessing. b. Set up the alternative hypothesis to test whether the probability of a correct selection differs from random guessing. c. Set up the alternative hypothesis to test whether the probability of a correct selection is greater than with random guessing. d. In one test with 83 Stage I lung cancer samples, the dogs correctly identified the cancer sample 81 times. The test statistic for the alternative hypothesis in part \(\mathrm{c}\) was \(z=17.7 .\) Report the \(\mathrm{P}\) -value to three decimal places and interpret. (The success of dogs in this study made researchers wonder whether dogs can detect cancer at an earlier stage than conventional methods such as MRI scans.

Short Answer

Expert verified
a. \( H_0: p = 0.2 \) b. \( H_a: p \neq 0.2 \) c. \( H_a: p > 0.2 \) d. P-value: 0.000, reject \( H_0 \).

Step by step solution

01

Null Hypothesis

We define the null hypothesis \( H_0 \) as the scenario where the dog's ability to detect a cancer sample is equivalent to random guessing. In this context, since there is 1 cancer sample among 5 samples (4 controls), the probability of randomly guessing correctly is \( p = \frac{1}{5} = 0.2 \). Thus, the null hypothesis is: \( H_0: p = 0.2 \).
02

Alternative Hypothesis (Different from Guessing)

For part (b), we consider the alternative hypothesis \( H_a \) where the probability of a correct detection by the dog differs from random guessing. This can be expressed as \( H_a: p eq 0.2 \). This is a two-tailed hypothesis test.
03

Alternative Hypothesis (Greater than Guessing)

For part (c), we set up a one-tailed alternative hypothesis to test if the dogs' success rate in detecting cancer is greater than the random chance. The alternative hypothesis is: \( H_a: p > 0.2 \).
04

Calculating P-value for the Test Statistic

In part (d), the test statistic for the hypothesis \( H_a: p > 0.2 \) is given as \( z = 17.7 \). Using a standard normal distribution table or a statistical software, we can find the p-value for \( z = 17.7 \). Since this \( z \) value is very large, indicating a high success rate compared to random guessing, the p-value is near 0. Therefore, to three decimal places, the p-value is calculated to be \( p = 0.000 \).
05

Interpretation of the P-value

A p-value this low suggests that the observed results would be extremely unlikely under the null hypothesis. Therefore, we reject the null hypothesis \( H_0: p = 0.2 \) in favor of the alternative hypothesis \( H_a: p > 0.2 \). This indicates that the dogs' ability to detect cancer samples is significantly better than random guessing.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis serves as a baseline or a default assumption that there is no effect or no difference. In the context of the dogs detecting cancer by scent, the null hypothesis assumes that the dogs' success rate in detecting cancer is purely by random chance.
Here, if there is a set of five samples, one with cancer and four without, the probability of a dog correctly selecting the cancer sample randomly is 0.2, or 20%. This is because there is one favorable outcome in five possible outcomes. Therefore, the null hypothesis is mathematically specified as \( H_0: p = 0.2 \).
By accepting the null hypothesis, we assume that the dogs do not actually detect cancer, but rather their sitting in front of the cancer sample is equivalent to a random guess.
Alternative Hypothesis
The alternative hypothesis presents a scenario that contrasts with the null hypothesis. It posits that any observed effects or differences are not due to randomness. There are two types of alternative hypotheses in our context:
  • **Two-tailed hypothesis**: This tests whether the probability of the dogs detecting cancer is different from random guessing (not equal to 0.2). The formulation is \( H_a: p eq 0.2 \). This can lead to results on two sides; just identifying any difference without directionality.

  • **One-tailed hypothesis**: This more specifically tests if dogs have a greater probability of success than random guessing. We express this alternative hypothesis as \( H_a: p > 0.2 \). It focuses on proving if the trained ability of dogs is indeed higher than random chance.

Selecting the correct type of hypothesis depends on the specific research question or interest, guiding how the results are interpreted.
P-value
A p-value quantifies the probability of obtaining a test statistic as extreme as the observed one, assuming the null hypothesis is true. It helps in determining the strength of evidence against the null hypothesis.
For the exercise, with a test statistic of \( z = 17.7 \) for the alternative hypothesis \( H_a: p > 0.2 \), the p-value is calculated to be close to 0.000. This extremely low p-value suggests that the probability of observing such a significant result under the null hypothesis is virtually zero.
Therefore, with a small p-value, we reject the null hypothesis in favor of the alternative hypothesis. Such a result implies that the dogs' ability to detect cancer by scent is significantly better than random guessing, bolstering the notion that dogs can be trained to detect cancer more reliably than by random chance.
In hypothesis testing, a common threshold for p-values is 0.05, meaning if the p-value is below 0.05, the results are statistically significant. Here, with a p-value of 0.000, the conclusion is profoundly significant.

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Most popular questions from this chapter

When the 583 female workers in the 2012 GSS were asked how many hours they worked in the previous week, the mean was 37.0 hours, with a standard deviation of 15.1 hours. Does this suggest that the population mean work week for females is significantly different from 40 hours? Answer by: a. Identifying the relevant variable and parameter. b. Stating null and alternative hypotheses. c. Reporting and interpreting the P-value for the test statistic value. d. Explaining how to make a decision for the significance level of 0.01

In the webcomic on the link http://xkcd.com/882/, a girl claims that jelly beans cause acne. Scientists investigate and find no link between the two \((p>0.05)\). They are asked to check if jelly beans of a particular color cause acne. They test 20 different colors each at a significance level of \(5 \%\) and find a link between green jelly beans and acne. This leads to a newspaper headline, "Green Jellybeans Cause Acne" where the \(5 \%\) chance of the link is mentioned as \(95 \%\) confidence. When the scientists repeat the same experiment, they are unable to find any link between acne and color of jelly beans. They conclude that the earlier result might be coincidental. Using this example, explain why you need to have some skepticism when research suggests that some therapy or drug has an impact in treating a disease.

In 2004 , New York Attorney General Eliot Spitzer filed a lawsuit against GlaxoSmithKline pharmaceutical company, claiming that the company failed to publish results of one of its studies that showed that an antidepressant drug (Paxil) may make adolescents more likely to commit suicide. Partly as a consequence, editors of 11 medical journals agreed to a new policy to make researchers and companies register all clinical trials when they begin, so that negative results cannot later be covered up. The International Journal of Medical Journal Editors wrote, "Unfortunately, selective reporting of trials does occur, and it distorts the body of evidence available for clinical decision-making." Explain why this controversy relates to the argument that it is misleading to report results only if they are "statistically significant." (Hint: See the subsection of this chapter on misinterpretations of significance tests.)

Example 8 tested a therapy for anorexia, using hypotheses \(\mathrm{H}_{0}: \mu=0\) and \(\mathrm{H}_{a}: \mu \neq 0\) about the population mean weight change \(\mu .\) In the words of that example, what would be (a) a Type I error and (b) a Type II error?

A study considers whether the mean score on a college entrance exam for students in 2010 is any different from the mean score of 500 for students who took the same exam in \(1985 .\) Let \(\mu\) represent the mean score for all students who took the exam in 2010\. For a random sample of 25,000 students who took the exam in \(2010, \bar{x}=498\) and \(s=100\). a. Show that the test statistic is \(t=-3.16\). b. Find the P-value for testing \(\mathrm{H}_{0}: \mu=500\) against \(\mathrm{H}_{a}: \mu \neq 500\) c. Explain why the test result is statistically significant but not practically significant.

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