91Ó°ÊÓ

Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section. The drug Eliquis (apixaban) is used to help prevent blood clots in certain patients. In clinical trials, among 5924 patients treated with Eliquis, 153 developed the adverse reaction of nausea (based on data from Bristol-Myers Squibb Co.). Use a 0.05 significance level to test the claim that \(3 \%\) of Eliquis users develop nausea. Does nausea appear to be a problematic adverse reaction?

Step by step solution

01

Identify the null and alternative hypotheses

The null hypothesis (H0) represents the status quo or a statement of no effect. The alternative hypothesis (H1) is what you aim to support.H0: p = 0.03 (The proportion of patients who develop nausea is 3%)H1: p ≠ 0.03 (The proportion of patients who develop nausea is not 3%)

02

Calculate the test statistic

To compute the test statistic, use the formula for the z-score in the context of proportions: \[z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}\]Where:\hat{p} = \frac{153}{5924}\approx 0.0258p_0 = 0.03n = 5924Plugging in the values:\[z = \frac{0.0258 - 0.03}{\sqrt{\frac{0.03(1 - 0.03)}{5924}}}\approx -1.050\]

03

Find the P-value

The next step is to find the P-value corresponding to the calculated z-score. Since it's a two-tailed test, we consider both extremes of the normal distribution:For z = -1.050,The P-value ≈ 2 * P(Z < -1.050)Consulting standard normal distribution tables, we find:P(Z < -1.050) ≈ 0.1475Thus, the P-value ≈ 2 * 0.1475 = 0.295

04

Compare the P-value to the significance level

Compare the P-value to the significance level (α = 0.05):P-value (0.295) > α (0.05)

05

State the conclusion about the null hypothesis

Because the P-value is greater than the significance level, we fail to reject the null hypothesis.

06

State the final conclusion that addresses the original claim

There is not enough evidence to reject the claim that 3% of Eliquis users develop nausea. Therefore, nausea does not appear to be a problematic adverse reaction at the 0.05 significance level.

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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 (denoted as H0) is a statement that indicates no effect or no difference. It serves as the default or baseline assumption that the test seeks to challenge. In our exercise, the null hypothesis states that 3% of Eliquis users develop nausea. This can be written mathematically as: H0: p = 0.03. The null hypothesis is what we test against and will only reject if the data provides sufficient evidence.

alternative hypothesis

Contrary to the null hypothesis is the alternative hypothesis (denoted as H1 or Ha). This hypothesis represents what we aim to support or prove. It suggests that there is an effect or a difference. In the same exercise, the alternative hypothesis claims that the proportion of Eliquis users who develop nausea is not 3%. This can be mathematically stated as: H1: p ≠ 0.03. The alternative hypothesis encompasses the possibility that the true proportion either exceeds or falls short of the 3% mark.

P-value

The P-value is a key measure in hypothesis testing, representing the probability of obtaining test results at least as extreme as the observed data, assuming the null hypothesis is true. Essentially, it helps in determining the significance of your results. In our example, the calculated P-value is approximately 0.295. To interpret it, we compare the P-value to the significance level (α). If the P-value is less than or equal to α, we reject the null hypothesis. Here, because 0.295 > 0.05 (significance level), we fail to reject H0. The higher the P-value, the stronger the evidence for failing to reject the null hypothesis.

test statistic

A test statistic is a standardized value that is derived from sample data during a hypothesis test. It quantifies the degree of evidence against the null hypothesis. In hypothesis testing for proportions, we commonly use the z-score.
The format of the z-score is: \(z = \frac{\text{\textasciicircum{p}} - p0}{\frac{\text{\textasciicircum{p}}0(1 - p0)}{n}} \)
where \( \text{\textasciicircum{p}} \) is the sample proportion,
\( p0 \) is the hypothesized proportion, and \( n \) is the sample size.
In our solution:
\( \text{z} \approx \frac{0.0258 - 0.03}{\frac{0.03(1 - 0.03)}{5924}} \frac{} \approx -1.050 \)
This z-score tells us how many standard deviations the sample proportion is away from the hypothesized population proportion. The value can then be used to find the P-value.

significance level

The significance level (denoted as α) is a threshold set by the researcher before conducting a hypothesis test. It specifies the probability of rejecting the null hypothesis when it is, in fact, true (Type I error). Commonly used levels are 0.05, 0.01, and 0.10. In our specific scenario, a significance level of 0.05 was chosen. At α = 0.05, we accept a 5% chance of erroneously rejecting the null hypothesis.
This means that if the P-value is less than 0.05, we would reject H0; otherwise, we fail to reject it. In our calculation, the P-value of 0.295 was greater than 0.05, leading to the conclusion that there is insufficient evidence to argue against the null hypothesis. Therefore, we accept that the proportion of patients experiencing nausea is indeed 3%, suggesting nausea is not a problematic adverse reaction.