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Chapter 10: Problem 9

Lipitor The drug Lipitor is meant to reduce total cholesterol and LDL cholesterol. In clinical trials, 19 out of 863 patients taking \(10 \mathrm{mg}\) of Lipitor daily complained of flulike symptoms. Suppose that it is known that \(1.9 \%\) of patients taking competing drugs complain of flulike symptoms. Is there sufficient evidence to conclude that more than \(1.9 \%\) of Lipitor users experience flulike symptoms as a side effect at the \(\alpha=0.01\) level of signifi-cance?

Short Answer

Expert verified
Do not reject \(H_0\). There isn't sufficient evidence to show that more than 1.9% of Lipitor users experience flulike symptoms.

Step by step solution

01

Define the Hypotheses

First, identify the null hypothesis (H_0) and the alternative hypothesis (H_1). H_0: p = 0.019 (The proportion of people experiencing flulike symptoms is 1.9%) H_1: p > 0.019 (The proportion of people experiencing flulike symptoms is greater than 1.9%)
02

Set the Significance Level

The significance level \(\alpha\) is given as 0.01. This is the probability of rejecting the null hypothesis when it is actually true.
03

Calculate the Test Statistic

Use the sample data to compute the test statistic. Formula for the test statistic (z):equation\[z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}}\]where \(\hat{p} = \frac{x}{n}\) \(\hat{p}\) is the sample proportion,\(x = 19\) and \(n = 863\).First, calculate the sample proportion \(\hat{p}\)equation\[\hat{p} = \frac{19}{863} \approx 0.022\]Then, substitute into the test statistic formula:equation\[z = \frac{0.022 - 0.019}{\sqrt{\frac{0.019 (1 - 0.019)}{863}}} \approx 0.474\]
04

Determine the Critical Value and Compare

Since this is a one-tailed test, determine the critical value for \(\alpha = 0.01\) using a Z table. The critical value is approximately 2.33. Compare the test statistic with the critical value. If the test statistic is greater than the critical value, reject the null hypothesis.
05

Make a Decision

In this case, the test statistic \(0.474\) is less than the critical value \(2.33\). Therefore, do not reject the null hypothesis.
06

State Conclusion

There is not enough evidence to conclude that more than 1.9% of Lipitor users experience flulike symptoms at the \(\alpha = 0.01\) significance level.

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Key Concepts

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

null hypothesis
The null hypothesis, denoted as \(H_0\), is a crucial concept in hypothesis testing. It represents a statement of no effect or no difference and serves as the default or starting assumption. In the given exercise, the null hypothesis is that the proportion of Lipitor users experiencing flulike symptoms is equal to the proportion found in users of competing drugs, which is 1.9%. Mathematically, this is expressed as: \[H_0: p = 0.019\] The null hypothesis acts like a safeguard, assuming there is no change unless strong evidence suggests otherwise.
alternative hypothesis
Opposite to the null hypothesis, the alternative hypothesis, denoted as \(H_1\) or \(H_a\), suggests there is evidence of an effect or a difference. In our case study, the alternative hypothesis posits that the proportion of Lipitor users experiencing flulike symptoms is greater than the 1.9% observed in users of competing drugs. Mathematically, this is expressed as: \[H_1: p > 0.019\]
This hypothesis covers what we aim to prove through our statistical test. If the evidence strongly supports the alternative hypothesis, we may reject the null hypothesis.
significance level
The significance level, denoted as \(\alpha\), represents the threshold for rejecting the null hypothesis. It is the probability of making a Type I error, which occurs when a true null hypothesis is incorrectly rejected. In this exercise, the significance level is set at 0.01, meaning there is a 1% risk of rejecting the null hypothesis if it is actually true. A lower significance level means we require stronger evidence to reject the null hypothesis. This helps control the likelihood of making a Type I error.
test statistic
The test statistic is a standardized value that helps compare the sample data against the null hypothesis. In our example, we use the z-test for proportions. The test statistic formula for a proportion is: \[z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}}\] Here, \(\hat{p}\) is the sample proportion (0.022), \(p_0\) is the null hypothesis proportion (0.019), and \(n\) is the sample size (863). Substituting these values gives us a test statistic of: \[z = \frac{0.022 - 0.019}{\sqrt{\frac{0.019 (1 - 0.019)}{863}}} \approx 0.474\] This test statistic is then compared with the critical value to make a decision.
critical value
The critical value is a point on the test distribution that helps decide whether to reject the null hypothesis. For a one-tailed z-test at a significance level of 0.01, the critical value can be found using a z-table. In this exercise, the critical value is approximately 2.33. If the test statistic exceeds the critical value, we reject the null hypothesis. Since our calculated test statistic (0.474) is less than the critical value (2.33), we do not have enough evidence to reject the null hypothesis. Therefore, we conclude there is insufficient evidence to support that more than 1.9% of Lipitor users experience flulike symptoms.

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Most popular questions from this chapter

To test \(H_{0}: \mu=105\) versus \(H_{1}: \mu \neq 105,\) a simple random sample of size \(n=35\) is obtained. (a) Does the population have to be normally distributed to test this hypothesis by using the methods presented in this section? (b) If \(\bar{x}=101.9\) and \(s=5.9,\) compute the test statistic. (c) Draw a \(t\) -distribution with the area that represents the \(P\) -value shaded. (d) Determine and interpret the \(P\) -value. (e) If the researcher decides to test this hypothesis at the \(\alpha=0.01\) level of significance, will the researcher reject the null hypothesis? Why?

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Oil Output An energy official claims that the oil output per well in the United States has declined from the 1998 level of 11.1 barrels per day. He randomly samples 50 wells throughout the United States and determines the mean output to be 10.7 barrels per day. Assume that \(\sigma=1.3\) barrels. Test the researcher's claim at the \(\alpha=0.05\) level of significance.
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