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Chapter 10: Problem 75

Although arsenic is known to be a poison, it also has some beneficial medicinal uses. In one study of the use of arsenic to treat acute promyelocytic leukemia (APL), a rare type of blood cell cancer, APL patients were given an arsenic compound as part of their treatment. Of those receiving arsenic, \(42 \%\) were in remission and showed no signs of leukemia in a subsequent examination (Washington Post, November 5,1998 ). It is known that \(15 \%\) of APL patients go into remission after the conventional treatment. Suppose that the study had included 100 randomly selected patients (the actual number in the study was much smaller). Is there sufficient evidence to conclude that the proportion in remission for the arsenic treatment is greater than \(.15\), the remission proportion for the conventional treatment? Test the relevant hypotheses using a. 01 significance level.

Short Answer

Expert verified
The answer requires to conduct a hypothesis test using the given parameters. The null hypothesis is \(H_0: p = 0.15\), and the alternative hypothesis is \(H_1: p > 0.15\). The Z score is calculated using the given data, and the p-value associated with this Z score is determined. The p-value is then compared to the significance level 0.01 to determine if there is sufficient evidence to reject the null hypothesis. The result of the hypothesis test is then interpreted in the context of the question: whether the arsenic treatment has a higher remission rate than the conventional treatment.

Step by step solution

01

State the hypotheses

The null hypothesis (\(H_0\)) is that the proportion of remission in the arsenic treatment group and the conventional treatment group are the same: \(p = 0.15\). The alternative hypothesis (\(H_1\)) is that the proportion of remission in the arsenic treatment group is greater than that in the conventional treatment group: \(p > 0.15\).
02

State the significance level

The significance level, \(\alpha\), is provided in the problem and it is 0.01.
03

Find the test statistic

The test statistic for this problem is a Z score. It can be calculated with the following formula: \( Z = (p_\text{sample} - p_\text{hypothesized}) / \sqrt{(p_\text{hypothesized}(1 - p_\text{hypothesized}) / n}) \). Substituting the given values in, we get \( Z = (0.42 - 0.15) / \sqrt{(0.15 * 0.85) / 100} \). Calculate this to get the Z statistic.
04

Determine the p-value

Check the z-table (or use a statistical program) to find the probability associated with the calculated Z statistic. This is the p-value.
05

Make a decision

If the p-value is less than the significance level, the null hypothesis is rejected, indicating that there is significant evidence to conclude that the proportion in remission for the arsenic treatment is greater than 0.15. If the p-value is more than the significance level, then the null hypothesis is not rejected, indicating that there is not significant evidence.
06

State the conclusion

Interpret the decision in the context of the problem.

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Key Concepts

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

Arsenic Treatment Efficacy
Understanding the effectiveness of arsenic in treating acute promyelocytic leukemia (APL) requires analyzing clinical data to determine if the treatment leads to a higher remission rate compared to conventional methods. To ensure that our analysis is statistically robust, we perform hypothesis testing.

Hypothesis testing in this context helps to validate whether the remission proportion of patients after receiving arsenic treatment significantly exceeds the known remission rate of 15% from conventional treatment. By comparing these remission proportions, healthcare professionals can better assess the potential of arsenic as an effective treatment for APL.
Remission Proportion Comparison
To compare the efficacy of arsenic treatment with traditional therapies for APL, we look closely at the proportion of patients in remission after receiving each type of treatment. In this case, the claim is that the arsenic treatment is more successful in inducing remission.

Formulating Comparisons

We approach this by setting up two hypotheses - a null hypothesis which posits no difference between the treatment effects, and an alternative hypothesis suggesting that arsenic treatment has a higher remission rate. By comparing these hypotheses through statistical testing, we can ascertain whether there is enough evidence to support the superiority of the arsenic treatment's remission rates over the conventional approach.
Test Statistic Calculation
The test statistic is at the heart of hypothesis testing, serving to measure how extreme the observed results are, should the null hypothesis be true. For our problem, we calculate the Z score to serve as our test statistic.

