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91Ó°ÊÓ

Side effects 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
Reject the null hypothesis if the p-value is less than 0.05; there's evidence for the claim.

Step by step solution

01

Define the Hypotheses

To test the manufacturer's claim, we need to define the null and alternative hypotheses. The null hypothesis \( H_0 \) states that the proportion of patients experiencing nausea is 10% (or 0.10), while the alternative hypothesis \( H_a \) states that the proportion is less than 10% (or 0.10). Mathematically, this is expressed as:\[ H_0: p = 0.10 \] \[ H_a: p < 0.10 \] where \( p \) is the true proportion of patients experiencing nausea.
02

Calculate the Test Statistic

Using the sample data, calculate the test statistic for the sample proportion. The sample proportion \( \hat{p} \) is calculated as the number of patients experiencing nausea divided by the total sample size, which is \( \hat{p} = \frac{25}{300} = 0.0833 \). The test statistic is a standard normal variable \( z \), calculated by:\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} \] where \( p_0 = 0.10 \) and \( n = 300 \). Substituting the values, we have:\[ z = \frac{0.0833 - 0.10}{\sqrt{\frac{0.10 \times 0.90}{300}}} \]
03

Determine the P-value

Using the calculated \( z \)-value, determine the p-value from the standard normal distribution table. The p-value represents the probability of observing a sample proportion as extreme or more extreme than \( \hat{p} \) under the null hypothesis. Look up the \( z \)-value in the z-table, which corresponds to the left-tail probability (since \( H_a \) is a 'less than' hypothesis).
04

Compare with Significance Level

Compare the p-value with the significance level \( \alpha = 0.05 \). If the p-value is less than \( \alpha \), we reject the null hypothesis. This indicates sufficient evidence to support the claim that the proportion of nauseous patients is less than 10%. If the p-value is greater than \( \alpha \), we fail to reject the null hypothesis.
05

Make a Conclusion

Based on our calculations, if the p-value is less than 0.05, conclude that there is enough evidence to support the manufacturer's claim that less than 10% of patients experience nausea. If the p-value is greater than or equal to 0.05, conclude that there is not enough evidence to support the claim.

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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 denoted by \( \alpha \), is a crucial part of hypothesis testing. It represents the threshold we set for deciding whether to reject the null hypothesis. In a typical test, a common choice for the significance level is \( \alpha = 0.05 \), which corresponds to a 5% risk of concluding that the null hypothesis is false when it is indeed true. This means there is a 5% chance of a Type I error—falsely rejecting the null hypothesis.When you set \( \alpha \) to 0.05, you need to consider this level before you start your test. It's about balancing risks:
  • Lowering \( \alpha \) reduces the risk of a Type I error but increases the risk of a Type II error—failing to reject a false null hypothesis.
  • Choosing the right \( \alpha \) is a subjective choice often influenced by the context of your research and the consequences of making errors.
By deciding on a significance level upfront, you define how stringent the test needs to be. It's crucial because it sets the criteria for deciding between the null and alternative hypotheses.
Sample Proportion
Sample proportion is a key concept in hypothesis testing, especially when dealing with proportions, as in this exercise. It is denoted by \( \hat{p} \) and calculated by dividing the number of successes by the total size of the sample. In the given exercise:
  • Number of patients with nausea = 25
  • Total sample size = 300
The sample proportion is calculated as:\[ \hat{p} = \frac{25}{300} = 0.0833 \]Sample proportion gives us an initial sense of how the sample compares to the claimed proportion.The sample proportion is essential because it helps compute the test statistic, which will eventually lead us to the p-value. Understanding the sample's proportion aids decision-making by providing evidence to either support or contradict the manufacturer's claim about the drug's side effects. It reveals what's potentially going on in the larger population from which your sample is drawn.
P-value
The p-value is a fundamental concept that tells us the probability of observing our sample data, or something more extreme, if the null hypothesis is true. It helps you gauge the strength of your evidence against the null hypothesis. In hypothesis testing, once you calculate the test statistic (often a \( z \)-score for proportion tests), you use this to find the p-value through standard normal distribution tables.To interpret the p-value:
  • If the p-value is less than the significance level \( \alpha \), there's enough evidence to reject the null hypothesis. This means the sample provides sufficient evidence that the true proportion of patients with nausea is less than 10%.
  • If the p-value is higher than \( \alpha \), you do not have enough evidence to reject the null hypothesis. It suggests that the data does not strongly support the claim that the proportion is less than 10%.
Thus, the p-value works as a bridge connecting your statistical findings to real-world decisions based on the evidence from your data. It is a way to determine statistical significance and guides your conclusion about the drug manufacturer's claim.

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Most popular questions from this chapter

Do you have ESP? A researcher looking for evidence of extrasensory perception (ESP) tests 500 subjects. Four of these subjects do significantly better (P 0.01) than random guessing. (a) Is it proper to conclude that these four people have ESP? Explain your answer. (b) What should the researcher now do to test whether any of these four subjects have ESP?

A researcher plans to conduct a significance test at the \(\alpha=0.01\) significance level. She designs her study to have a power of 0.90 at a particular alternative value of the parameter of interest. The probability that the researcher will commit a Type II error for the particular alternative value of the parameter at which she computed the power is (a) 0.01. (b) 0.10. (c) 0.89. (d) 0.90. (e) 0.99.

You are testing \(H_{0} : \mu=10\) against \(H_{a} : \mu \neq 10\) based on an SRS of 15 observations from a Normal population. What values of the \(t\) statistic are statistically significant at the \(\alpha=0.005\) level? $$ \begin{array}{ll}{\text { (a) } t>3.326} & {\text { (d) } t<-3.326 \text { or } t>3.326} \\ {\text { (b) } t>3.286} & {\text { (e) } t<-3.286 \text { or } t>3.286}\end{array} $$ (c) \(t > 2.977\)

After checking that conditions are met, you perform a significance test of \(H_{0} : \mu=1\) versus \(H_{a} : \mu \neq 1 .\) You obtain a P-value of 0.022 . Which of the following is true? (a) A 95\(\%\) confidence interval for \(\mu\) will include the value \(1 .\) (b) A 95\(\%\) confidence interval for \(\mu\) will include the value \(0 .\) (c) A 999\(\%\) confidence interval for \(\mu\) will include the value \(1 .\) (d) A 99\(\%\) confidence interval for \(\mu\) will include the value \(0 .\) (e) None of these is necessarily true.

Explain why we aren’t safe carrying out a one-sample z test for the population proportion p. No test You toss a coin 10 times to test the hypothesis \(H_{0} : p=0.5\) that the coin is balanced.

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