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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 α = 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 α = 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 α = 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 α = 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

(a) Determine the critical value for a right-tailed test of a population standard deviation with 18 degrees of freedom at the \(\alpha=0.05\) level of significance. (b) Determine the critical value for a left-tailed test of a population standard deviation for a sample of size \(n=23\) at the \(\alpha=0.1\) level of significance. (c) Determine the critical values for a two-tailed test of a population standard deviation for a sample of size \(n=30\) at the \(\alpha=0.05\) level of significance.

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