/*! 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 40 A drug manufacturer claims that ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A drug manufacturer claims that less than \(10 \%\) of patients who take its new drug for treating Alzheimer's disease will experience nausea. To test this claim, researchers conduct an experiment. They give the new drug to a random sample of 300 out of 5000 Alzheimer's patients whose families have given informed consent for the patients to participate in the study. In all, 25 of the subjects experience nausea. Use these data to perform a test of the drug manufacturer's claim at the \(\alpha=0.05\) significance level.

Short Answer

Expert verified
There is not enough evidence to support that less than 10% experience nausea.

Step by step solution

01

Set Up Hypotheses

First, we need to define the null hypothesis (H0) and the alternative hypothesis (H1). The null hypothesis is that the proportion of patients experiencing nausea is equal to 10%, and the alternative hypothesis is that the proportion is less than 10%.- Null Hypothesis (H0): \( p = 0.10 \)- Alternative Hypothesis (H1): \( p < 0.10 \)
02

Determine Sample Proportion

Calculate the sample proportion of patients who experienced nausea:\[ \hat{p} = \frac{25}{300} = \frac{25}{300} \approx 0.0833 \]
03

Calculate Standard Deviation

Use the standard deviation formula for a proportion:\[ \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.10 (1-0.10)}{300}} = \sqrt{\frac{0.10 \times 0.90}{300}} = \sqrt{\frac{0.09}{300}} \approx 0.0173 \]
04

Compute Test Statistic

Find the z-score using the formula:\[ z = \frac{\hat{p} - p}{\sigma_{\hat{p}}} = \frac{0.0833 - 0.10}{0.0173} \approx \frac{-0.0167}{0.0173} \approx -0.966 \]
05

Determine Critical Value and Decision

At \(\alpha = 0.05\) significance level for a one-tailed test, the critical z-value is approximately -1.645 (from standard normal distribution tables). Compare the computed z-score to the critical value.- If \( z < -1.645 \), reject H0.- Since \(-0.966 > -1.645\), we fail to reject H0.
06

State Conclusion

We failed to reject the null hypothesis that the proportion of patients experiencing nausea is 10%. Therefore, there is insufficient evidence to support the drug manufacturer's claim that less than 10% of patients will experience nausea.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Proportion
A proportion is a type of ratio that compares a part to a whole. In hypothesis testing, proportions are used to understand the fraction or percentage of a population that displays a certain characteristic. For instance, in our drug test example, we're interested in the proportion of patients who experience nausea after taking a new drug.
The proportion represents the ratio of the number of favorable outcomes (e.g., patients experiencing nausea) to the total number of observations or trials (e.g., total patients tested). In our example, the sample proportion, denoted as \( \hat{p} \), is calculated using the formula:
  • \( \hat{p} = \frac{\text{Number of patients experiencing nausea}}{\text{Total number of patients in the sample}} \)
  • \( \hat{p} = \frac{25}{300} = 0.0833 \).
This tells us that approximately 8.33% of the sample population experienced nausea.
Significance Level
The significance level, denoted by \( \alpha \), is a critical component in hypothesis testing. It is the probability of rejecting the null hypothesis when it is actually true, essentially representing the risk of making a Type I error. In simpler terms, it is the threshold we set for deciding whether an observed effect is statistically significant.
For the drug example, we use a significance level of \( \alpha = 0.05 \). This means we have a 5% risk of concluding that the proportion of patients experiencing nausea is different from the claimed \(10\%\), even if it isn't true. It helps us determine the critical value against which we'll compare our test statistic.
  • If our test statistic is more extreme than the critical value, we reject the null hypothesis.
  • Otherwise, we do not reject the null hypothesis.
In our case, the critical z-value is approximately -1.645 for a one-tailed test at the 0.05 significance level.
Standard Deviation
Standard deviation measures the amount of variation or dispersion in a set of values. When calculating proportions, the standard deviation is crucial for assessing how much the sample proportion might differ from the population proportion.
To compute the standard deviation for a sample proportion, we use the formula:
  • \( \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} \)
In the context of our drug test experiment, \( p \) is the hypothesized population proportion (0.10) and \( n \) is the sample size (300). Plugging these numbers into the formula gives us:
  • \( \sigma_{\hat{p}} = \sqrt{\frac{0.10 \times 0.90}{300}} \approx 0.0173 \).
This tells us that the standard deviation of our sample proportion is approximately 0.0173, providing insight into the expected variability of our sample proportion around the hypothesized population proportion.
Null Hypothesis
The null hypothesis, often referred to as \( H_0 \), is a crucial starting point for hypothesis testing. It represents the statement we aim to test, usually positing that there is no effect or no difference. In many cases, as with the drug test example, the null hypothesis serves as a baseline claim that we want to challenge.
For the manufacturer's claim about the drug, the null hypothesis \( H_0 \) is:
  • \( p = 0.10 \)
This proposes that the true proportion of patients experiencing nausea is exactly 10%. The alternative hypothesis \( H_1 \) here is:\( p < 0.10 \). This outlines that the proportion is actually less than 10%, supporting the manufacturer's claim.
In hypothesis testing, the objective is to gather enough evidence to either reject the null hypothesis or fail to reject it, based on the sample data. In this case, since the test statistic did not pass the critical threshold, we failed to reject \( H_0 \), indicating that we did not find significant evidence against the assumption that 10% of patients experience nausea.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Of the 24,611 degrees in mathematics given by U.S. colleges and universities in a recent year, \(70 \%\) were bachelor's degrees, \(24 \%\) were master's degrees, and the rest were doctorates. Moreover, women earned \(43 \%\) of the bachelor's degrees, \(41 \%\) of the master's degrees, and \(29 \%\) of the doctorates. (a) How many of the mathematics degrees given in this year were earned by women? Justify your answer. (b) Are the events "degree earned by a woman" and "degree was a bachelor's degree" independent? Justify your answer using appropriate probabilities. (c) If you choose 2 of the 24,61 l mathematics degrees at random, what is the probability that at least 1 of the 2 degrees was earned by a woman? Show your work.

The \(z\) statistic for a test of \(H_{0}: p=0.4\) versus \(H_{a}: p \neq 0.4\) is \(z=2.43 .\) This test is (a) not significant at either \(\alpha=0.05\) or \(\alpha=0.01\). (b) significant at \(\alpha=0.05\) but not at \(\alpha=0.01\). (c) significant at \(\alpha=0.01\) but not at \(\alpha=0.05\). (d) significant at both \(\alpha=0.05\) and \(\alpha=0.01\). (e) inconclusive because we don't know the value of \(\hat{p}\).

You read that a statistical test at the \(\alpha=0.01\) level has probability 0.14 of making a Type II error when a specific alternative is true. What is the power of the test against this alternative?

A study with 5000 subjects reported a result that was statistically significant at the \(5 \%\) level. Explain why this result might not be particularly large or important.

Experiments on learning in animals sometimes measure how long it takes mice to find their way through a maze. The mean time is 18 seconds for one particular maze. A researcher thinks that a loud noise will cause the mice to complete the maze faster. She measures how long each of 10 mice takes with a noise as stimulus. The appropriate hypotheses for the significance test are (a) \(H_{0}: \mu=18 ; H_{a}: \mu \neq 18\). (b) \(H_{0}: \mu=18 ; H_{a}: \mu>18\). (c) \(H_{0}: \mu<18 ; H_{a}: \mu=18\) (d) \(H_{0}: \mu=18 ; H_{a}: \mu<18\). (e) \(H_{0}: \bar{x}=18 ; H_{a}: \bar{x}<18\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.