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Chapter 10: Problem 17

The drug Lipitor is meant to reduce 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 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 significance?

Short Answer

Expert verified
Fail to reject the null hypothesis. There is not enough evidence to conclude that more than 1.9% of Lipitor users experience flulike symptoms at the 伪 = 0.01 level.

Step by step solution

01

State the Hypotheses

We need to set up our null and alternative hypotheses. The null hypothesis (H_0) states that the proportion of Lipitor users experiencing flulike symptoms is equal to the proportion of users taking competing drugs experiencing the same symptoms. The alternative hypothesis (H_a) states that the proportion of Lipitor users experiencing flulike symptoms is greater than the proportion of users taking competing drugs experiencing the same symptoms.\[ H_0: p = 0.019 \]\[ H_a: p > 0.019 \]
02

Determine the Test Statistic

We use a one-sample z-test for proportions. First, calculate the sample proportion (\hat{p}).\[ \hat{p} = \frac{19}{863} \approx 0.022 \]Next, the test statistic can be calculated using the formula:\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \]where \hat{p} is the sample proportion, p_0 is the null hypothesis proportion (0.019), and n is the sample size (863).
03

Calculate the Test Statistic

Substitute the given values into the formula from Step 2:\[ z = \frac{0.022 - 0.019}{\sqrt{\frac{0.019(1 - 0.019)}{863}}} \]Compute these values:\[z = \frac{0.003}{\sqrt{\frac{0.019 \times 0.981}{863}}} \approx \frac{0.003}{0.00452} \approx 0.664 \]
04

Determine the Critical Value and Make a Decision

At the 伪 = 0.01 level of significance, we compare the calculated test statistic to the critical z-value from the standard normal distribution. The critical value for a one-tailed test at 伪 = 0.01 is approximately 2.33.Since 0.664 < 2.33, we fail to reject the null hypothesis.
05

Conclusion

There is not enough evidence to conclude that more than 1.9% of Lipitor users experience flulike symptoms as a side effect at the 伪 = 0.01 level of significance.

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Key Concepts

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

One-Sample Z-Test
A one-sample z-test is a statistical method used to determine whether there is a significant difference between the sample proportion and a known population proportion.
This type of test is particularly useful when we want to compare the proportion of a specific outcome in a sample to a known value.
For instance, in the provided exercise, we want to compare the proportion of Lipitor users experiencing flu-like symptoms to the known proportion of 1.9% for users of competing drugs.

The formula for the test statistic in a one-sample z-test for proportions is as follows:
\[z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}\]
  • \( \hat{p} \) is the sample proportion.
  • \( p_0 \) is the population proportion (null hypothesis).
  • \( n \) is the sample size.

This formula calculates the number of standard deviations the sample proportion is from the population proportion, allowing us to understand if the observed difference is statistically significant.
Proportion Testing
Proportion testing involves comparing the proportion of a specific outcome in a sample to a known value or another sample.
It's a common technique used when working with categorical data.
In the exercise, we test the proportion of Lipitor users with flu-like symptoms against the known 1.9%.

A few key points about proportion testing:
  • It's usually conducted using a z-test when the sample size is large (n > 30).
  • The sample proportion (\(\hat{p}\)) is calculated by dividing the number of successes by the total sample size.

Example calculation from the exercise: \[\hat{p} = \frac{19}{863} \approx 0.022\]
This calculation helps us understand the observed proportion of Lipitor users experiencing flu-like symptoms in our sample.
Significance Level
The significance level, denoted by \( \alpha \), is a threshold used to determine whether the test statistic is extreme enough to reject the null hypothesis.
It's often set at common values like 0.05, 0.01, or 0.10.

In the provided exercise, the significance level is set at \( \alpha = 0.01 \), meaning that we have a 1% risk of incorrectly rejecting the null hypothesis.

The critical value for a one-tailed test at \(\alpha = 0.01 \) from the z-distribution is roughly 2.33.
We compare our test statistic to this critical value:
\[\text{Critical value} = 2.33\]
If the test statistic is greater than 2.33, we reject the null hypothesis.

However, in our exercise:
  • The test statistic was 0.664.
  • Since 0.664 < 2.33, we fail to reject the null hypothesis.

This indicates that there's not enough evidence to conclude that more than 1.9% of Lipitor users experience flu-like symptoms at the \(\alpha = 0.01\) significance level.

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Most popular questions from this chapter

What happens to the power of the test as the true value of the parameter gets closer to the value of the parameter stated in the null hypothesis? Why is this result reasonable?

According to the National Highway and Traffic Safety Administration, the proportion of fatal traffic accidents in the United States in which the driver had a positive blood alcohol concentration (BAC) is 0.36. Suppose a random sample of 105 traffic fatalities in the state of Hawaii results in 51 that involved a positive BAC. Does the sample evidence suggest that Hawaii has a higher proportion of traffic fatalities involving a positive BAC than the United States at the \(\alpha=0.05\) level of significance?

Test the hypothesis using (a) the classical approach and (b) the P-value approach. Be sure to verify the requirements of the test. $$\begin{array}{l}H_{0}: p=0.25 \text { versus } H_{1}: p<0.25 \\\n=400 ; x=96 ; \alpha=0.1\end{array}$$

Based on historical birthing records, the proportion of males born worldwide is \(0.51 .\) In other words, the commonly held belief that boys are just as likely as girls is false. Systematic lupus erythematosus (SLE), or lupus for short, is a disease in which one's immune system attacks healthy cells and tissue by mistake. It is well known that lupus tends to exist more in females than in males Researchers wondered, however, if families with a child who had lupus had a lower ratio of males to females than the general population. If this were true, it would suggest that something happens during conception that causes males to be conceived at a lower rate when the SLE gene is present. To determine if this hypothesis is true, the researchers obtained records of families with a child who had SLE A total of 23 males and 79 females were found to have SLE. The 23 males with SL \(E\) had \(a\) total of 23 male siblings and 22 female siblings The 79 females with SLE had a total of 69 male siblings and 80 female siblings Source L.N. Moorthy, M.G.E. Peterson, K.B. Onel, and T.J.A. Lehman. "Do Children with Lupus Have Fewer Male Siblings" Laprus 2008 \(17: 128-131,2008\) (a) Explain why this is an observational study. (b) Is the study retrospective or prospective? Why? (c) There are a total of \(23+69=92\) male siblings in the study How many female siblings are in the study? (d) Draw a relative frequency bar graph of gender of the siblings. (e) Find a point estimate for the proportion of male siblings in families where one of the children has SLE (f) Does the sample evidence suggest that the proportion of male siblings in families where one of the children has SLE is less than 0.51 , the accepted proportion of males born in the general population? Use the \(\alpha=0.05\) level of significance. (g) Construct a \(95 \%\) confidence interval for the proportion of male siblings in a family where one of the children has SLE.

In Problems 23-29, decide whether the problem requires a confidence interval or hypothesis test, and determine the variable of interest. For any problem requiring a confidence interval, state whether the confidence interval will be for a population proportion or population mean. For any problem requiring a hypothesis test, write the null and alternative hypothesis. An investigator with the Food and Drug Administration wanted to determine whether a typical bag of potato chips contained less than the 16 ounces claimed by the manufacturer.
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