How to Calculate

It involves the difference between the sample proportion and hypothesized proportion under the null hypothesis, normalized by the standard error. The formula \( Z = (p_\text{sample} - p_\text{hypothesized}) / \sqrt{(p_\text{hypothesized}(1 - p_\text{hypothesized}) / n}) \) provides a standardized value that allows comparison across different studies or experiments.
P-value Determination
Once the test statistic is calculated, the next step is to determine the p-value, which reflects the probability of obtaining results at least as extreme as the observed ones, given that the null hypothesis is true.

By using the Z-table or statistical software, we determine the p-value associated with our Z statistic. If the data leads to a p-value that is lower than the significance level set for our test, it indicates a low probability that our results are coincidental, thus lending support to the alternative hypothesis.
Significance Level Analysis
The significance level (\(\alpha\)) is a threshold that decides whether or not the null hypothesis can be rejected, and is chosen prior to conducting the test. This level of significance essentially quantifies our willingness to accept the risk of incorrectly rejecting the null hypothesis when it's true, known as a Type I error.

Our given significance level of 0.01 is strict, meaning we require strong evidence before we reject the null hypothesis and accept that the arsenic treatment's remission rate is greater than 15%. This analysis forms a fundamental part of hypothesis testing, maintaining a balance between sensitivity and reliability of the results.

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Most popular questions from this chapter

Let \(\mu\) denote the true average lifetime for a certain type of pen under controlled laboratory conditions. A test of \(H_{0}: \mu=10\) versus \(H_{a}: \mu<10\) will be based on a sample of size 36. Suppose that \(\sigma\) is known to be \(0.6\), from which \(\sigma_{x}=0.1\). The appropriate test statistic is then $$ z=\frac{\bar{x}-10}{0.1} $$ a. What is \(\alpha\) for the test procedure that rejects \(H_{0}\) if \(z \leq\) \(-1.28 ?\) b. If the test procedure of Part (a) is used, calculate \(\beta\) when \(\mu=9.8\), and interpret this error probability. c. Without doing any calculation, explain how \(\beta\) when \(\mu=9.5\) compares to \(\beta\) when \(\mu=9.8\). Then check your assertion by computing \(\beta\) when \(\mu=9.5\). d. What is the power of the test when \(\mu=9.8 ?\) when \(\mu=9.5 ?\)

Water samples are taken from water used for cooling as it is being discharged from a power plant into a river. It has been determined that as long as the mean temperature of the discharged water is at most \(150^{\circ} \mathrm{F}\), there will be no negative effects on the river's ecosystem. To investigate whether the plant is in compliance with regulations that prohibit a mean discharge water temperature above \(150^{\circ} \mathrm{F}\), researchers will take 50 water samples at randomly selected times and record the temperature of each sample. The resulting data will be used to test the hypotheses \(H_{0}: \mu=150^{\circ} \mathrm{F}\) versus \(H_{a}: \mu>150^{\circ} \mathrm{F}\). In the context of this example, describe Type I and Type II errors. Which type of error would you consider more serious? Explain.

The poll referenced in the previous exercise ("Military Draft Study," AP- Ipsos, June 2005) also included the following question: "If the military draft were reinstated, would you favor or oppose drafting women as well as men?" Forty-three percent of the 1000 people responding said that they would favor drafting women if the draft were reinstated. Using a \(.05\) significance level, carry out a test to determine if there is convincing evidence that fewer than half of adult Americans would favor the drafting of women.

In a study of computer use, 1000 randomly selected Canadian Internet users were asked how much time they spend using the Internet in a typical week (Ipsos Reid, August 9,2005 ). The mean of the 1000 resulting observations was \(12.7\) hours. a. The sample standard deviation was not reported, but suppose that it was 5 hours. Carry out a hypothesis test with a significance level of \(.05\) to decide if there is convincing evidence that the mean time spent using the Internet by Canadians is greater than \(12.5\) hours. b. Now suppose that the sample standard deviation was 2 hours. Carry out a hypothesis test with a significance level of . 05 to decide if there is convincing evidence that the mean time spent using the Internet by Canadians is greater than \(12.5\) hours.

Let \(\pi\) denote the proportion of grocery store customers that use the store's club card. For a large sample \(z\) test of \(H_{0}: \pi=.5\) versus \(H_{a}: \pi>.5\), find the \(P\) -value associated with each of the given values of the test statistic: a. \(1.40\) d. \(2.45\) b. \(0.93\) e. \(-0.17\) c. \(1.96\)
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